We now apply our results to certain multiple Γ-type functions and multiple \(\log \Gamma \)-type functions that are known to be well-studied special functions, namely: the gamma function, the digamma function, the polygamma functions, the q-gamma function, the Barnes G-function, the Hurwitz zeta function and its higher order derivatives, the generalized Stieltjes constants, and the Catalan number function. For recent background on some of these functions, see, e.g., Srivastava and Choi [93].

Each of these examples is examined and studied systematically by following the steps and results given in the previous chapter. When algebraic computations become tedious, a computer algebra system can be of great assistance in executing the details. Further examples will be discussed in the next two chapters.

In this chapter and the next, we occasionally address and solve some secondary but interesting issues. They are then presented and numbered in a Project environment.

Most of the applications we consider in this work illustrate how powerful is our theory to produce formulas and identities methodically. Although many of these formulas and identities are already known, to our knowledge they had never been derived from such a general and unified setting.

10.1 The Gamma Function

Since the Euler gamma function was the starting point of this theory and therefore also Webster’s motivating example in his introduction of the Γ-type functions, it is natural to test our results on this function first.

The following investigation of the gamma function does not reveal quite new formulas. However, it can be regarded as a tutorial that clearly demonstrates how our results can be used to carry out this investigation in a systematic way.

In addition to the remarkable book by Artin [11], the interested reader can also find a very good expository tour of the gamma function in Srinivasan’s paper [92].

ID Card

The following table summarizes the ID card corresponding to the log and log-gamma functions.

g(x)

Membership

deg g

Σg(x)

\(\mathstrut ^{\mathstrut }_{\mathstrut } \ln x\)

\(\mathcal {C}^{\infty }\cap \mathcal {D}^1\cap \mathcal {K}^{\infty }\)

0

\(\ln \Gamma (x)\)

Bohr-Mollerup’s Theorem

A characterization of the gamma function is given in Bohr-Mollerup’s theorem (see Theorem 1.1 and Example 3.2). In the additive notation, we have the following statement.

All eventually convex or concave solutions \(f\colon \mathbb {R}_+\to \mathbb {R}\) to the equation

$$\displaystyle \begin{aligned} f(x+1)-f(x) ~=~ \ln x \end{aligned}$$

are of the form \(f(x)=c+\ln \Gamma (x)\), where \(c\in \mathbb {R}\).

Using Proposition 3.9, we can also derive the following alternative characterization of the gamma function (see Example 3.11).

All solutions \(f\colon \mathbb {R}_+\to \mathbb {R}\) to the equation

$$\displaystyle \begin{aligned} f(x+1)-f(x) ~=~ \ln x \end{aligned}$$

that satisfy the asymptotic condition that, for each x > 0,

$$\displaystyle \begin{aligned} f(x+n)-f(n)-x\ln n ~\to ~0\qquad \mbox{as }n\to_{\mathbb{N}}\infty \end{aligned}$$

are of the form \(f(x)=c+\ln \Gamma (x)\), where \(c\in \mathbb {R}\).

Extended ID Card

The value of σ[g] has been discussed in Example 6.5. More precisely, we also have the following values:

  • Inequality

    $$\displaystyle \begin{aligned} |\sigma[g]| ~\leq ~ \ln 4-\frac{5}{4} ~\approx ~ 0.14. \end{aligned}$$
  • Alternative representations of σ[g] = γ[g]

    $$\displaystyle \begin{aligned} \begin{array}{rcl} \sigma[g] & =&\displaystyle \int_1^{\infty}\left(\{t\}\ln\frac{1+\lfloor t\rfloor}{t}+(1-\{t\})\ln\frac{\lfloor t\rfloor}{t}\right)dt{\,},\\ \sigma[g] & =&\displaystyle \lim_{n\to\infty}\left(\ln n! +n-1-\left(n+\frac{1}{2}\right)\ln n\right),\\ \sigma[g] & =&\displaystyle \sum_{k=1}^{\infty}\left(1-\left(k+\frac{1}{2}\right)\ln\left(1+\frac{1}{k}\right)\right),\\ \sigma[g] & =&\displaystyle \int_1^{\infty}\left(\frac{1}{2}\ln(\lfloor t\rfloor^2 +\lfloor t\rfloor)-\ln t\right)dt{\,},\\ \sigma[g] & =&\displaystyle \int_1^{\infty}\frac{\{t\}-\frac{1}{2}}{t}{\,}dt{\,},\\ \sigma[g] & =&\displaystyle \int_0^1\ln\Gamma(t+1){\,}dt. \end{array} \end{aligned} $$
  • Binet’s function

    $$\displaystyle \begin{aligned} J^2[\ln\circ\Gamma](x) ~=~ J(x) ~=~ \ln\Gamma(x)-\frac{1}{2}\ln(2\pi)+x-\left(x-\frac{1}{2}\right)\ln x{\,},\qquad x>0. \end{aligned}$$
  • Raabe’s formula

    $$\displaystyle \begin{aligned} \int_x^{x+1}\ln\Gamma(t){\,}dt ~=~ \frac{1}{2}\,\ln(2\pi)+x\ln x-x{\,},\qquad x>0. \end{aligned}$$
  • Alternative characterization. The function \(f(x)=\ln \Gamma (x)\) is the unique solution lying in \(\mathcal {C}^0\cap \mathcal {K}^1\) to the equation

    $$\displaystyle \begin{aligned} \int_x^{x+1}f(t){\,}dt ~=~ \frac{1}{2}\,\ln(2\pi)+x\ln x-x{\,}, \qquad x>0. \end{aligned}$$

\(\overline {\sigma }[g]\)

σ[g]

γ[g]

\(\mathstrut ^{\mathstrut }_{\mathstrut } \frac {1}{2}\ln (2\pi )\)

\(-1+\frac {1}{2}\ln (2\pi )\)

γ[g] = σ[g]

Inequalities

The following inequalities hold for any x > 0, any a ≥ 0, and any \(n\in \mathbb {N}^*\).

  • Symmetrized generalized Wendel’s inequality (equality if a ∈{0, 1})

    $$\displaystyle \begin{aligned} \big|\ln\Gamma(x+a)-\ln\Gamma(x)-a\ln x\big| ~\leq ~ |a-1|\,\ln\left(1+\frac{a}{x}\right), \end{aligned}$$
    $$\displaystyle \begin{aligned} \left(1+\frac{a}{x}\right)^{-\left|a-1\right|} \leq ~ \frac{\Gamma(x+a)}{\Gamma(x){\,}x^a} ~\leq ~ \left(1+\frac{a}{x}\right)^{\left|a-1\right|}. \end{aligned}$$
  • Symmetrized generalized Wendel’s inequality (discrete version)

    $$\displaystyle \begin{aligned} \left|\ln\Gamma(x)-\sum_{k=1}^{n-1}\ln k+\sum_{k=0}^{n-1}\ln(x+k)-x\ln n\right| ~\leq ~ |x-1|\,\ln\left(1+\frac{x}{n}\right). \end{aligned}$$
    $$\displaystyle \begin{aligned} \left(1+\frac{x}{n}\right)^{-\left|x-1\right|} \leq ~ \Gamma(x)\,\frac{x(x+1)\cdots (x+n-1)}{(n-1)!{\,}n^x} ~ \leq ~ \left(1+\frac{x}{n}\right)^{\left|x-1\right|}. \end{aligned}$$
  • Symmetrized Stirling’s formula-based inequality

    $$\displaystyle \begin{aligned} \left|J(x)\right| ~\leq ~ \frac{(x+1)^2}{2}\ln\left(1+\frac{1}{x}\right)-\frac{x}{2}-\frac{3}{4} ~\leq ~ \frac{1}{2}\,\ln\left(1+\frac{1}{x}\right), \end{aligned}$$
    $$\displaystyle \begin{aligned} \left(1+\frac{1}{x}\right)^{-\frac{1}{2}} \leq ~ \frac{\Gamma(x)}{\sqrt{2\pi}{\,}e^{-x}{\,}x^{x-\frac{1}{2}}} ~\leq ~ \left(1+\frac{1}{x}\right)^{\frac{1}{2}}. \end{aligned}$$
  • Burnside’s formula-based inequality

    $$\displaystyle \begin{aligned} \left|\ln\Gamma\left(x+\frac{1}{2}\right)-\frac{1}{2}\ln(2\pi)+x-x\ln x\right| ~\leq ~ \left|J(x)\right|. \end{aligned}$$
  • Generalized Gautschi’s inequality

    $$\displaystyle \begin{aligned} (x+\lceil a\rceil)^{a-\lceil a\rceil} ~\leq ~ e^{(a-\lceil a\rceil)\,\psi(x+\lceil a\rceil)} ~\leq ~ \frac{\Gamma(x+a)}{\Gamma(x+\lceil a\rceil)} ~\leq ~ (x+\lfloor a\rfloor)^{a-\lceil a\rceil}{\,}. \end{aligned}$$

Stirling’s and Related Formulas

For any a ≥ 0, we have the following limits and asymptotic equivalences as x →,

$$\displaystyle \begin{aligned} \ln\Gamma(x+a)-\ln\Gamma(x)-a\ln x ~\to ~0, \end{aligned}$$
$$\displaystyle \begin{aligned} \ln\Gamma(x)-\frac{1}{2}\ln(2\pi)+x-\left(x-\frac{1}{2}\right)\ln x ~\to ~0, \end{aligned}$$
$$\displaystyle \begin{aligned} \ln\Gamma\left(x+\frac{1}{2}\right)-\frac{1}{2}\ln(2\pi)+x-x\ln x ~\to ~0, \end{aligned}$$
$$\displaystyle \begin{aligned} \Gamma(x+a) ~\sim ~ x^a\,\Gamma(x),\qquad \ln\Gamma(x+a) ~\sim ~ x\ln x, \end{aligned}$$
$$\displaystyle \begin{aligned} \Gamma(x) ~\sim ~ \sqrt{2\pi}{\,}e^{-x}x^{x-\frac{1}{2}},\qquad \Gamma(x+1) ~\sim ~ \sqrt{2\pi x}{\,}e^{-x}x^x{\,}. \end{aligned}$$

Burnside’s approximation (better than Stirling’s approximation)

$$\displaystyle \begin{aligned} \Gamma(x) ~\sim ~ \sqrt{2\pi}\left(\frac{x-\frac{1}{2}}{e}\right)^{x-\frac{1}{2}}. \end{aligned}$$

Further results (obtained by differentiation)

$$\displaystyle \begin{aligned} \psi(x+a)-\psi(x) ~\to ~0,\qquad \psi(x)-\ln x ~\to ~ 0,\qquad \psi(x+a) ~\sim ~ \ln x{\,}, \end{aligned}$$
$$\displaystyle \begin{aligned} \psi_k(x+a) ~\sim ~ (-1)^{k-1}\,\frac{(k-1)!}{x^k}{\,},\qquad \psi_k(x) ~\to ~0,\qquad k\in\mathbb{N}^*. \end{aligned}$$

Asymptotic Expansions

For any \(m,q\in \mathbb {N}^*\) we have the following expansion as x →

$$\displaystyle \begin{aligned} \begin{array}{rcl} \frac{1}{m}\sum_{j=0}^{m-1}\ln\Gamma\left(x+\frac{j}{m}\right) & =&\displaystyle \frac{1}{2}\ln(2\pi)+x\ln x-x-\frac{1}{2m}\ln x\\ & &\displaystyle +\sum_{k=1}^q\frac{B_{k+1}}{k(k+1){\,}x^k{\,}m^{k+1}}+O\left(x^{-q-1}\right){\,}. \end{array} \end{aligned} $$

Setting m = 1 in this formula, we retrieve the known asymptotic expansion of the log-gamma function \(\ln \Gamma (x)\) as x → (see, e.g., [93, p. 7])

$$\displaystyle \begin{aligned} \ln\Gamma(x) ~=~ \frac{1}{2}\ln(2\pi)-x+\left(x-\frac{1}{2}\right)\ln x+\sum_{k=1}^q\frac{B_{k+1}}{k(k+1){\,}x^k}+O\left(x^{-q-1}\right), \end{aligned} $$
(10.1)

or equivalently,

$$\displaystyle \begin{aligned} J(x) ~=~ \sum_{k=1}^q\frac{B_{k+1}}{k(k+1){\,}x^k}+O\left(x^{-q-1}\right). \end{aligned}$$

For instance, setting q = 4 in (10.1) we get

$$\displaystyle \begin{aligned} \ln\Gamma(x) ~=~ \frac{1}{2}\ln(2\pi)-x+\left(x-\frac{1}{2}\right)\ln x+\frac{1}{12x}-\frac{1}{360x^3}+O\left(x^{-5}\right). \end{aligned}$$

Generalized Liu’s Formula

For any x > 0 we have

$$\displaystyle \begin{aligned} \ln\Gamma(x) ~=~ \frac{1}{2}\ln(2\pi)-x+\left(x-\frac{1}{2}\right)\ln x+\int_0^{\infty}\frac{\frac{1}{2}-\{t\}}{t+x}{\,}dt, \end{aligned}$$

or equivalently,

$$\displaystyle \begin{aligned} J(x) ~=~ \int_0^{\infty}\frac{\frac{1}{2}-\{t\}}{t+x}{\,}dt. \end{aligned}$$

Limit, Series, and Integral Representations

We now consider various representations of \(\ln \Gamma (x)\), including the Eulerian and Weierstrassian forms.

  • Eulerian form and related identities. We have

    $$\displaystyle \begin{aligned} \ln\Gamma(x) ~=~ -\ln x-\sum_{k=1}^{\infty}\left(\ln(x+k)-\ln k-x\ln\left(1+\frac{1}{k}\right)\right), \end{aligned}$$
    $$\displaystyle \begin{aligned} \Gamma(x) ~=~ \frac{1}{x}\,\prod_{k=1}^{\infty}\frac{(1+\frac{1}{k})^x}{1+\frac{x}{k}}{\,}. \end{aligned}$$

    Upon differentiation and integration, we obtain (cf. Example 8.3)

    $$\displaystyle \begin{aligned} \psi(x) ~=~ -\frac{1}{x}-\sum_{k=1}^{\infty}\left(\frac{1}{x+k}-\ln\left(1+\frac{1}{k}\right)\right), \end{aligned}$$
    $$\displaystyle \begin{aligned} \psi_k(x) ~=~ (-1)^{k-1}{\,}k!\,\zeta(k+1,x),\qquad k\in\mathbb{N}^*, \end{aligned}$$
    $$\displaystyle \begin{aligned} \psi_{-2}(x) ~=~ x-x\ln x-\sum_{k=1}^{\infty}\left((x+k)\ln\left(1+\frac{x}{k}\right)-x-\frac{x^2}{2}\ln\left(1+\frac{1}{k}\right)\right). \end{aligned}$$
  • Weierstrassian form and related identities. We have

    $$\displaystyle \begin{aligned} \ln\Gamma(x) ~=~ -\gamma x-\ln x-\sum_{k=1}^{\infty}\left(\ln(x+k)-\ln k-\frac{x}{k}\right), \end{aligned}$$
    $$\displaystyle \begin{aligned} \Gamma(x) ~=~ \frac{e^{-\gamma x}}{x}\,\prod_{k=1}^{\infty}\frac{e^{\frac{x}{k}}}{1+\frac{x}{k}}{\,}. \end{aligned}$$

    Upon differentiation and integration, we obtain (cf. Example 8.8)

    $$\displaystyle \begin{aligned} \psi(x) ~=~ -\gamma-\frac{1}{x}-\sum_{k=1}^{\infty}\left(\frac{1}{x+k}-\frac{1}{k}\right), \end{aligned}$$
    $$\displaystyle \begin{aligned} \psi_{-2}(x) ~=~ -\gamma\,\frac{x^2}{2}+x-x\ln x -\sum_{k=1}^{\infty}\left((x+k)\ln\left(1+\frac{x}{k}\right)-x-\frac{x^2}{2k}\right). \end{aligned}$$
  • Gauss’ limit and related identities. The Gauss limit is

    $$\displaystyle \begin{aligned} \ln\Gamma(x) ~=~ \lim_{n\to\infty}\left(\ln(n-1)! -\sum_{k=0}^{n-1}\ln(x+k)+x\ln n\right). \end{aligned}$$

    Upon differentiation and integration, we obtain

    $$\displaystyle \begin{aligned} \psi(x) ~=~ \lim_{n\to\infty}\left(\ln n-\sum_{k=0}^{n-1}\frac{1}{x+k}\right), \end{aligned}$$
    $$\displaystyle \begin{aligned} \psi_k(x) ~=~ (-1)^{k+1}{\,}k!\,\zeta(k+1,x),\qquad k\in\mathbb{N}^*, \end{aligned} $$
    $$\displaystyle \begin{aligned} \psi_{-2}(x) ~=~ \lim_{n\to\infty}\left(nx-x\ln x+(\ln n)\frac{x^2}{2}-\sum_{k=1}^{n-1}(x+k)\ln\left(1+\frac{x}{k}\right)\right). \end{aligned} $$
    (10.2)

    The multiplicative version of Gauss’ limit reduces to the following formula (just replace n with n + 1 and note that (n + 1)x ∼ n x as n →)

    $$\displaystyle \begin{aligned} \Gamma(x) ~=~ \lim_{n\to\infty}\frac{n!{\,}n^x}{x(x+1){\,}\cdots{\,}(x+n)} \end{aligned}$$

    as stated in (1.6). We also have the following alternative form of Gauss’ limit, which immediately follows from the Weierstrassian form

    $$\displaystyle \begin{aligned} \Gamma(x) ~=~ \frac{e^{-\gamma x}}{x}\,\lim_{n\to\infty}\prod_{k=1}^n\frac{e^{\frac{x}{k}}}{1+\frac{x}{k}} ~=~ \lim_{n\to\infty}\frac{n!{\,}e^{x\psi(n)}}{x(x+1){\,}\cdots{\,}(x+n)}{\,}. \end{aligned}$$

    This latter limit can also be derived immediately from Gauss’ limit and the well-known fact that \(\psi (x)-\ln x\to 0\) as x →.

  • Integral representation. Considering the antiderivative of the digamma function φ = ψ as the solution to the equation Δφ = g′ (using the elevator method), we obtain

    $$\displaystyle \begin{aligned} \ln\Gamma(x) ~=~ \psi_{-1}(x) ~=~ \int_1^x\psi(t){\,}dt. \end{aligned}$$
  • Gregory’s formula-based series representation. For any x > 0 we have the series representation (see Example 8.12)

    $$\displaystyle \begin{aligned} \begin{array}{rcl} \ln\Gamma(x) & =&\displaystyle \frac{1}{2}\ln(2\pi)-x+x\ln x-\sum_{n=0}^{\infty}G_{n+1}\Delta^n\ln(x){}\\ & =&\displaystyle \frac{1}{2}\ln(2\pi)-x+x\ln x-\sum_{n=0}^{\infty}|G_{n+1}|\,\sum_{k=0}^n(-1)^k{\textstyle{{{n}\choose{k}}}}{\,}\ln(x+k). \end{array} \end{aligned} $$
    (10.3)

    Setting x = 1 in this identity yields the following analogue of Fontana-Mascheroni series

    $$\displaystyle \begin{aligned} \sum_{n=0}^{\infty}|G_{n+1}|\,\sum_{k=0}^n(-1)^k{\textstyle{{{n}\choose{k}}}}{\,}\ln(k+1) ~=~ -1+\frac{1}{2}\ln(2\pi). \end{aligned}$$

Gauss’ Multiplication Formula

For any \(m\in \mathbb {N}^*\) and any x > 0, we have

$$\displaystyle \begin{aligned} \prod_{j=0}^{m-1}\Gamma\left(\frac{x+j}{m}\right) ~=~ (2\pi)^{\frac{m-1}{2}}m^{\frac{1}{2}-x}\,\Gamma(x). \end{aligned}$$

Corollary 8.33 provides the following asymptotic equivalence for any x > 0

$$\displaystyle \begin{aligned} \Gamma(mx)^{\frac{1}{m}} ~\sim ~ e^{-x}x^xm^x\qquad \mbox{as }m\to_{\mathbb{N}}\infty, \end{aligned}$$

which also follows from Stirling’s formula.

Wallis’s Product Formula

We have the following limits

$$\displaystyle \begin{aligned} \lim_{n\to\infty}\frac{1\cdot 3{\,}\cdots{\,}(2n-1)}{2\cdot 4{\,}\cdots{\,}(2n)}{\,}\sqrt{n} ~=~ \frac{1}{\sqrt{\pi}}{\,}, \end{aligned}$$
$$\displaystyle \begin{aligned} \lim_{n\to\infty}\left(\frac{1}{2}\,\ln(\pi n)+\sum_{k=1}^{2n}(-1)^{k-1}\,\ln k\right) ~=~ 0. \end{aligned}$$

Restriction to the Natural Integers

We have the well-known identity

$$\displaystyle \begin{aligned} \Gamma(n+1) ~=~ n!{\,},\qquad n\in\mathbb{N}. \end{aligned}$$

Gregory’s formula states that for any \(n\in \mathbb {N}^*\) and any \(q\in \mathbb {N}\) we have

$$\displaystyle \begin{aligned} \ln n! ~=~ 1-n+(n+1)\ln n-\sum_{j=1}^qG_j\left((\Delta^{j-1}\ln)(n)-(\Delta^{j-1}\ln)(1)\right)-R^q_n{\,}, \end{aligned}$$

with

$$\displaystyle \begin{aligned} |R^q_n| ~\leq ~ \overline{G}_q{\,}|(\Delta^q\ln)(n)-(\Delta^q\ln)(1)|. \end{aligned}$$

Moreover, Eq. (10.1) yields the following asymptotic expansion as x →. For any \(q\in \mathbb {N}^*\), we have

$$\displaystyle \begin{aligned} \ln n! ~=~ \frac{1}{2}\ln(2\pi n)-n+n\ln n+\sum_{k=1}^q\frac{B_{k+1}}{k(k+1){\,}n^k}+O\left(n^{-q-1}\right). \end{aligned}$$

Similarly, Eq. (10.3) yields the following series representation

$$\displaystyle \begin{aligned} \ln n! ~=~ \frac{1}{2}\ln(2\pi)-n+(n+1)\ln n-\sum_{k=0}^{\infty}G_{k+1}\Delta^kg(n){\,},\quad n\in\mathbb{N}^*. \end{aligned}$$

We also have Liu’s formula

$$\displaystyle \begin{aligned} \ln n! ~=~ \frac{1}{2}\ln(2\pi n)-n+n\ln n+\int_n^{\infty}\frac{\frac{1}{2}-\{t\}}{t}{\,}dt{\,}. \end{aligned}$$

Many other representations of \(\ln n!\) can be derived from, e.g., the limit and series representations of the log-gamma function described above.

Generalized Webster’s Functional Equation

For any \(m\in \mathbb {N}^*\) and any a > 0, there is a unique solution \(f\colon \mathbb {R}_+\to \mathbb {R}_+\) to the equation

$$\displaystyle \begin{aligned} \prod_{j=0}^{m-1}f(x+a j) ~=~ x \end{aligned}$$

such that \(\ln f\) lies in \(\mathcal {K}^0\) (or in \(\mathcal {K}^1\)), namely

$$\displaystyle \begin{aligned} f(x) ~=~ (am)^{\frac{1}{m}}\frac{\Gamma(\frac{x+a}{am})}{\Gamma(\frac{x}{am})}{\,}. \end{aligned}$$

Analogue of Euler’s Series Representation of γ

The Taylor series expansion of \(\ln \Gamma (x+1)\) about x = 0 is

$$\displaystyle \begin{aligned} \ln\Gamma(x+1) ~=~ -\gamma x+\sum_{k=2}^{\infty}\frac{\zeta(k)}{k}{\,}(-x)^k{\,},\qquad |x|<1. \end{aligned}$$

Integrating both sides of this equation on (0, 1), we obtain (see Example 7.16)

$$\displaystyle \begin{aligned} \sum_{k=2}^{\infty}(-1)^k\frac{1}{k(k+1)}\,\zeta(k) ~=~ \frac{1}{2}\,\gamma -1+\frac{1}{2}\ln(2\pi){\,}. \end{aligned}$$

Reflection Formula

For any \(x\in \mathbb {R}\setminus \mathbb {Z}\), we have \(\Gamma (x)\Gamma (1-x)=\pi \csc {}(\pi x)\).

10.2 The Digamma and Harmonic Number Functions

Let us now see what we get if we apply our results to both the digamma function xψ(x) and the harmonic number function xH x. Recall first that the identity

$$\displaystyle \begin{aligned} H_{x-1} ~=~ \psi(x)+\gamma \end{aligned}$$

holds for any x > 0.

ID Card

We have the following data about the functions 1∕x and ψ(x):

g(x)

Membership

deg g

Σg(x)

\(\mathstrut ^{\mathstrut }_{\mathstrut } 1/x\)

\(\mathcal {C}^{\infty }\cap \mathcal {D}^0\cap \mathcal {K}^{\infty }\)

− 1

H x−1 = ψ(x) + γ

Analogue of Bohr-Mollerup’s Theorem

The digamma function can be characterized as follows.

All eventually monotone solutions \(f\colon \mathbb {R}_+\to \mathbb {R}\) to the equation

$$\displaystyle \begin{aligned} f(x+1)-f(x) ~=~ \frac{1}{x} \end{aligned}$$

are of the form f(x) = c + ψ(x), where \(c\in \mathbb {R}\).

It is noteworthy that this characterization immediately follows from the basic version when p = 0 of our Theorem 1.4, which was established by John [49].

Interestingly, this characterization enables us to establish almost instantly the following identities for every x > 0,

$$\displaystyle \begin{aligned} H_{x-1} ~=~ \psi(x)+\gamma ~=~ \int_0^1\frac{1-t^{x-1}}{1-t}{\,}dt{\,}. \end{aligned}$$

Indeed, each of the three expressions above vanishes at x = 1 and is an eventually increasing solution to the equation f(x + 1) − f(x) = 1∕x. Hence, they must coincide on \(\mathbb {R}_+\). We can actually prove many other representations similarly; for instance, the following Gauss and Dirichlet integral representations (see, e.g., [93, p. 26])

$$\displaystyle \begin{aligned} \psi(x) ~=~ \int_0^{\infty}\left(\frac{e^{-t}}{t}-\frac{e^{-xt}}{1-e^{-t}}\right)dt{\,},\qquad x>0, \end{aligned}$$
$$\displaystyle \begin{aligned} \psi(x) ~=~ \int_0^{\infty}\left(e^{-t}-\frac{1}{(t+1)^x}\right)\frac{dt}{t}{\,},\qquad x>0. \end{aligned}$$

Kairies [51] obtained a variant of the characterization of the digamma function above by replacing the eventual monotonicity with the convexity property. This variant is also immediate from our results since g also lies in \(\mathcal {D}^1\cap \mathcal {K}^1\).

Using Proposition 3.9, we can also derive the following alternative characterization of the digamma function.

All solutions \(f\colon \mathbb {R}_+\to \mathbb {R}\) to the equation

$$\displaystyle \begin{aligned} f(x+1)-f(x) ~=~ \frac{1}{x} \end{aligned}$$

that satisfy the asymptotic condition that, for each x > 0,

$$\displaystyle \begin{aligned} f(x+n)-f(n) ~\to ~0\qquad \mbox{as }n\to_{\mathbb{N}}\infty \end{aligned}$$

are of the form f(x) = c + ψ(x), where \(c\in \mathbb {R}\).

Extended ID Card

We already know that σ[g] = γ (see Example 8.19). Hence we have the following table:

  • Alternative representations of σ[g] = γ[g] = γ

    $$\displaystyle \begin{aligned} \begin{array}{rcl} \gamma & =&\displaystyle \lim_{n\to\infty}\left(\sum_{k=1}^n\frac{1}{k}-\ln n\right) ~=~ \sum_{k=1}^{\infty}\left(\frac{1}{k}-\ln\left(1+\frac{1}{k}\right)\right),\\ \gamma & =&\displaystyle \int_1^{\infty}\left(\frac{1}{\lfloor t\rfloor}-\frac{1}{t}\right){\,}dt ~=~ \frac{1}{2}-\int_1^{\infty}\frac{\{t\}-\frac{1}{2}}{t^2}{\,}dt{\,},\\ \gamma & =&\displaystyle \int_0^1H_t{\,}dt{\,}. \end{array} \end{aligned} $$
  • Generalized Binet’s function. For any \(q\in \mathbb {N}\) and any x > 0

    $$\displaystyle \begin{aligned} J^{q+1}[\psi](x) ~=~ \psi(x)-\ln x+\sum_{j=1}^q|G_j|{\,}\mathrm{B}(x,j), \end{aligned}$$

    where (x, y)↦B(x, y) is the beta function.

  • Analogue of Raabe’s formula (see Example 8.19)

    $$\displaystyle \begin{aligned} \int_x^{x+1}\psi(t){\,}dt ~=~ \ln x{\,}, \qquad x>0. \end{aligned}$$
  • Alternative characterization. The function f = ψ is the unique solution lying in \(\mathcal {C}^0\cap \mathcal {K}^0\) to the equation

    $$\displaystyle \begin{aligned} \int_x^{x+1}f(t){\,}dt ~=~ \ln x{\,}, \qquad x>0. \end{aligned}$$

\(\overline {\sigma }[g]\)

σ[g]

γ[g]

\(\mathstrut ^{\mathstrut }_{\mathstrut } \infty \)

γ

γ

Inequalities

The following inequalities hold for any x > 0, any a ≥ 0, and any \(n\in \mathbb {N}^*\).

  • Symmetrized generalized Wendel’s inequality (equality if a ∈{0, 1})

    $$\displaystyle \begin{aligned} |\psi(x+a)-\psi(x)| ~\leq ~ \lceil a\rceil\,\frac{1}{x}{\,}. \end{aligned}$$
  • Symmetrized generalized Wendel’s inequality (discrete version)

    $$\displaystyle \begin{aligned} \left|\psi(x)+\gamma-\sum_{k=1}^{n-1}\frac{1}{k}+\sum_{k=0}^{n-1}\frac{1}{x+k}\right| ~\leq ~ \lceil x\rceil\,\frac{1}{n}{\,}. \end{aligned}$$
  • Symmetrized Stirling’s and Burnside’s formulas-based inequalities

    $$\displaystyle \begin{aligned} \left|\psi\left(x+\frac{1}{2}\right)-\ln x\right| ~\leq ~ |\psi(x)-\ln x| ~\leq ~ \frac{1}{x}{\,}. \end{aligned}$$

    Considering for instance the value p = 1 in Corollary 6.12, we see that the latter inequality can be refined into

    $$\displaystyle \begin{aligned} \frac{1}{2(x+1)}-\frac{1}{x} ~\leq ~ \psi(x)-\ln x ~\leq ~ -\frac{1}{2(x+1)}{\,}. \end{aligned}$$
  • Generalized Gautschi’s inequality

    $$\displaystyle \begin{aligned} \frac{a-\lceil a\rceil}{x+\lfloor a\rfloor} ~\leq ~ \psi(x+a)-\psi(x+\lceil a\rceil) ~\leq ~ (a-\lceil a\rceil)\,\psi_1(x+\lceil a\rceil) ~\leq ~ \frac{a-\lceil a\rceil}{x+\lceil a\rceil} {\,}. \end{aligned}$$

Generalized Stirling’s and Related Formulas

For any a ≥ 0, we have the following limits and asymptotic equivalence as x →,

$$\displaystyle \begin{aligned} \psi(x+a)-\psi(x) ~\to ~0,\qquad \psi(x)-\ln x ~\to ~ 0,\qquad \psi(x+a) ~\sim ~ \ln x. \end{aligned}$$

Burnside-like approximation (better than Stirling-like approximation)

$$\displaystyle \begin{aligned} \psi(x)-\ln\left(x-\frac{1}{2}\right) ~\to ~ 0. \end{aligned}$$

Further results (obtained by differentiation)

$$\displaystyle \begin{aligned} \psi_k(x+a) ~\sim ~ (-1)^{k-1}\,\frac{(k-1)!}{x^k}{\,},\qquad \psi_k(x) ~\to ~0,\qquad k\in\mathbb{N}^*. \end{aligned}$$

Asymptotic Expansions

For any \(m,q\in \mathbb {N}^*\) we have the following expansion as x →

$$\displaystyle \begin{aligned} \frac{1}{m}\sum_{j=0}^{m-1}\psi\left(x+\frac{j}{m}\right) ~=~ \ln x + \sum_{k=1}^q\frac{(-1)^{k-1}{\,}B_k}{k{\,}(mx)^k}+O\left(x^{-q-1}\right){\,}. \end{aligned} $$
(10.4)

Setting m = 1 in this formula, we retrieve the known asymptotic expansion of ψ(x) as x → (see, e.g., [93, p. 36])

$$\displaystyle \begin{aligned} \psi(x) ~=~ \ln x + \sum_{k=1}^q\frac{(-1)^{k-1}{\,}B_k}{k{\,}x^k}+O\left(x^{-q-1}\right), \end{aligned}$$

or equivalently,

$$\displaystyle \begin{aligned} J^1[\psi](x) ~=~ \sum_{k=1}^q\frac{(-1)^{k-1}{\,}B_k}{k{\,}x^k}+O\left(x^{-q-1}\right). \end{aligned}$$

For instance, setting q = 5 we get

$$\displaystyle \begin{aligned} \psi(x) ~=~ \ln x -\frac{1}{2x}-\frac{1}{12x^2}+\frac{1}{120x^4}+O\left(x^{-6}\right). \end{aligned}$$

Generalized Liu’s Formula

For any x > 0 we have

$$\displaystyle \begin{aligned} \psi(x) ~=~ \ln x-\frac{1}{2x}+\int_0^{\infty}\frac{\{t\}-\frac{1}{2}}{(t+x)^2}{\,}dt. \end{aligned}$$

Limit and Series Representations

Let us now examine the main limit and series representations of the digamma function that we obtain from our results.

  • Eulerian and Weierstrassian forms. We have

    $$\displaystyle \begin{aligned} \psi(x) ~=~ -\gamma-\frac{1}{x}+\sum_{k=1}^{\infty}\left(\frac{1}{k}-\frac{1}{x+k}\right), \end{aligned}$$
    $$\displaystyle \begin{aligned} \psi(x) ~=~ -\frac{1}{x}+\sum_{k=1}^{\infty}\left(\ln\left(1+\frac{1}{k}\right)-\frac{1}{x+k}\right). \end{aligned}$$

    Upon differentiation, we obtain

    $$\displaystyle \begin{aligned} \psi_k(x) ~=~ (-1)^{k-1}{\,}k!\,\zeta(k+1,x),\qquad k\in\mathbb{N}^*. \end{aligned}$$

    Moreover, integrating the Eulerian (resp. Weierstrassian) form of the digamma function on (0, x), we retrieve the Weierstrassian (resp. Eulerian) form of the log-gamma function.

  • The analogue of Gauss’ limit coincides with the Eulerian form.

  • Gregory’s formula-based series representation. For any x > 0 we have the series representation

    $$\displaystyle \begin{aligned} \psi(x) ~=~ \ln x-\sum_{n=1}^{\infty}|G_n|{\,}\mathrm{B}(x,n) ~=~ \ln x-\sum_{n=1}^{\infty}\frac{|G_n|}{n{x+n-1\choose n}}{\,}. \end{aligned}$$

    Setting x = 1 in this identity, we retrieve the Fontana-Mascheroni series (see, e.g., Blagouchine [20, p. 379])

    $$\displaystyle \begin{aligned} \gamma ~=~ \sum_{n=1}^{\infty}\frac{|G_n|}{n}{\,}. \end{aligned}$$

    Setting x = 2, we get

    $$\displaystyle \begin{aligned} 1-\ln 2 ~=~ \sum_{n=1}^{\infty}\frac{|G_n|}{n+1}{\,}, \end{aligned}$$

    which is consistent with the identities given in Example 8.16.

Analogue of Gauss’ Multiplication Formula

For any \(m\in \mathbb {N}^*\) and any x > 0, we have (see, e.g., Berndt [18, p. 5])

$$\displaystyle \begin{aligned} \sum_{j=0}^{m-1}\psi\left(x+\frac{j}{m}\right) ~=~ m(\psi(mx)-\ln m) \end{aligned} $$
(10.5)

and

$$\displaystyle \begin{aligned} \sum_{j=0}^{m-1}H_{x+j/m} ~=~ m(H_{mx+m-1}-\ln m){\,}. \end{aligned}$$

Corollary 8.33 provides the following formula for any x > 0

$$\displaystyle \begin{aligned} \lim_{m\to\infty}(H_{mx-1}-H_{m-1}) ~=~ \ln x. \end{aligned}$$

Analogue of Wallis’s Product Formula

The analogue of Wallis’s formula reduces to the classical identity

$$\displaystyle \begin{aligned} \sum_{k=1}^{\infty}(-1)^{k-1}\frac{1}{k} ~=~ \ln 2{\,}. \end{aligned}$$

Project 10.1

Find the analogue of Wallis’s formula for the function g(x) = ψ(x). We apply our method (see Sect. 9.7) to the function

$$\displaystyle \begin{aligned} \tilde{g}(x) ~=~ \Delta g(2x) ~=~ \frac{1}{2x}{\,}. \end{aligned}$$

Thus, we get

$$\displaystyle \begin{aligned} h(x) ~=~ \psi(2n)-\psi(1)-\frac{1}{2}\,\gamma-\frac{1}{2}\ln n ~=~ \frac{1}{2}(\gamma +\ln(4n))+O\left(n^{-1}\right), \end{aligned}$$

and the analogue of Wallis’s formula for g(x) = ψ(x) is

$$\displaystyle \begin{aligned} \lim_{n\to\infty}\left(-\ln(4n)+2\sum_{k=1}^{2n}(-1)^k\psi(k)\right) ~=~ \gamma{\,}. \end{aligned}$$

This provides yet another formula to define Euler’s constant γ. \(\lozenge \)

Restriction to the Natural Integers

For any \(n\in \mathbb {N}\) we have

$$\displaystyle \begin{aligned} H_n ~=~ \sum_{k=1}^n\frac{1}{k}{\,}. \end{aligned}$$

Gregory’s formula states that for any \(n\in \mathbb {N}^*\) and any \(q\in \mathbb {N}\) we have

$$\displaystyle \begin{aligned} H_{n-1} ~=~ \ln n-\sum_{j=1}^q|G_j|\left(\mathrm{B}(n,j)-\frac{1}{j}\right)-R^q_n{\,}, \end{aligned}$$

with

$$\displaystyle \begin{aligned} |R^q_n| ~\leq ~\overline{G}_q \left|\mathrm{B}(n,q+1)-\frac{1}{q}\right|. \end{aligned}$$

Many representations of H n can be derived from, e.g., the limit and series representations of the digamma function described above. For instance, using the generalized Liu formula, we get (see also Remark 8.47)

$$\displaystyle \begin{aligned} H_n ~=~ \ln n+\gamma +\frac{1}{2n}+\int_n^{\infty}\frac{\{t\}-\frac{1}{2}}{t^2}{\,}dt ~=~ \ln n+\frac{1}{2}+\frac{1}{2n}-\int_1^n\frac{\{t\}-\frac{1}{2}}{t^2}{\,}dt{\,}. \end{aligned}$$

Generalized Webster’s Functional Equation

For any \(m\in \mathbb {N}^*\) and any a > 0, there is a unique eventually monotone solution \(f\colon \mathbb {R}_+\to \mathbb {R}\) to the equation

$$\displaystyle \begin{aligned} \sum_{j=0}^{m-1}f(x+a j) ~=~ \frac{1}{x}{\,}, \end{aligned}$$

namely

$$\displaystyle \begin{aligned} f(x) ~=~ \frac{1}{am}\,\psi\left(\frac{x+a}{am}\right)-\frac{1}{am}\,\psi\left(\frac{x}{am}\right){\,}. \end{aligned}$$

Analogue of Euler’s Series Representation of γ

We have ψ(1) = −γ and

$$\displaystyle \begin{aligned} \psi_k(1) ~=~ (-1)^{k-1} k!\,\zeta(k+1){\,},\qquad k\in\mathbb{N}^*. \end{aligned}$$

Thus, the Taylor series expansion of ψ(x + 1) about x = 0 is

$$\displaystyle \begin{aligned} H_x ~=~ \psi(x+1) + \gamma ~=~ \sum_{k=1}^{\infty}(-1)^{k-1}\zeta(k+1){\,}x^k{\,},\qquad |x|< 1. \end{aligned}$$

Integrating both sides of this equation on (0, 1), we retrieve Euler’s series representation of γ

$$\displaystyle \begin{aligned} \gamma ~=~ \sum_{k=2}^{\infty}(-1)^k\,\frac{\zeta(k)}{k}{\,}. \end{aligned}$$

Analogue of the Reflection Formula

For any \(x\in \mathbb {R}\setminus \mathbb {Z}\), we have

$$\displaystyle \begin{aligned} \psi(x)-\psi(1-x) ~=~ -\pi\cot{}(\pi x). \end{aligned}$$

10.3 The Polygamma Functions

We now investigate the polygamma functions ψ ν for any \(\nu \in \mathbb {Z}\). In this context, our results will prove to be particularly interesting when ν ≤−2, that is, when the function ψ ν has a strictly positive asymptotic degree.

For any \(\nu \in \mathbb {Z}\), we set g ν =  Δψ ν; hence we have \(g^{\prime }_{\nu }=g_{\nu +1}\) and \(\psi ^{\prime }_{\nu }=\psi _{\nu +1}\). It follows immediately that

$$\displaystyle \begin{aligned} \Sigma g_{\nu}(x) ~=~ \psi_{\nu}(x)-\psi_{\nu}(1). \end{aligned}$$

(The cases ν = 0 and ν = −1 correspond to the functions ψ(x) and \(\ln \Gamma (x)\), respectively, and have been already considered in the previous sections.) We will often deal with the cases ν ≥ 1 and ν ≤−1 separately. In the latter case, we will often consider the value ν = −2 for simplicity and brevity.

ID Card When ν 1

Here we clearly have

$$\displaystyle \begin{aligned} g_{\nu}(x) ~=~ D^{\nu}_x\frac{1}{x} ~=~ (-1)^{\nu}\,\frac{\nu!}{x^{\nu+1}} \end{aligned}$$

and (see Example 7.6)

$$\displaystyle \begin{aligned} \psi_{\nu}(1) ~=~ (-1)^{\nu+1}\nu!{\,}\zeta(\nu+1). \end{aligned}$$

Hence we have the following table.

g ν(x)

Membership

deg g ν

Σg ν(x)

\(\mathstrut ^{\mathstrut }_{\mathstrut } (-1)^{\nu }\nu !{\,}x^{-\nu -1}\)

\(\mathcal {C}^{\infty }\cap \mathcal {D}^{-1}\cap \mathcal {K}^{\infty }\)

− 1

ψ ν(x) − ψ ν(1)

ID Card When ν ≤−1

Using (8.9), we obtain the following recurrence to compute the functions g ν. For any integer ν ≤−1, we have

$$\displaystyle \begin{aligned} \begin{array}{rcl} g_{\nu-1}(x) & =&\displaystyle \int_x^{x+1}\psi_{\nu}(t){\,}dt ~=~ \int_0^x g_{\nu}(t){\,}dt + \int_0^1\psi_{\nu}(t){\,}dt\\ & =&\displaystyle \int_0^x g_{\nu}(t){\,}dt + \psi_{\nu -1}(1). \end{array} \end{aligned} $$

In particular,

$$\displaystyle \begin{aligned} \lim_{x\to 0}g_{\nu -1}(x) ~=~ \psi_{\nu -1}(1) ~=~ \int_0^1\psi_{\nu}(t){\,}dt. \end{aligned}$$

Unfolding this recurrence, we obtain \(g_{-1}(x)=\ln x\) and, for any integer ν ≤−1,

$$\displaystyle \begin{aligned} g_{\nu -1}(x) ~=~ \int_0^x\frac{(x-t)^{-\nu -1}}{(-\nu -1)!}{\,}\ln t{\,}dt+\sum_{j=0}^{-\nu -1} \psi_{\nu +j-1}(1)\,\frac{x^j}{j!}{\,}, \end{aligned} $$
(10.6)

which is precisely the (−ν − 1)th order Taylor expansion of g ν−1(x).

Thus, we have

$$\displaystyle \begin{aligned} \begin{array}{rcl} g_{-1}(x) & =&\displaystyle \ln x{\,},\\ g_{-2}(x) & =&\displaystyle x\ln x-x+\frac{1}{2}\ln(2\pi){\,},\\ g_{-3}(x) & =&\displaystyle \frac{1}{2}{\,}x^2\ln x-\frac{3}{4}{\,}x^2+\left(\frac{1}{2}{\,}x+\frac{1}{4}\right)\ln(2\pi)+\ln A. \end{array} \end{aligned} $$

Hence the following ID card

g ν(x)

Membership

deg g ν

Σg ν(x)

\(\mathstrut ^{\mathstrut }_{\mathstrut } \mbox{Eq. (10.6)}\)

\(\mathcal {C}^{\infty }\cap \mathcal {D}^{-\nu }\cap \mathcal {K}^{\infty }\)

− ν − 1

ψ ν(x) − ψ ν(1)

Analogue of Bohr-Mollerup’s Theorem

The function ψ ν can be characterized as follows.

All solutions \(f\colon \mathbb {R}_+\to \mathbb {R}\) to the equation f(x + 1) − f(x) = g ν(x) that lie in \(\mathcal {K}^{(-\nu )_+}\) are of the form f(x) = c ν + ψ ν(x), where \(c_{\nu }\in \mathbb {R}\).

When ν ≥ 1, this characterization enables us to prove easily the following integral representation of ψ ν

$$\displaystyle \begin{aligned} \psi_{\nu}(x) ~=~ (-1)^{\nu -1}\int_0^{\infty}\frac{t^{\nu}{\,}e^{-xt}}{1-e^{-t}}{\,}dt{\,},\qquad x>0. \end{aligned}$$

Indeed, both sides of this identity coincide at x = 1 and are eventually monotone solutions to the equation Δf = g ν. Hence they must coincide on \(\mathbb {R}_+\).

Extended ID Card

The asymptotic constant σ[g ν] satisfies the following identity

$$\displaystyle \begin{aligned} \sigma[g_{\nu}] ~=~ \int_0^1\psi_{\nu}(t+1){\,}dt -\psi_{\nu}(1) ~=~ g_{\nu-1}(1)-\psi_{\nu}(1). \end{aligned}$$

Moreover, if ν ≥ 1 we also have

$$\displaystyle \begin{aligned} \begin{array}{rcl} \sigma[g_{\nu}] ~=~ \gamma[g_{\nu}] & =&\displaystyle \sum_{k=1}^{\infty}g_{\nu}(k)-\int_1^{\infty}g_{\nu}(t){\,}dt\\ & =&\displaystyle (-1)^{\nu}\,\Gamma(\nu){\,}(\nu\,\zeta(\nu+1)-1) \end{array} \end{aligned} $$

and hence the following values

\(\overline {\sigma }[g_{\nu }]\)

σ[g ν]

γ[g ν]

\(\mathstrut ^{\mathstrut }_{\mathstrut } \infty \)

(−1)ν  Γ(ν) (ν ζ(ν + 1) − 1)

γ[g ν] = σ[g ν]

For ν ≤−1 we have the values

\(\overline {\sigma }[g_{\nu }]\)

σ[g ν]

γ[g ν]

\(\mathstrut ^{\mathstrut }_{\mathstrut } \psi _{\nu -1}(1)-\psi _{\nu }(1)\)

g ν−1(1) − ψ ν(1)

\(\sigma [g_{\nu }]-\sum _{j=1}^{-\nu }G_j\Delta ^{j-1}g_{\nu }(1)\)

For instance we have

$$\displaystyle \begin{aligned} \overline{\sigma}[g_{-2}] ~=~ \ln A-\frac{1}{4}\ln(2\pi),\qquad \sigma[g_{-2}] ~=~ \ln A+\frac{1}{4}\ln(2\pi)-\frac{3}{4}{\,}, \end{aligned}$$

and

$$\displaystyle \begin{aligned} \gamma[g_{-2}] ~=~ \ln A+\frac{1}{6}\ln 2-\frac{1}{3}{\,}. \end{aligned}$$

We also have the following identities.

  • Alternative representations of σ[g]

    $$\displaystyle \begin{aligned} \begin{array}{rcl} \sigma[g_{\nu}] & =&\displaystyle \sum_{j=1}^{(-\nu)_+}G_j\,\Delta^{j-1}g_{\nu}(1) -\sum_{k=1}^{\infty}\left(\Delta g_{\nu -1}(k)-\sum_{j=0}^{(-\nu)_+}G_j\,\Delta^jg_{\nu}(k)\right),\\ \sigma[g_{\nu}] & =&\displaystyle \lim_{n\to\infty}\left(\sum_{k=1}^{n-1}g_{\nu}(k)+g_{\nu -1}(1)-g_{\nu -1}(n)+\sum_{j=1}^{(-\nu)_+}G_j\,\Delta^{j-1}g_{\nu}(n)\right), \\ \sigma[g_{\nu}] & =&\displaystyle \lim_{n\to\infty}\left(\sum_{k=1}^{n-1}g_{\nu}(k)+g_{\nu -1}(1)-g_{\nu -1}(n)-\sum_{j=1}^{(-\nu)_+}\frac{B_j}{j!}{\,}g_{\nu +j-1}(n)\right). \end{array} \end{aligned} $$

    If ν ≥ 1, then

    $$\displaystyle \begin{aligned} \sigma[g_{\nu}] ~=~ (-1)^{\nu}{\,}\nu !\left(\frac{1}{2}-(\nu +1)\int_1^{\infty}\frac{\{t\}-\frac{1}{2}}{t^{\nu +2}}{\,}dt\right). \end{aligned}$$

    If ν ≤−1, then for any integer q ≥⌈−ν∕2⌉,

    $$\displaystyle \begin{aligned} \sigma[g_{\nu}] ~=~ \frac{1}{2}{\,}g_{\nu}(1)-\sum_{k=1}^q\frac{B_{2k}}{(2k)!}{\,}g_{\nu +2k-1}(1) - \int_1^{\infty}\frac{B_{2q}(\{t\})}{(2q)!}{\,}g_{\nu + 2q}(t){\,}dt. \end{aligned}$$
  • Representations of γ[g]

    $$\displaystyle \begin{aligned} \begin{array}{rcl} \gamma[g_{\nu}] & =&\displaystyle \sigma[g_{\nu}]-\sum_{j=1}^{(-\nu)_+}G_j\Delta^{j-1}g_{\nu}(1){\,},\\ \gamma[g_{\nu}] & =&\displaystyle \int_1^{\infty}\left(\sum_{j=0}^{(-\nu)_+}G_j\,\Delta^jg_{\nu}(\lfloor t\rfloor)-g_{\nu}(t)\right)dt{\,},\\ \gamma[g_{\nu}] & =&\displaystyle \int_1^{\infty}\left(\sum_{j=0}^{(-\nu)_+}{\{t\}\choose j}\,\Delta^jg_{\nu}(\lfloor t\rfloor)-g_{\nu}(t)\right)dt{\,}. \end{array} \end{aligned} $$
  • Generalized Binet’s function. For any \(q\in \mathbb {N}\) and any x > 0

    $$\displaystyle \begin{aligned} J^{q+1}[\psi_{\nu}](x) ~=~ \psi_{\nu}(x)-g_{\nu -1}(x)+\sum_{j=1}^qG_j\,\Delta^{j-1}g_{\nu}(x). \end{aligned}$$

    For instance,

    $$\displaystyle \begin{aligned} \begin{array}{rcl} J^3[\psi_{-2}](x) & =&\displaystyle \psi_{-2}(x)-\frac{1}{12}{\,}(x+1)\ln(x+1)+\frac{1}{12}{\,}(3x-1)^2\\ & &\displaystyle -\frac{1}{12}{\,}x(6x-7)\ln x-\frac{1}{2}{\,}x\ln(2\pi)-\ln A. \end{array} \end{aligned} $$
  • Analogue of Raabe’s formula

    $$\displaystyle \begin{aligned} \int_x^{x+1}\psi_{\nu}(t){\,}dt ~=~ g_{\nu -1}(x){\,},\qquad x>0. \end{aligned}$$
  • Alternative characterization. The function f = ψ ν is the unique solution lying in \(\mathcal {C}^0\cap \mathcal {K}^{(-\nu )_+}\) to the equation

    $$\displaystyle \begin{aligned} \int_x^{x+1}f(t){\,}dt ~=~ g_{\nu -1}(x){\,}, \qquad x>0. \end{aligned}$$

Inequalities When ν 1

The following inequalities hold for any x > 0, any a ≥ 0, and any \(n\in \mathbb {N}^*\).

  • Symmetrized generalized Wendel’s inequality (equality if a ∈{0, 1})

    $$\displaystyle \begin{aligned} \left|\psi_{\nu}(x+a)-\psi_{\nu}(x)\right| ~\leq ~ \lceil a\rceil{\,}\frac{\nu !}{x^{\nu +1}}{\,}. \end{aligned}$$
  • Symmetrized generalized Wendel’s inequality (discrete version)

    $$\displaystyle \begin{aligned} \left|\psi_{\nu}(x)-\psi_{\nu}(1)-\sum_{k=1}^{n-1}g_{\nu}(k)+\sum_{k=0}^{n-1}g_{\nu}(x+k)\right| ~\leq ~ \lceil x\rceil{\,}\frac{\nu !}{n^{\nu +1}}{\,}. \end{aligned}$$
  • Symmetrized Stirling’s and Burnside’s formulas-based inequalities

    $$\displaystyle \begin{aligned} \textstyle{\left|\psi_{\nu}\left(x+\frac{1}{2}\right)-g_{\nu -1}(x)\right| ~\leq ~ \left|\psi_{\nu}(x)-g_{\nu -1}(x)\right| ~\leq ~ |g_{\nu}(x)|}{\,}. \end{aligned}$$

    Considering for instance the value p = 1 in Corollary 6.12, we see that the latter inequality can be refined into

    $$\displaystyle \begin{aligned} \left|\psi_{\nu}(x)-g_{\nu -1}(x)+\frac{1}{2}{\,}g_{\nu}(x)\right| ~\leq ~ \frac{1}{2}{\,}|\Delta g_{\nu}(x)|. \end{aligned}$$
  • Additional inequality

    $$\displaystyle \begin{aligned} |\psi_{\nu}(x+n)| ~=~ \left|\sum_{k=n}^{\infty}g_{\nu}(x+k)\right| ~\leq ~ \left|\psi_{\nu}(n)\right|. \end{aligned}$$
  • Generalized Gautschi’s inequality

    $$\displaystyle \begin{aligned} \begin{array}{rcl} (-1)^{\nu -1}(a-\lceil a\rceil)\,\psi_{\nu +1}(x+\lceil a\rceil) & \leq &\displaystyle (-1)^{\nu -1}(\psi_{\nu}(x+a)-\psi_{\nu}(x+\lceil a\rceil))\\ & \leq &\displaystyle (-1)^{\nu -1}(a-\lceil a\rceil){\,}g_{\nu}(x+\lfloor a\rfloor){\,}. \end{array} \end{aligned} $$

Inequalities When ν ≤−1

The following inequalities hold for any x > 0, any a ≥ 0, and any \(n\in \mathbb {N}^*\).

  • Symmetrized generalized Wendel’s inequality (equality if a ∈{0, 1, …, −ν})

    $$\displaystyle \begin{aligned} \begin{array}{rcl} & &\displaystyle {\left|\psi_{\nu}(x+a)-\psi_{\nu}(x)-\sum_{j=1}^{-\nu}{\textstyle{{{a}\choose{j}}}}\,\Delta^{j-1}g_{\nu}(x)\right|}\\ & \leq &\displaystyle \left|{\textstyle{{{a-1}\choose{-\nu}}}}\right|\left|\Delta^{-\nu -1}g_{\nu}(x+a)-\Delta^{-\nu -1}g_{\nu}(x)\right|\\ & \leq &\displaystyle \lceil a\rceil\left|{\textstyle{{{a-1}\choose{-\nu}}}}\right|\left|\Delta^{-\nu}g_{\nu}(x)\right|. \end{array} \end{aligned} $$
  • Symmetrized generalized Wendel’s inequality (discrete version)

    $$\displaystyle \begin{aligned} \begin{array}{rcl} \left|\psi_{\nu}(x)-\psi_{\nu}(1)-f_n^{-\nu}[g_{\nu}](x)\right| & \leq &\displaystyle \left|{\textstyle{{{x-1}\choose{-\nu}}}}\right|\left|\Delta^{-\nu -1}g_{\nu}(x+n)-\Delta^{-\nu -1}g_{\nu}(n)\right|\\ & \leq &\displaystyle \lceil x\rceil\left|{\textstyle{{{x-1}\choose{-\nu}}}}\right|\left|\Delta^{-\nu}g_{\nu}(n)\right|, \end{array} \end{aligned} $$

    where

    $$\displaystyle \begin{aligned} f_n^{-\nu}[g_{\nu}](x) ~=~ \sum_{k=1}^{n-1}g_{\nu}(k)-\sum_{k=0}^{n-1} g_{\nu}(x+k)+\sum_{j=1}^{-\nu}{\textstyle{{{x}\choose{j}}}}\,\Delta^{j-1}g_{\nu}(n){\,}. \end{aligned}$$
  • Symmetrized Stirling’s formula-based inequality

    $$\displaystyle \begin{aligned} \begin{array}{rcl} & &\displaystyle {\left|\psi_{\nu}(x)-g_{\nu -1}(x)+\sum_{j=1}^{-\nu}G_j\Delta^{j-1}g_{\nu}(x)\right|}\\ & \leq &\displaystyle \int_0^1\left|{\textstyle{{{t-1}\choose{-\nu}}}}\right|\left|\Delta^{-\nu -1}g_{\nu}(x+t)-\Delta^{-\nu -1}g_{\nu}(x)\right|dt\\ & \leq &\displaystyle \overline{G}_{-\nu}\left|\Delta^{-\nu}g_{\nu}(x)\right| \end{array} \end{aligned} $$
  • Generalized Gautschi’s inequality

    Considering the function ψ −2, we obtain

    $$\displaystyle \begin{aligned} \begin{array}{rcl} (a-\lceil a\rceil)\,\psi_{-1}(x+\lceil a\rceil) & \leq &\displaystyle \psi_{-2}(x+a)-\psi_{-2}(x+\lceil a\rceil)\\ & \leq &\displaystyle (a-\lceil a\rceil){\,}g_{-2}(x+\lfloor a\rfloor), \end{array} \end{aligned} $$

    for any x + ⌊a⌋≥ x 0, where x 0 = 1.461… is the unique positive zero of the digamma function.

Generalized Stirling’s and Related Formulas When ν 1

For any a ≥ 0, we have the following limit and asymptotic equivalence as x →,

$$\displaystyle \begin{aligned} \psi_{\nu}(x+a) ~\sim ~ g_{\nu -1}(x) ~=~ (-1)^{\nu -1}\,\frac{(\nu -1)!}{x^{\nu}}{\,},\qquad \psi_{\nu}(x)~\to ~ 0. \end{aligned}$$

Burnside-like approximation (better than Stirling-like approximation)

$$\displaystyle \begin{aligned} \textstyle{\psi_{\nu}(x)-g_{\nu -1}(x-\frac{1}{2})} ~\to ~ 0{\,}. \end{aligned}$$

Generalized Stirling’s and Related Formulas When ν ≤−1

For any a ≥ 0, we have the following limits and asymptotic equivalence as x →,

$$\displaystyle \begin{aligned} \psi_{\nu}(x+a)-\psi_{\nu}(x)-\sum_{j=1}^{-\nu}{\textstyle{{{a}\choose{j}}}}\,\Delta^{j-1}g_{\nu}(x) ~ \to ~ 0, \end{aligned}$$
$$\displaystyle \begin{aligned} \psi_{\nu}(x)-g_{\nu -1}(x)+\sum_{j=1}^{-\nu}G_j\Delta^{j-1}g_{\nu}(x) ~\to ~ 0, \end{aligned}$$
$$\displaystyle \begin{aligned} \psi_{\nu}(x)-\sum_{k=0}^{-\nu}\frac{B_k}{k!}{\,}g_{\nu +k-1}(x) ~\to ~ 0, \end{aligned}$$
$$\displaystyle \begin{aligned} \psi_{\nu}(x+a) ~\sim ~ g_{\nu -1}(x) ~\sim ~ \frac{1}{(-\nu)!}{\,}x^{-\nu}\ln x. \end{aligned}$$

When ν = −2 for instance, these limits reduce to

$$\displaystyle \begin{aligned} \int_x^{x+a}\ln\Gamma(t){\,}dt -a \ln\left(\sqrt{2\pi}\,\frac{x^x}{e^x}\right)-{\textstyle{{{a}\choose{2}}}}\ln\left(\frac{(x+1)^{x+1}}{e{\,}x^x}\right) ~\to ~0 {\,}, \end{aligned} $$
$$\displaystyle \begin{aligned} \begin{array}{rcl} & &\displaystyle {\psi_{-2}(x)-\frac{1}{12}{\,}(x+1)\ln(x+1)+\frac{1}{12}{\,}(3x-1)^2}\\ & &\displaystyle -\frac{1}{12}{\,}x(6x-7)\ln x-\frac{1}{2}{\,}x\ln(2\pi) ~\to ~ \ln A{\,}, \end{array} \end{aligned} $$
$$\displaystyle \begin{aligned} \psi_{-2}(x) - \frac{1}{12}(6x^2-6x+1)\ln x +\frac{1}{4}(3x-2)x-\frac{1}{2}{\,}x\ln(2\pi) ~\to ~ \ln A{\,}, \end{aligned}$$
$$\displaystyle \begin{aligned} \psi_{-2}(x+a) ~\sim ~ \frac{1}{2}{\,}x^2\ln x{\,}. \end{aligned}$$

Asymptotic Expansions

For any \(m,q\in \mathbb {N}^*\) we have the following expansion as x →

$$\displaystyle \begin{aligned} \frac{1}{m}\sum_{j=0}^{m-1}\psi_{\nu}\left(x+\frac{j}{m}\right) ~=~ \sum_{k=0}^q\frac{B_k}{m^k{\,}k!}{\,}g_{\nu +k-1}(x)+O(g_{\nu +q}(x)){\,}. \end{aligned}$$

Setting m = 1 in this formula, we obtain

$$\displaystyle \begin{aligned} \psi_{\nu}(x) ~=~ \sum_{k=0}^q\frac{B_k}{k!}{\,}g_{\nu +k-1}(x)+O(g_{\nu +q}(x)){\,}. \end{aligned}$$

For instance the asymptotic expansion of ψ −2 is

$$\displaystyle \begin{aligned} \begin{array}{rcl} \psi_{-2}(x) & =&\displaystyle \frac{1}{12}(6x^2-6x+1)\ln x -\frac{1}{4}(3x-2)x+\frac{1}{2}{\,}x\ln(2\pi)+\ln A\\ & &\displaystyle +\frac{1}{720x^2}+O\left(x^{-4}\right). \end{array} \end{aligned} $$

Generalized Liu’s Formula

For any ν ≥ 1 and any x > 0 we have

$$\displaystyle \begin{aligned} \psi_{\nu}(x) ~=~ (-1)^{\nu -1}\,\Gamma(\nu)\left(\frac{2x+\nu}{2x^{\nu +1}}+\nu (\nu +1)\int_0^{\infty}\frac{\frac{1}{2}-\{t\}}{(t+x)^{\nu +2}}{\,}dt\right). \end{aligned}$$

For ν = −2 and any x > 0 we have

$$\displaystyle \begin{aligned} \begin{array}{rcl} \psi_{-2}(x) & =&\displaystyle \frac{1}{12}(6x^2-6x+1)\ln x -\frac{1}{4}(3x-2)x+\frac{1}{2}{\,}x\ln(2\pi)+\ln A\\ & &\displaystyle +\int_0^{\infty}\frac{B_2(\{t\})}{2(x+t)}{\,}dt. \end{array} \end{aligned} $$

Limit and Series Representations When ν 1

The Eulerian and Weierstrassian forms of ψ ν reduce to

$$\displaystyle \begin{aligned} \psi_{\nu}(x) ~=~ -\sum_{k=0}^{\infty}g_{\nu}(x+k) ~=~ (-1)^{\nu -1}{\,}\nu !\,\zeta(\nu +1,x) \end{aligned}$$

and this series converges uniformly on \(\mathbb {R}_+\).

Limit and Series Representations When ν ≤−1

The analogue of Gauss’ limit is

$$\displaystyle \begin{aligned} \psi_{\nu}(x) ~=~ \psi_{\nu}(1)+\lim_{n\to\infty}f^{-\nu}_n[g_{\nu}](x) \end{aligned}$$

and both sides can be integrated on any bounded subset of [0, ) (the limit and the integral commute). They can also be differentiated infinitely many times (the limit and the derivative operator commute).

For instance, when ν = −2 we obtain

$$\displaystyle \begin{aligned} \begin{array}{rcl} \psi_{-2}(x) & =&\displaystyle \lim_{n\to\infty}\Bigg(\sum_{k=1}^{n-1}k\ln k-\sum_{k=0}^{n-1}(x+k)\ln(x+k)+x\left(n\ln n+\frac{1}{2}\ln(2\pi)\right) \\ & &\displaystyle + {x\choose 2}\left((n+1)\ln\left(1+\frac{1}{n}\right)+\ln n-1\right)\Bigg). \end{array} \end{aligned} $$

Comparing this formula with that of (10.2), we see that the latter is less complicated, since it was produced from less terms in its polynomial part. Now, differentiating the formula above, we obtain a limit representation for \(\ln \Gamma (x)\), but the Gauss limit is less complicated. In this context, finding the simplest limit representations seems to be an interesting problem.

The Eulerian and Weistrassian representations of ψ ν take the following forms

$$\displaystyle \begin{aligned} \begin{array}{rcl} \psi_{\nu}(x)-\psi_{\nu}(1) & =&\displaystyle -g_{\nu}(x)+\sum_{j=1}^{-\nu}{\textstyle{{{x}\choose{j}}}}\Delta^{j-1}g_{\nu}(1)\\ & &\displaystyle + \sum_{k=1}^{\infty}\left(-g_{\nu}(x+k)+\sum_{j=0}^{-\nu}{\textstyle{{{x}\choose{j}}}}\,\Delta^j g_{\nu}(k)\right) \end{array} \end{aligned} $$

and

$$\displaystyle \begin{aligned} \begin{array}{rcl} \psi_{\nu}(x)-\psi_{\nu}(1) & =&\displaystyle -g_{\nu}(x)+\sum_{j=1}^{-\nu-1}{\textstyle{{{x}\choose{j}}}}\Delta^{j-1}g_{\nu}(1)-\gamma{\textstyle{{{x}\choose{-\nu}}}}\\ & &\displaystyle + \sum_{k=1}^{\infty}\left(-g_{\nu}(x+k)+\sum_{j=0}^{-\nu -1}{\textstyle{{{x}\choose{j}}}}\,\Delta^j g_{\nu}(k)+{\textstyle{{{x}\choose{-\nu}}}}\frac{1}{k}\right), \end{array} \end{aligned} $$

respectively. These series can be integrated term by term on any bounded subset of [0, ). They can also be differentiated term by term infinitely many times.

For instance, when ν = −2, both identities above reduce to

$$\displaystyle \begin{aligned} \psi_{-2}(x) ~=~ \ln\left(\frac{(2\pi)^{\frac{1}{2}x}(\frac{4}{e})^{{x\choose 2}}}{x^x}{\,} \prod_{k=1}^{\infty}\frac{(1+2/k)^{(k+2){x\choose 2}}}{(1+x/k)^{x+k}{\,}(1+1/k)^{(k+1)x(x-2)}}\right) \end{aligned}$$

and

$$\displaystyle \begin{aligned} \psi_{-2}(x) ~=~ \ln\left(\frac{(2\pi)^{\frac{1}{2}x}e^{-\gamma{x\choose 2}}}{x^x}{\,} \prod_{k=1}^{\infty}\frac{e^{\frac{1}{k}{x\choose 2}}{\,}(1+1/k)^{(k+1)x}}{(1+x/k)^{x+k}}\right). \end{aligned}$$

Integrating both the Eulerian and Weierstrassian forms of \(\ln \Gamma (x)\), we obtain the following representations (which are simpler than the previous ones since less terms are involved; see also Examples 8.3 and 8.8)

$$\displaystyle \begin{aligned} \psi_{-2}(x) ~=~ \ln\left(\frac{e^x}{x^x}\,\prod_{k=1}^{\infty}\frac{e^x(1+1/k)^{x^2/2}}{(1+x/k)^{x+k}}\right) \end{aligned}$$

and

$$\displaystyle \begin{aligned} \psi_{-2}(x) ~=~ \ln\left(e^{-\gamma x^2/2}{\,}\frac{e^x}{x^x}{\,}\prod_{k=1}^{\infty}\frac{e^{x+x^2/(2k)}}{(1+x/k)^{x+k}}\right). \end{aligned}$$

Here again, finding the simplest Eulerian and Weierstrassian forms remains an interesting problem.

Integral Representation

For any \(\nu \in \mathbb {Z}\), we have

$$\displaystyle \begin{aligned} \psi_{\nu}(x) ~=~ \psi_{\nu}(1)+\int_1^x\psi_{\nu +1}(t){\,}dt. \end{aligned}$$

If ν ≥ 1, then ψ ν is not integrable at x = 0 (since g ν is not). If ν ≤−1, then ψ ν is integrable at 0 by definition and we have

$$\displaystyle \begin{aligned} \psi_{\nu-1}(x) ~=~ \int_0^x\psi_{\nu}(t){\,}dt ~=~ \int_0^x\frac{(x-t)^{-\nu-1}}{(-\nu-1)!}{\,}\ln\Gamma(t){\,}dt. \end{aligned}$$

Gregory’s Formula-Based Series Representation

Proposition 8.11 gives the following series representation: for any x > 0 we have

$$\displaystyle \begin{aligned} \begin{array}{rcl} \psi_{\nu}(x) & =&\displaystyle g_{\nu -1}(x)-\sum_{n=0}^{\infty}G_{n+1}\,\Delta^ng_{\nu}(x)\\ & =&\displaystyle g_{\nu -1}(x)-\sum_{n=0}^{\infty}|G_{n+1}|\,\sum_{k=0}^n(-1)^k{\textstyle{{{n}\choose{k}}}}g_{\nu}(x+k){\,}. \end{array} \end{aligned} $$

Setting x = 1 in this identity yields the analogue of Fontana-Mascheroni series. For instance, taking ν = 1, we derive the identity (see, e.g., Merlini et al. [72, p. 1920])

$$\displaystyle \begin{aligned} \sum_{n=1}^{\infty}|G_n|\,\frac{H_n}{n} ~=~ \frac{\pi^2}{6}-1{\,}. \end{aligned}$$

Taking ν = 2, we obtain

$$\displaystyle \begin{aligned} \sum_{n=1}^{\infty}|G_n|\,\frac{\psi_1(n+1)-H_n^2}{n} ~=~ 1-2\,\zeta(3)+\gamma\,\frac{\pi^2}{6}{\,}. \end{aligned}$$

Analogue of Gauss’ Multiplication Formula

Assume first that ν ≥ 1. Differentiating repeatedly both sides of the multiplication formula (10.5) for the digamma function ψ, we obtain the following formula. For any \(m\in \mathbb {N}^*\) and any x > 0, we have

$$\displaystyle \begin{aligned} \sum_{j=0}^{m-1}\psi_{\nu}\left(\frac{x+j}{m}\right) ~=~ m^{\nu +1}\,\psi_{\nu}(x). \end{aligned}$$

Moreover, Corollary 8.33 provides the following limit

$$\displaystyle \begin{aligned} \lim_{m\to\infty}m^{\nu}\psi_{\nu}(mx) ~=~ g_{\nu -1}(x),\qquad x>0. \end{aligned}$$

Assume now that ν ≤−1. Applying Theorem 8.27 to the function g ν, we obtain that for any \(m\in \mathbb {N}^*\) and any x > 0

$$\displaystyle \begin{aligned} \sum_{j=0}^{m-1}\psi_{\nu}\left(\frac{x+j}{m}\right) ~=~ \sum_{j=1}^{m-1}\psi_{\nu}\left(\frac{j}{m}\right)+\psi_{\nu}(1)+\Sigma_x{\,}g_{\nu}\left(\frac{x}{m}\right){\,}. \end{aligned}$$

Let us expand this formula in the special case when ν = −2. First, we have

$$\displaystyle \begin{aligned} g_{-2}\left(\frac{x}{m}\right) ~=~ \frac{1}{m}{\,}g_{-2}(x)-x\frac{\ln m}{m}+\frac{m-1}{m}\,\psi_{-2}(1) \end{aligned}$$

and hence

$$\displaystyle \begin{aligned} \Sigma_x{\,}g_{-2}\left(\frac{x}{m}\right) ~=~ \frac{1}{m}{\,}\psi_{-2}(x)-{x\choose 2}\,\frac{\ln m}{m}+\left(\frac{m-1}{m}{\,}x-1\right)\,\psi_{-2}(1). \end{aligned}$$

Using Proposition 8.28, after some algebra we also obtain

$$\displaystyle \begin{aligned} \sum_{j=1}^{m-1}\psi_{-2}\left(\frac{j}{m}\right) ~=~ \left(1-\frac{1}{m}\right)\ln A-\frac{\ln m}{12{\,}m}+(m-1)\ln((2\pi)^{\frac{1}{4}}A). \end{aligned}$$

Now, collecting terms, we finally get the following multiplication formula for ψ −2

$$\displaystyle \begin{aligned} \begin{array}{rcl} \sum_{j=0}^{m-1}\psi_{-2}\left(\frac{x+j}{m}\right) & =&\displaystyle \frac{1}{m}\,\psi_{-2}(x)-\frac{1}{12m}{\,}(6x^2-6x+1)\ln m \\ & &\displaystyle + (m-1)\ln(2\pi)\left(\frac{x}{2m}+\frac{1}{4}\right)+\left(m-\frac{1}{m}\right)\ln A. \end{array} \end{aligned} $$

Setting m = 2 in the formula above, we obtain the following analogue of Legendre’s duplication formula

$$\displaystyle \begin{aligned} \begin{array}{rcl} \psi_{-2}\left(\frac{x}{2}\right)+\psi_{-2}\left(\frac{x+1}{2}\right) & =&\displaystyle \frac{1}{2}\,\psi_{-2}(x)-\frac{1}{24}{\,}(6x^2-6x+1)\ln 2\\ & &\displaystyle +\frac{1}{4}\ln(2\pi){\,}(x+1)+\frac{3}{2}\ln A . \end{array} \end{aligned} $$

Taking x = 0 in this latter identity, we obtain

$$\displaystyle \begin{aligned} \psi_{-2}\left(\frac{1}{2}\right) ~=~ \frac{5}{24}\ln 2+\frac{3}{2}\ln A+\frac{1}{4}\ln\pi{\,}. \end{aligned}$$

Moreover, Corollary 8.33 provides the following limit

$$\displaystyle \begin{aligned} \lim_{m\to\infty}\left(\frac{1}{m^2}\,\psi_{-2}(mx)-\frac{x^2}{2}\ln m\right) ~=~ \frac{1}{2}{\,}x^2\ln x-\frac{3}{4}{\,}x^2{\,},\qquad x>0. \end{aligned}$$

Analogue of Wallis’s Product Formula

If ν ≥ 1, then the analogue of Wallis’s formula is simply

$$\displaystyle \begin{aligned} \sum_{k=1}^{\infty}(-1)^{k-1}g_{\nu}(k) ~=~ (-1)^{\nu}(1-2^{-\nu}){\,}\nu!{\,}\zeta(\nu +1), \end{aligned}$$

or equivalently,

$$\displaystyle \begin{aligned} \sum_{k=1}^{\infty}(-1)^{k-1}g_{\nu}(k) ~=~ (-1)^{\nu}{\,}\nu!{\,}\eta(\nu +1), \end{aligned}$$

where η is Dirichlet’s eta function. In the case when ν = −2, after a bit of calculus we obtain the following analogue of Wallis’s formula

$$\displaystyle \begin{aligned} \lim_{n\to\infty}\left(h(n)+\sum_{k=1}^{2n}(-1)^{k-1}g_{-2}(k)\right) ~=~ \frac{1}{12}\ln 2-3\ln A. \end{aligned}$$

where

$$\displaystyle \begin{aligned} h(n) ~=~ \left(n+\frac{1}{4}\right)\ln n - n(1-\ln 2). \end{aligned}$$

Project 10.2

Find the analogue of Wallis’s formula for the function g(x) = ψ −2(x). After some algebra, we obtain

$$\displaystyle \begin{aligned} \lim_{n\to\infty}\left(h(n)+\sum_{k=1}^{2n}(-1)^{k-1}\psi_{-2}(k)\right) ~=~ \ln A-\frac{1}{12}\,\ln 2{\,}, \end{aligned}$$

where

$$\displaystyle \begin{aligned} h(n) ~=~ n^2\ln(2n)-\frac{3}{2}{\,}n^2+\frac{1}{2}{\,}n\,\ln(2\pi)-\frac{1}{12}\,\ln n. \end{aligned}$$

This formula is a little harder to obtain than the former one; it requires the computation of both functions Σψ −2(x) and 2 Σx ψ −2(2x) using the elevator method (Corollary 7.20) with r = 2. That is,

$$\displaystyle \begin{aligned} \begin{array}{rcl} \Sigma\psi_{-2}(x) & =&\displaystyle -\frac{1}{12}{\,}x(x-1)(2x-1)+\frac{1}{4}{\,}x(x+1)\ln(2\pi)\\ & &\displaystyle +2x\ln A+(x-1)\,\psi_{-2}(x)-2\,\psi_{-3}(x){} \end{array} \end{aligned} $$
(10.7)

and

$$\displaystyle \begin{aligned} \begin{array}{rcl} 2{\,}\Sigma_x\psi_{-2}(2x) & =&\displaystyle -\frac{1}{6}{\,}x(2x-1)(4x-1)+(4x+3)\ln A\\ & &\displaystyle +\frac{1}{12}{\,}(-24x^2+48x+5)\ln 2-4\,\psi_{-2}(x)\\ & &\displaystyle +2x\,\psi_{-2}(2x)-2\,\psi_{-2}\left(x+\frac{1}{2}\right)-2\,\psi_{-3}(2x). \end{array} \end{aligned} $$

These formulas can also be verified using the difference operator. \(\lozenge \)

Restriction to the Natural Integers When ν 1

For any \(n\in \mathbb {N}^*\), we have

$$\displaystyle \begin{aligned} \psi_{\nu}(n)-\psi_{\nu}(1) ~=~ \sum_{k=1}^{n-1}g_{\nu}(k) ~=~ (-1)^{\nu}\nu!{\,}\sum_{k=1}^{n-1}\frac{1}{k^{\nu +1}}{\,}. \end{aligned}$$

In particular,

$$\displaystyle \begin{aligned} \psi_{\nu}(1) ~=~ -\sum_{k=1}^{\infty}g_{\nu}(k). \end{aligned}$$

Gregory’s formula states that for any \(n\in \mathbb {N}^*\) and any \(q\in \mathbb {N}\) we have

$$\displaystyle \begin{aligned} \begin{array}{rcl} \sum_{k=1}^{n-1}g_{\nu}(k) & =&\displaystyle g_{\nu -1}(n)-g_{\nu -1}(1)\\ & &\displaystyle -\sum_{j=1}^qG_j\left(\Delta^{j-1}g_{\nu}(n)-\Delta^{j-1}g_{\nu}(1)\right)-R^q_n{\,}, \end{array} \end{aligned} $$

with

$$\displaystyle \begin{aligned} |R^q_n| ~\leq ~\overline{G}_q\left|\Delta^qg_{\nu}(n)-\Delta^qg_{\nu}(1)\right|. \end{aligned}$$

Generalized Webster’s Functional Equation

For any \(m\in \mathbb {N}^*\), there is a unique solution \(f\colon \mathbb {R}_+\to \mathbb {R}\) to the equation

$$\displaystyle \begin{aligned} \sum_{j=0}^{m-1}f\left(x+\frac{j}{m}\right) ~=~ g_{\nu}(x) \end{aligned}$$

that lies in \(\mathcal {K}^{(-\nu )_+}\), namely

$$\displaystyle \begin{aligned} f(x) ~=~ \psi_{\nu}\left(x+\frac{1}{m}\right)-\psi_{\nu}(x){\,}. \end{aligned}$$

Analogue of Euler’s Series Representation of γ

Assume first that ν ≥ 1. In this case, for any \(k\in \mathbb {N}\) we have

$$\displaystyle \begin{aligned} \psi_{\nu}^{(k)}(1) ~=~ \psi_{\nu +k}(1) ~=~ (-1)^{\nu +k-1} (\nu +k)!\,\zeta(\nu +k+1). \end{aligned}$$

Thus, the Taylor series expansion of ψ ν(x + 1) about x = 0 is

$$\displaystyle \begin{aligned} \psi_{\nu}(x+1) ~=~ \sum_{k=0}^{\infty}(-1)^{\nu +k-1}\frac{(\nu +k)!}{k!}\,\zeta(\nu +k+1){\,}x^k,\qquad |x|<1. \end{aligned}$$

Integrating both sides of this equation on (0, 1), we obtain the identity

$$\displaystyle \begin{aligned} g_{\nu -1}(1) ~=~ \sum_{k=0}^{\infty}(-1)^{\nu +k-1}\frac{(\nu +k)!}{(k+1)!}\,\zeta(\nu +k+1){\,}. \end{aligned}$$

We proceed similarly when ν ≤−1. To keep the computations simple, let us assume that ν = −2. We then have

$$\displaystyle \begin{aligned} \psi_{-2}(1) ~=~ \frac{1}{2}\ln(2\pi),\quad \psi^{\prime}_{-2}(1) ~=~ \psi_{-1}(1) ~=~ 0,\quad \psi^{\prime\prime}_{-2}(1) ~=~ \psi_0(1) ~=~ -\gamma, \end{aligned}$$

and for any integer k ≥ 3,

$$\displaystyle \begin{aligned} \psi_{-2}^{(k)}(1) ~=~ \psi_{k-2}(1) ~=~ (-1)^{k-1}(k-2)!\,\zeta(k-1). \end{aligned}$$

Thus, the Taylor series expansion of ψ −2(x + 1) about x = 0 is

$$\displaystyle \begin{aligned} \psi_{-2}(x+1) ~=~ \frac{1}{2}\ln(2\pi)-\gamma\,\frac{x^2}{2}+\sum_{k=3}^{\infty}(-1)^{k-1}\frac{\zeta(k-1)}{(k-1)k}{\,}x^k,\qquad |x|<1. \end{aligned}$$

Integrating both sides of this equation on (0, 1), we obtain

$$\displaystyle \begin{aligned} \sum_{k=2}^{\infty}(-1)^k\,\frac{\zeta(k)}{k(k+1)(k+2)} ~=~ \frac{1}{6}\,\gamma-\frac{3}{4}+\frac{1}{4}\ln(2\pi)+\ln A{\,}. \end{aligned}$$

Analogue of the Reflection Formula

Assume first that ν ≥ 1. Differentiating the reflection formula for ψ repeatedly, we obtain the following formula. For any \(x\in \mathbb {R}\setminus \mathbb {Z}\), we have

$$\displaystyle \begin{aligned} \psi_{\nu}(x)-(-1)^{\nu}\psi_{\nu}(1-x) ~=~ -\pi{\,}D^{\nu}\cot{}(\pi x). \end{aligned}$$

When ν ≤−1, a reflection formula on (0, 1) can be obtained by integrating both sides of the identity

$$\displaystyle \begin{aligned} \ln\Gamma(x)+\ln\Gamma(1-x) ~=~ \ln\pi-\ln\sin{}(\pi x). \end{aligned}$$

For example, for any x ∈ (0, 1) we have

$$\displaystyle \begin{aligned} \psi_{-2}(x)-\psi_{-2}(1-x) ~=~ x\,\ln\pi-\frac{1}{2}\ln(2\pi)-\int_0^x\ln\sin{}(\pi t){\,}dt. \end{aligned}$$

As a byproduct, we obtain

$$\displaystyle \begin{aligned} \int_0^{\frac{1}{2}}\ln\sin{}(\pi t){\,}dt ~=~ -\frac{1}{2}\ln 2. \end{aligned}$$

10.4 The q-Gamma Function

For any 0 < q < 1, the q-gamma function \(\Gamma _q\colon \mathbb {R}_+\to \mathbb {R}_+\) is defined by the equation (see, e.g., [93, p. 490])

$$\displaystyle \begin{aligned} \Gamma_q(x) ~=~ (1-q)^{1-x}\,\prod_{k=0}^{\infty}\frac{1-q^{k+1}}{1-q^{x+k}} ~=~ (1-q)^{1-x}\,\frac{(q;q)_{\infty}}{(q^x;q)_{\infty}}\qquad \mbox{for }x>0. \end{aligned} $$
(10.8)

Here we use the standard notation

$$\displaystyle \begin{aligned} (a;q)_{\infty} ~=~ \prod_{k=0}^{\infty}\left(1-aq^k\right). \end{aligned}$$

Note that these functions should not to be confused with the multiple gamma functions discussed in Sect. 5.2 (although the same symbols are used).

The function \(f_q(x)=\ln \Gamma _q(x)\) is a convex solution satisfying f q(1) = 0 to the equation Δf q = g q on \(\mathbb {R}_+\), where \(g_q\colon \mathbb {R}_+\to \mathbb {R}\) is the function defined by the equation

$$\displaystyle \begin{aligned} g_q(x) ~=~ \ln\frac{1-q^x}{1-q}\qquad \mbox{for }x>0. \end{aligned}$$

Since g q lies in \(\mathcal {D}^1\cap \mathcal {K}^1\) (and deg g q = 0), by the uniqueness theorem we must have

$$\displaystyle \begin{aligned} \ln\Gamma_q(x) ~=~ \Sigma g_q(x),\qquad x>0. \end{aligned} $$
(10.9)

Askey [13] proved an analogue of the Bohr-Mollerup theorem for Γq. However, as Webster [98, p. 615] already observed, this is actually an immediate consequence of the uniqueness Theorem 3.1 in the special case when p = 1.

Let us now investigate this function in the light of our results.

Remark 10.3

When q > 1, the q-gamma function \(\Gamma _q\colon \mathbb {R}_+\to \mathbb {R}_+\) is also defined by Eq. (10.9). In this case, using L’Hospital’s rule we can readily see that \(\Delta g_q(x)\to \ln q\) as x →, and hence deg g q = 1. An analogue of the Bohr-Mollerup characterization for Γq was established by Moak [74]. We can see now that this characterization is a trivial consequence of our uniqueness Theorem 3.1 in the special case when p = 2. The complete analysis of Γq through our results is similar to the case when 0 < q < 1 and is left to the reader. \(\lozenge \)

ID Card

As discussed above, the function Γq is a Γ-type function and we immediately derive the following basic information.

g q(x)

Membership

deg g q

Σg q(x)

\(\mathstrut ^{\mathstrut }_{\mathstrut } \ln \frac {1-q^x}{1-q_{\mathstrut }}\)

\(\mathcal {C}^{\infty }\cap \mathcal {D}^1\cap \mathcal {K}^{\infty }\)

0

\(\ln \Gamma _q(x)\)

Analogue of Bohr-Mollerup’s Theorem

The q-gamma function can be characterized as follows.

All eventually convex or concave solutions \(f_q\colon \mathbb {R}_+\to \mathbb {R}\) to the equation

$$\displaystyle \begin{aligned} f_q(x+1)-f_q(x) ~=~ \ln\frac{1-q^x}{1-q} \end{aligned}$$

are of the form \(f_q(x)=c_q+\ln \Gamma _q(x)\), where \(c_q\in \mathbb {R}\).

Using Proposition 3.9, we can also derive the following alternative characterization of the q-gamma function.

All solutions \(f_q\colon \mathbb {R}_+\to \mathbb {R}\) to the equation

$$\displaystyle \begin{aligned} f_q(x+1)-f_q(x) ~=~ \ln\frac{1-q^x}{1-q} \end{aligned}$$

that satisfy the asymptotic condition that, for each x > 0,

$$\displaystyle \begin{aligned} f_q(x+n)-f_q(n)-x\ln\frac{1-q^n}{1-q} ~\to ~0\qquad \mbox{as }n\to_{\mathbb{N}}\infty \end{aligned}$$

are of the form \(f_q(x)=c_q+\ln \Gamma _q(x)\), where \(c_q\in \mathbb {R}\).

Extended ID Card

Interestingly, El Bachraoui [35] recently established the following analogue of Raabe’s formula

$$\displaystyle \begin{aligned} \int_x^{x+1}\ln\Gamma_q(t){\,}dt ~=~ \left(\frac{1}{2}-x\right)\ln(1-q)-\frac{1}{\ln q}\,\mathrm{Li}_2(q^x)+\ln(q;q)_{\infty},\qquad x\geq 0, \end{aligned}$$

where

$$\displaystyle \begin{aligned} \mathrm{Li}_s(z) ~=~ \sum_{k=1}^{\infty}\frac{z^k}{k^s} \end{aligned}$$

is the polylogarithm function. This formula provides immediately the following values

$$\displaystyle \begin{aligned} \begin{array}{rcl} \overline{\sigma}[g_q] & =&\displaystyle \frac{1}{2}\ln(1-q)-\frac{\zeta(2)}{\ln q}+\ln(q;q)_{\infty}{\,},{} \end{array} \end{aligned} $$
(10.10)
$$\displaystyle \begin{aligned} \begin{array}{rcl} \sigma[g_q] & =&\displaystyle -\frac{1}{2}\ln(1-q)-\frac{1}{\ln q}\,\mathrm{Li}_2(q)+\ln(q;q)_{\infty}{\,},{} \end{array} \end{aligned} $$
(10.11)

and the integral

$$\displaystyle \begin{aligned} \int_1^x g_q(t){\,}dt ~=~ (1-x)\ln(1-q)-\frac{1}{\ln q}\left(\mathrm{Li}_2(q^x)-\mathrm{Li}_2(q)\right). \end{aligned}$$

We then have the following values

  • Alternative representations of σ[g q] = γ[g q]

    $$\displaystyle \begin{aligned} \begin{array}{rcl} \sigma[g_q] & =&\displaystyle \int_0^1\ln\Gamma_q(t+1){\,}dt{\,},\\ \sigma[g_q] & =&\displaystyle \log[q]\,\int_1^{\infty}\left(\frac{1}{2}-\{t\}\right)\frac{q^t}{1-q^t}{\,}dt{\,},\\ \sigma[g_q] & =&\displaystyle \int_1^{\infty}\ln\frac{(1-q^{\lfloor t\rfloor})^{1/2}(1-q^{\lfloor t+1\rfloor})^{1/2}}{1-q^t}{\,}dt{\,},\\ \sigma[g_q] & =&\displaystyle \frac{1}{2}\sum_{k=1}^{\infty}\ln\left((1-q^k)(1-q^{k+1})\right)-\frac{1}{\ln q}\,\mathrm{Li}_2(q){\,}. \end{array} \end{aligned} $$
  • Generalized Binet’s function

    $$\displaystyle \begin{aligned} \begin{array}{rcl} J^2[\ln\circ\Gamma_q](x) & =&\displaystyle \ln\Gamma_q(x)+(x-1)\ln(1-q)+\frac{1}{\ln q}\,\mathrm{Li}_2(q^x)+\frac{1}{2}\ln(1-q^x)\\ & &\displaystyle -\ln(q;q)_{\infty}{\,}. \end{array} \end{aligned} $$
  • Alternative characterization. The function \(f_q(x) = \ln \Gamma _q(x)\) is the unique solution lying in \(\mathcal {C}^0\cap \mathcal {K}^1\) to the equation

    $$\displaystyle \begin{aligned} \int_x^{x+1}f_q(t){\,}dt ~=~ \left(\frac{1}{2}-x\right)\ln(1-q)-\frac{1}{\ln q}\,\mathrm{Li}_2(q^x)+\ln(q;q)_{\infty},\qquad x>0. \end{aligned}$$

\(\overline {\sigma }[g_q]\)

σ[g q]

γ[g q]

\(\mathstrut ^{\mathstrut }_{\mathstrut } Eq.\,\mbox{ (10.10)}\)

Eq. (10.11)

γ[g q] = σ[g q]

Inequalities

The following inequalities hold for any x > 0 and any a ≥ 0.

  • Symmetrized generalized Wendel’s inequality (equality if a ∈{0, 1})

    $$\displaystyle \begin{aligned} \begin{array}{rcl} \left|\ln\Gamma_q(x+a)-\ln\Gamma_q(x)-a{\,}g_q(x)\right| & \leq &\displaystyle |a-1|{\,}\left|g_q(x+a)-g_q(x)\right|\\ & \leq &\displaystyle \lceil a\rceil{\,}|a-1|{\,}|\Delta g_q(x)|{\,}, \end{array} \end{aligned} $$
    $$\displaystyle \begin{aligned} \left(\frac{1-q^{x+a}}{1-q^x}\right)^{-|a-1|} \leq ~\frac{\Gamma_q(x+a)}{\Gamma_q(x)\left(\frac{1-q^x}{1-q}\right)^a} ~\leq ~ \left(\frac{1-q^{x+a}}{1-q^x}\right)^{|a-1|}. \end{aligned}$$
  • Symmetrized Stirling’s formula-based inequality

    $$\displaystyle \begin{aligned} |J^2[\ln\circ\Gamma_q](x)| ~\leq ~ \frac{1}{2}\left(g_q(x+1)-g_q(x)\right), \end{aligned}$$
    $$\displaystyle \begin{aligned} \left(\frac{1-q^{x+1}}{1-q^x}\right)^{-\frac{1}{2}} \leq ~ \frac{\Gamma_q(x){\,}(1-q)^{x-1}(1-q^x)^{\frac{1}{2}}}{(q;q)_{\infty}{\,}\exp\left(-\frac{1}{\ln q}\mathrm{Li}_2(q^x)\right)} ~\leq ~ \left(\frac{1-q^{x+1}}{1-q^x}\right)^{\frac{1}{2}}. \end{aligned}$$
  • Burnside’s formula-based inequality

    $$\displaystyle \begin{aligned}\displaystyle \left|\ln\Gamma_q\left(x+\frac{1}{2}\right)+\left(x-\frac{1}{2}\right)\ln(1-q)+\frac{1}{\ln q}\,\mathrm{Li}_2(q^x)-\ln(q;q)_{\infty}\right|\\\displaystyle \leq ~ |J^2[\ln\circ\Gamma_q](x)|. \end{aligned} $$
  • Generalized Gautschi’s inequality

    $$\displaystyle \begin{aligned} e^{(a-\lceil a\rceil)\,\psi_{q,0}(x+\lceil a\rceil)} ~\leq ~ \frac{\Gamma_q(x+a)}{\Gamma_q(x+\lceil a\rceil)} ~\leq ~ \left(\frac{1-q^{x+\lfloor a\rfloor}}{1-q}\right)^{a-\lceil a\rceil}, \end{aligned}$$

    where \(\psi _{q,0}(x)=D\ln \Gamma _q(x)\).

Generalized Stirling’s and Related Formulas

For any a ≥ 0, we have the following limits and asymptotic equivalences as x →,

$$\displaystyle \begin{aligned} \ln\Gamma_q(x+a)-\ln\Gamma_q(x) ~\to ~ -a\ln(1-q), \end{aligned}$$
$$\displaystyle \begin{aligned} \frac{\Gamma_q(x)}{\Gamma_q(x+a)} ~\sim ~ (1-q)^a,\qquad \ln\Gamma_q(x+a) ~\sim ~ -x\ln(1-q){\,}, \end{aligned}$$
$$\displaystyle \begin{aligned} \ln\Gamma_q(x)+(x-1)\ln(1-q)-\ln(q;q)_{\infty} ~\to ~ 0{\,}, \end{aligned}$$
$$\displaystyle \begin{aligned} \Gamma_q(x) ~\sim ~ (q;q)_{\infty}{\,}(1-q)^{1-x}{\,}. \end{aligned}$$

The generalized Stirling formula simply shows that \(\ln \Gamma _q(x)\) has the oblique asymptote

$$\displaystyle \begin{aligned} y ~=~ (1-x)\ln(1-q)+\ln(q;q)_{\infty}. \end{aligned}$$

Burnside-like approximation (better than Stirling-like approximation)

$$\displaystyle \begin{aligned} \Gamma_q(x) ~\sim ~ (q;q)_{\infty}{\,}(1-q)^{1-x}\,\exp\left(-\frac{1}{\ln q}\mathrm{Li}_2(q^{x-\frac{1}{2}})\right). \end{aligned}$$

Further results (obtained by differentiation). For any 0 < q < 1 and any \(\nu \in \mathbb {N}\), let the function \(\psi _{q,\nu }\colon \mathbb {R}_+\to \mathbb {R}\) denote the q-polygamma function defined by the equation

$$\displaystyle \begin{aligned} \psi_{q,\nu}(x) ~=~ D^{\nu +1}\ln\Gamma_q(x)\qquad \mbox{for }x>0. \end{aligned}$$

We then have the following limits and asymptotic equivalences as x →,

$$\displaystyle \begin{aligned} \psi_{q,0}(x+a)-\psi_{q,0}(x) ~\to ~ 0,\qquad \psi_{q,0}(x) ~\to ~ -\ln(1-q){\,}, \end{aligned}$$
$$\displaystyle \begin{aligned} \psi_{q,0}(x+a) ~\sim ~ -\ln(1-q),\qquad \psi_{q,\nu}(x) ~\to ~ 0,\qquad \nu\in\mathbb{N}^*. \end{aligned}$$

Project 10.4

Find the generalized Stirling formula when q > 1. In the case when q > 1, we have deg g q = 1 and hence the generalized Stirling formula is

$$\displaystyle \begin{aligned} \ln\Gamma_q(x)-\int_x^{x+1}\ln\Gamma_q(t){\,}dt+\frac{1}{2}{\,}g_q(x)-\frac{1}{12}\Delta g_q(x) ~\to ~ 0\qquad \mbox{as }x\to\infty, \end{aligned}$$

where \(\Delta g_q(x)\to \ln q\) as x →. However, here the integral takes the following more complicated form (see El Bachraoui [35] and the references therein)

$$\displaystyle \begin{aligned} \begin{array}{rcl} \int_x^{x+1}\ln\Gamma_q(t){\,}dt & =&\displaystyle \ln C_q-\frac{1}{2q^x\ln q}\Bigg(\frac{1-q^x}{1-q^{-x}}{\,}(2\,\mathrm{Li}_2(q^{-x})+(\ln(1-q^{-x}))^2)\\ & &\displaystyle - 2\,\frac{1-q^x}{1-q^{-x}}\ln\frac{1-q^x}{1-q}\ln(1-q^{-x})-q^x\left(\ln\frac{1-q^x}{1-q}\right)^2\Bigg) \end{array} \end{aligned} $$

where

$$\displaystyle \begin{aligned} C_q ~=~ q^{-\frac{1}{12}}(q-1)^{\frac{1}{2}-\frac{\ln(q-1)}{2\ln q}}(q^{-1};q^{-1})_{\infty}{\,}. \end{aligned}$$

This is the analogue of Raabe’s formula for \(\ln \Gamma _q(x)\) when q > 1. \(\lozenge \)

Asymptotic Expansions

For any \(m,r\in \mathbb {N}^*\) we have the following expansion as x →

$$\displaystyle \begin{aligned} \begin{array}{rcl} \frac{1}{m}\sum_{j=0}^{m-1}\ln\Gamma_q\left(x+\frac{j}{m}\right) & =&\displaystyle \left(\frac{1}{2}-x\right)\ln(1-q)-\frac{1}{\ln q}\,\mathrm{Li}_2(q^x)+\ln(q;q)_{\infty}\\ & &\displaystyle + \sum_{k=1}^r\frac{B_k}{m^k{\,}k!}{\,}g_q^{(k-1)}(x)+O\left(g_q^{(r)}(x)\right). \end{array} \end{aligned} $$

Setting m = 1 in this formula, we obtain the expansion of the log-q-gamma function

$$\displaystyle \begin{aligned} \begin{array}{rcl} \ln\Gamma_q(x) & =&\displaystyle \left(\frac{1}{2}-x\right)\ln(1-q)-\frac{1}{\ln q}\,\mathrm{Li}_2(q^x)+\ln(q;q)_{\infty}\\ & &\displaystyle + \sum_{k=1}^r\frac{B_k}{k!}{\,}g_q^{(k-1)}(x)+O\left(g_q^{(r)}(x)\right). \end{array} \end{aligned} $$

Generalized Liu’s Formula

For any x > 0, we have

$$\displaystyle \begin{aligned} \begin{array}{rcl} \ln\Gamma_q(x) & =&\displaystyle \left(\frac{1}{2}-x\right)\ln(1-q)-\frac{1}{\ln q}\,\mathrm{Li}_2(q^x)+\ln(q;q)_{\infty}\\ & &\displaystyle -\frac{1}{2}\ln\frac{1-q^x}{1-q}+(\ln q)\int_0^{\infty}\left(\{t\}-\frac{1}{2}\right)\,\frac{q^{x+t}}{1-q^{x+t}}{\,}dt. \end{array} \end{aligned} $$

Limit and Series Representations

It is not difficult to see that both the Eulerian form of Σg q(x) and the analogue of Gauss’s limit reduce to the definition of the q-gamma function given in Eq. (10.8). Let us now examine the other series representations.

  • Weierstrassian form. For any x > 0, we have

    $$\displaystyle \begin{aligned} \ln\Gamma_q(x) ~=~ -\ln\frac{1-q^x}{1-q}+\psi_{q,0}(1){\,}x-\sum_{k=1}^{\infty}\left(\ln\frac{1-q^{x+k}}{1-q^k}+(\ln q)\,\frac{q^k}{1-q^k}{\,}x\right). \end{aligned}$$

    Differentiating this series term by term, we obtain

    $$\displaystyle \begin{aligned} \psi_{q,0}(x) ~=~ (\ln q)\,\frac{q^x}{1-q^x}+\psi_{q,0}(1)+(\ln q)\sum_{k=1}^{\infty}\left(\frac{1}{1-q^{x+k}}-\frac{1}{1-q^k}\right). \end{aligned}$$
  • Gregory’s formula-based series representation. For any x > 0 we have the series representation

    $$\displaystyle \begin{aligned} \begin{array}{rcl} \ln\Gamma_q(x) & =&\displaystyle \left(\frac{1}{2}-x\right)\ln(1-q)-\frac{1}{\ln q}\,\mathrm{Li}_2(q^x)+\ln(q;q)_{\infty}\\ & &\displaystyle -\sum_{n=0}^{\infty}|G_{n+1}|\,\sum_{k=0}^n(-1)^k{\textstyle{{{n}\choose{k}}}}g_q(x+k). \end{array} \end{aligned} $$

    Setting x = 1 in this identity yields the following analogue of Fontana-Mascheroni series

    $$\displaystyle \begin{aligned} \sum_{n=0}^{\infty}|G_{n+1}|\,\sum_{k=0}^n(-1)^k{\textstyle{{{n}\choose{k}}}}g_q(k+1) ~=~ -\frac{1}{2}\ln(1-q)-\frac{1}{\ln q}\,\mathrm{Li}_2(q)+\ln(q;q)_{\infty}. \end{aligned}$$

Analogue of Gauss’ Multiplication Formula

After first noting that

$$\displaystyle \begin{aligned} g_q\left(\frac{x}{m}\right) ~=~ g_{q^{\frac{1}{m}}}(x)+g_q\left(\frac{1}{m}\right),\qquad x>0, \end{aligned}$$

we immediately obtain the following identity

$$\displaystyle \begin{aligned} \sum_{j=0}^{m-1}\ln\Gamma_q\left(x+\frac{j}{m}\right) ~=~ \sum_{j=1}^m\ln\Gamma_q\left(\frac{j}{m}\right)+\ln\Gamma_{q^{\frac{1}{m}}}(mx)+(mx-1){\,}g_q\left(\frac{1}{m}\right). \end{aligned}$$

Now, using Proposition 8.28, we also obtain

$$\displaystyle \begin{aligned} \sum_{j=1}^m\ln\Gamma_q\left(\frac{j}{m}\right) ~=~ \frac{m-1}{2}\ln(1-q)+m\ln(q;q)_{\infty}-\ln\left(q^{\frac{1}{m}};q^{\frac{1}{m}}\right)_{\infty}. \end{aligned}$$

Thus, we get the following multiplication formula

$$\displaystyle \begin{aligned} \prod_{j=0}^{m-1}\Gamma_q\left(x+\frac{j}{m}\right) ~=~ (1-q)^{\frac{m-1}{2}}\frac{(q;q)^m_{\infty}}{\left(q^{\frac{1}{m}};q^{\frac{1}{m}}\right)_{\infty}} ~\Gamma_{q^{\frac{1}{m}}}(mx)\left(\frac{1-q^{\frac{1}{m}}}{1-q}\right)^{mx-1}{\,}, \end{aligned}$$

or equivalently, replacing q with q m,

$$\displaystyle \begin{aligned} \prod_{j=0}^{m-1}\Gamma_{q^m}\left(x+\frac{j}{m}\right) ~=~ (1-q^m)^{\frac{m-1}{2}}\frac{\left(q^m;q^m\right)^m_{\infty}}{(q;q)_{\infty}} ~\Gamma_q(mx)\left(\frac{1-q}{1-q^m}\right)^{mx-1}{\,}. \end{aligned}$$

(See also, e.g., Srivastava and Choi [93, p. 494] and Webster [98, p. 617].) For instance, when m = 2, we obtain the following analogue of Legendre’s duplication formula

$$\displaystyle \begin{aligned} \Gamma_{q^2}(x)\,\Gamma_{q^2}\left(x+\frac{1}{2}\right) ~=~ (1-q^2)^{\frac{1}{2}}\frac{\left(q^2;q^2\right)^2_{\infty}}{(q;q)_{\infty}}~\frac{\Gamma_q(2x)}{(1+q)^{2x-1}}{\,}. \end{aligned}$$

Analogue of Wallis’s Product Formula

Using Proposition 8.49 with

$$\displaystyle \begin{aligned} \tilde{g}_q(x) ~=~ 2g_q(2x) ~=~ 2(g_{q^2}(x)+g_q(2)), \end{aligned}$$

we obtain

$$\displaystyle \begin{aligned} \begin{array}{rcl} h(n) & =&\displaystyle \Sigma\tilde{g}_q(n+1)-\Sigma g_q(2n+1)\\ & =&\displaystyle 2\ln\Gamma_{q^2}(n+1)+2g_2(2)n-\ln\Gamma_q(2n+1). \end{array} \end{aligned} $$

Using the generalized Stirling formula, we then have

$$\displaystyle \begin{aligned} \lim_{n\to\infty}h(n) ~=~ 2\ln(q^2;q^2)_{\infty}-\ln(q;q)_{\infty}. \end{aligned}$$

Finally, we obtain the following analogue of Wallis’s formula

$$\displaystyle \begin{aligned} \lim_{n\to\infty}\sum_{k=1}^{2n}(-1)^{k-1}\ln\frac{1-q^k}{1-q} ~=~ \ln\frac{(q;q)_{\infty}}{(q^2;q^2)^2_{\infty}}{\,}. \end{aligned}$$

Generalized Webster’s Functional Equation

For any \(m\in \mathbb {N}^*\) and any a > 0, there is a unique solution \(f\colon \mathbb {R}_+\to \mathbb {R}_+\) to the equation

$$\displaystyle \begin{aligned} \prod_{j=0}^{m-1}f(x+a j) ~=~ \frac{1-q^x}{1-q} \end{aligned}$$

such that \(\ln f\) lies in \(\mathcal {K}^0\) (or in \(\mathcal {K}^1\)), namely

$$\displaystyle \begin{aligned} f(x) ~=~ \frac{\Gamma_{q^{am}}(\frac{x+a}{am})}{\Gamma_{q^{am}}(\frac{x}{am})}\left(\frac{1-q^{am}}{1-q}\right)^{\frac{1}{m}}. \end{aligned}$$

10.5 The Barnes G-Function

The Barnes function \(G\colon \mathbb {R}_+\to \mathbb {R}_+\) is the function G = 1∕ Γ2 as defined in Sect. 5.2. Hence, it can be defined by the equations

$$\displaystyle \begin{aligned} \ln G(x) ~=~ \Sigma\ln\Gamma(x) ~=~ \Sigma\psi_{-1}(x)\qquad \mbox{for }x>0. \end{aligned}$$

ID Card

We have the following basic information about the Barnes G-function:

g(x)

Membership

deg g

Σg(x)

\(\mathstrut ^{\mathstrut }_{\mathstrut } \ln \Gamma (x)\)

\(\mathcal {C}^{\infty }\cap \mathcal {D}^2\cap \mathcal {K}^{\infty }\)

1

\(\ln G(x)\)

Analogue of Bohr-Mollerup’s Theorem

The function G can be characterized in the multiplicative notation as follows.

All solutions \(f\colon \mathbb {R}_+\to \mathbb {R}_+\) to the equation f(x + 1) =  Γ(x)f(x) for which \(\ln f\) lies in \(\mathcal {K}^2\) are of the form f(x) = c G(x), where c > 0.

Interestingly, this characterization enables one to establish the following identity

$$\displaystyle \begin{aligned} \ln G(x) ~=~ -{\textstyle{{{x}\choose{2}}}}+(x-1)\ln\Gamma(x)+\textstyle{\frac{1}{2}}\ln(2\pi){\,}x-\psi_{-2}(x). \end{aligned} $$
(10.12)

Indeed, both sides vanish at x = 1 and are eventually 2-convex solutions to the equation

$$\displaystyle \begin{aligned} f(x+1)-f(x) ~=~ \ln\Gamma(x). \end{aligned}$$

Hence, they must coincide on \(\mathbb {R}_+\).

Using Proposition 3.9, we can also derive the following alternative characterization of the Barnes G-function.

All solutions \(f\colon \mathbb {R}_+\to \mathbb {R}_+\) to the equation f(x + 1) =  Γ(x)f(x) that satisfy the asymptotic condition that, for each x > 0,

$$\displaystyle \begin{aligned} f(x+n) ~\sim ~ \Gamma(n)^x{\,}n^{{x\choose 2}} f(n)\qquad \mbox{as }n\to_{\mathbb{N}}\infty \end{aligned}$$

are of the form f(x) = c G(x), where c > 0.

Extended ID Card

The value of the asymptotic constant σ[g] can be derived for instance from identity (10.12). One can show that (see, e.g., [93, p. 53])

$$\displaystyle \begin{aligned} \sigma[g] ~=~ \int_0^1\ln G(t+1){\,}dt ~=~ \frac{1}{12}+\frac{1}{4}\ln(2\pi)-2\ln A ~\approx ~ 0.045. \end{aligned}$$

We then have the following values:

  • Inequality

    $$\displaystyle \begin{aligned} |\sigma[g]| ~\leq ~ \frac{7}{3}\ln 2-\frac{109}{72} ~\approx ~ 0.10{\,}. \end{aligned}$$
  • Alternative representations of σ[g] = γ[g]

    $$\displaystyle \begin{aligned} \begin{array}{rcl} \sigma[g] & =&\displaystyle \frac{1}{2}\ln(2\pi)+\lim_{n\to\infty}\left(\sum_{k=1}^n\ln\Gamma(k) -\psi_{-2}(n)-\frac{1}{2}\ln\Gamma(n)-\frac{1}{12}\ln n\right), \\ \sigma[g] & =&\displaystyle \frac{1}{2}\ln(2\pi)+\lim_{n\to\infty}\left(\sum_{k=1}^n\ln\Gamma(k) -\psi_{-2}(n)-\frac{1}{2}\ln\Gamma(n)-\frac{1}{12}\,\psi(n)\right), \\ \sigma[g] & =&\displaystyle \int_1^{\infty}\left(\ln\frac{\Gamma(\lfloor t\rfloor)}{\Gamma(t)}+\{t\}\ln\lfloor t\rfloor+{\{t\}\choose 2}\ln\left(1+\frac{1}{\lfloor t\rfloor}\right)\right)dt{\,},\\ \sigma[g] & =&\displaystyle \int_1^{\infty}\left(\ln\frac{\Gamma(\lfloor t\rfloor)}{\Gamma(t)}+\ln\frac{\lfloor t\rfloor^{7/12}}{\lfloor t+1\rfloor^{1/12}}\right)dt{\,},\\ \sigma[g] & =&\displaystyle \frac{1}{12}\,\gamma -\frac{1}{2}\,\int_1^{\infty}B_2(\{t\})\,\psi_1(t){\,}dt{\,},\\ \sigma[g] & =&\displaystyle \ln\left(\prod_{k=1}^{\infty}\frac{\Gamma(k){\,}e^k{\,}\sqrt{k}}{\left(1+\frac{1}{k}\right)^{\frac{1}{12}}k^k{\,}\sqrt{2\pi}}\right). \end{array} \end{aligned} $$
  • Generalized Binet’s function. For any \(q\in \mathbb {N}\) and any x > 0

    $$\displaystyle \begin{aligned} J^{q+1}[\ln\circ G](x) ~=~ \ln G(x)-\psi_{-2}(x)-\overline{\sigma}[g]+\sum_{j=1}^qG_j\,\Delta^{j-1}\ln\Gamma(x). \end{aligned}$$

    For instance,

    $$\displaystyle \begin{aligned} J^3[\ln\circ G](x) ~=~ \ln G(x)-\psi_{-2}(x)-\overline{\sigma}[g]+\frac{1}{2}\ln\Gamma(x)-\frac{1}{12}\ln x. \end{aligned}$$
  • Analogue of Raabe’s formula

    $$\displaystyle \begin{aligned} \int_x^{x+1}\ln G(t){\,}dt ~=~ \overline{\sigma}[g]+\psi_{-2}(x){\,},\qquad x>0. \end{aligned} $$
    (10.13)
  • Alternative characterization. The function \(f(x)=\ln G(x)\) is the unique solution lying in \(\mathcal {C}^0\cap \mathcal {K}^2\) to the equation

    $$\displaystyle \begin{aligned} \int_x^{x+1}f(t){\,}dt ~=~ \overline{\sigma}[g]+\psi_{-2}(x){\,}, \qquad x>0. \end{aligned}$$

\(\overline {\sigma }[g]\)

σ[g]

γ[g]

\(\mathstrut ^{\mathstrut }_{\mathstrut } \frac {1}{12}-\frac {1}{4}\ln (2\pi )-2\ln A\)

\(\frac {1}{12}+\frac {1}{4}\ln (2\pi )-2\ln A\)

γ[g] = σ[g]

Project 10.5

Find a closed-form expression for the integral

$$\displaystyle \begin{aligned} \int_1^x\ln G(t){\,}dt. \end{aligned}$$

We apply Proposition 8.20. Using (10.13) and then (10.7) we obtain

$$\displaystyle \begin{aligned} \begin{array}{rcl} \int_1^x\ln G(t){\,}dt & =&\displaystyle \Sigma_x\int_x^{x+1}\ln G(t){\,}dt ~=~ \overline{\sigma}[g]{\,}(x-1)+\Sigma\psi_{-2}(x)\\ & =&\displaystyle 2\ln A+\frac{1}{4}{\,}(x^2+1)\ln(2\pi)-\frac{1}{12}{\,}(2x+1)(x-1)^2\\ & &\displaystyle +(x-1)\,\psi_{-2}(x)-2\,\psi_{-3}(x){\,}. \end{array} \end{aligned} $$

This expression could have been obtained also by integrating both sides of (10.12). \(\lozenge \)

Inequalities

The following inequalities hold for any x > 0, any a ≥ 0, and any \(n\in \mathbb {N}^*\).

  • Symmetrized generalized Wendel’s inequality (equality if a ∈{0, 1, 2})

    $$\displaystyle \begin{aligned} \left|\ln G(x+a)-\ln G(x)-a\ln\Gamma(x)-{\textstyle{{{a}\choose{2}}}}\ln x\right| ~\leq ~ \left|{\textstyle{{{a-1}\choose{2}}}}\right|\,\ln\left(1+\frac{a}{x}\right), \end{aligned}$$
    $$\displaystyle \begin{aligned} \left(1+\frac{a}{x}\right)^{-\left|{a-1\choose 2}\right|} ~\leq ~ \frac{G(x+a)}{G(x)\,\Gamma(x)^a{\,}x^{a\choose 2}} ~\leq ~ \left(1+\frac{a}{x}\right)^{\left|{a-1\choose 2}\right|}. \end{aligned}$$
  • Symmetrized generalized Wendel’s inequality (discrete version)

    $$\displaystyle \begin{aligned}\displaystyle \left|\ln G(x)-\sum_{k=1}^{n-1}\ln\Gamma(k)+\sum_{k=0}^{n-1}\ln\Gamma(x+k)-x\ln\Gamma(n)-{\textstyle{{{x}\choose{2}}}}\ln n\right|\\\displaystyle \leq ~ \left|{\textstyle{{{x-1}\choose{2}}}}\right|\,\ln\left(1+\frac{x}{n}\right), \end{aligned} $$
    $$\displaystyle \begin{aligned} \left(1+\frac{x}{n}\right)^{-\left|{x-1\choose 2}\right|} ~\leq ~ G(x){\,}\frac{\Gamma(x)\Gamma(x+1)\cdots\Gamma(x+n-1)}{\Gamma(1)\Gamma(2)\cdots\Gamma(n-1)\Gamma(n)^x n^{{x\choose 2}}}~\leq ~ \left(1+\frac{x}{n}\right)^{\left|{x-1\choose 2}\right|}. \end{aligned}$$
  • Symmetrized Stirling’s formula-based inequality

    $$\displaystyle \begin{aligned} \begin{array}{rcl} \left|J^3[\ln\circ G](x)\right| & \leq &\displaystyle \frac{1}{12}(x+1)^2(2x+5)\ln\left(1+\frac{1}{x}\right)-\frac{1}{72}(12x^2+48x+49)\\ & \leq &\displaystyle \frac{5}{12}\ln\left(1+\frac{1}{x}\right), \end{array} \end{aligned} $$
    $$\displaystyle \begin{aligned} \left(1+\frac{1}{x}\right)^{-5/12} \leq ~ \frac{G(x)\,\Gamma(x)^{1/2}}{x^{1/12}{\,}e^{\psi_{-2}(x)+\overline{\sigma}[g]}} ~\leq ~ \left(1+\frac{1}{x}\right)^{5/12}. \end{aligned}$$
  • Generalized Gautschi’s inequality

    $$\displaystyle \begin{aligned} \Gamma(x+\lceil a\rceil)^{a-\lceil a\rceil} ~\leq ~ e^{(a-\lceil a\rceil)D\ln G(x+\lceil a\rceil)} ~\leq ~\frac{G(x+a)}{G(x+\lceil a\rceil)} ~\leq ~ \Gamma(x+\lceil a\rceil)^{a-\lfloor a\rfloor}. \end{aligned}$$

    (These inequalities are valid only if x + ⌊a⌋≥ x 0, where x 0 = 1.92… is the unique positive zero of the function \(D^2\ln G(x)\).)

Remark 10.6

It is not difficult to see that the first inequality in Proposition 6.19 does not hold for large values of x when \(g(x)=\ln \Gamma (x)\). This shows that the analogue of Burnside’s formula does not hold in general when deg g ≥ 1. \(\lozenge \)

Generalized Stirling’s and Related Formulas

For any a ≥ 0, we have the following limits and asymptotic equivalences as x →,

$$\displaystyle \begin{aligned} \ln G(x+a)-\ln G(x)-a\ln\Gamma(x)-{\textstyle{{{a}\choose{2}}}}\ln x ~\to ~0, \end{aligned}$$
$$\displaystyle \begin{aligned} \ln G(x)-\psi_{-2}(x)+\frac{1}{2}\ln\Gamma(x)-\frac{1}{12}\ln x ~\to ~ \overline{\sigma}[g], \end{aligned}$$
$$\displaystyle \begin{aligned} \ln G(x)-\psi_{-2}(x)+\frac{1}{2}\ln\Gamma(x)-\frac{1}{12}\,\psi(x) ~\to ~ \overline{\sigma}[g], \end{aligned}$$
$$\displaystyle \begin{aligned} G(x+a) ~\sim ~ G(x)\,\Gamma(x)^a{\,}x^{{a\choose 2}},\qquad \ln G(x+a) ~\sim ~ \psi_{-2}(x), \end{aligned}$$
$$\displaystyle \begin{aligned} G(x) ~\sim ~ \exp(\psi_{-2}(x)+\overline{\sigma}[g]){\,}\Gamma(x)^{-\frac{1}{2}}{\,}x^{\frac{1}{12}}. \end{aligned}$$

Further results (obtained by differentiation)

$$\displaystyle \begin{aligned} x\,\psi(x+a)-x\,\psi(x) ~\to ~a,\quad x{\,}\psi_1(x) ~\to ~1,\quad x\,\psi(x+a) ~\sim ~ \ln\Gamma(x), \end{aligned}$$
$$\displaystyle \begin{aligned} \ln\Gamma(x)-\left(x-\frac{1}{2}\right)\psi(x)+x ~\to ~ \frac{1}{2}{\,}(1+\ln(2\pi)). \end{aligned}$$

Remark 10.7

Using one of the asymptotic equivalences above, we get

$$\displaystyle \begin{aligned} G(x+1) ~\sim ~ \exp(\psi_{-2}(x)+\overline{\sigma}[g]){\,}\Gamma(x)^{\frac{1}{2}}{\,}x^{\frac{1}{12}}\qquad \mbox{as }x\to\infty. \end{aligned}$$

Combining this latter equivalence with identity (10.12) and the Stirling formula for the gamma function, we also obtain the following simpler form

$$\displaystyle \begin{aligned} G(x+1) ~\sim ~ A^{-1}{\,}x^{\frac{1}{2}x^2-\frac{1}{12}}(2\pi)^{\frac{x}{2}}{\,}e^{-\frac{3}{4}x^2+\frac{1}{12}}\qquad \mbox{as }x\to\infty. \end{aligned}$$

\(\lozenge \)

Asymptotic Expansions

For any \(m,q\in \mathbb {N}^*\) we have the following expansion as x →

$$\displaystyle \begin{aligned} \frac{1}{m}\sum_{j=0}^{m-1}\ln G\left(x+\frac{j}{m}\right) ~=~ \overline{\sigma}[g]+ \sum_{k=0}^q\frac{B_k}{m^k{\,}k!}{\,}\psi_{k-2}(x)+O(\psi_{q-1}(x)){\,}. \end{aligned} $$
(10.14)

Setting m = 1 in this formula, we obtain

$$\displaystyle \begin{aligned} \ln G(x) ~=~ \overline{\sigma}[g]+ \sum_{k=0}^q\frac{B_k}{k!}{\,}\psi_{k-2}(x)+O(\psi_{q-1}(x)){\,}, \end{aligned}$$

or equivalently, if q ≥ 2,

$$\displaystyle \begin{aligned} J^3[\ln\circ G](x) ~=~ \frac{1}{12}\left(\psi(x)-\ln x\right)+ \sum_{k=3}^q\frac{B_k}{k!}{\,}\psi_{k-2}(x)+O(\psi_{q-1}(x)){\,}. \end{aligned}$$

Setting q = 4 for instance, we obtain the following expansion

$$\displaystyle \begin{aligned} \ln G(x) ~=~ \overline{\sigma}[g]+\psi_{-2}(x)-\frac{1}{2}\,\psi_{-1}(x)+\frac{1}{12}\,\psi(x) -\frac{1}{720}\,\psi_2(x)+O\left(x^{-4}\right). \end{aligned}$$

Generalized Liu’s Formula

For any x > 0 we have

$$\displaystyle \begin{aligned} \ln G(x) ~=~ \overline{\sigma}[g]+\psi_{-2}(x)-\frac{1}{2}\,\psi_{-1}(x) +\frac{1}{12}\,\psi(x)+\frac{1}{2}\,\int_0^{\infty}B_2(\{t\})\,\psi_1(x+t){\,}dt \end{aligned}$$

or equivalently,

$$\displaystyle \begin{aligned} J^3[\ln\circ G](x) ~=~ \frac{1}{12}\left(\psi(x)-\ln x\right)+\frac{1}{2}\,\int_0^{\infty}B_2(\{t\})\,\psi_1(x+t){\,}dt. \end{aligned}$$

Limit, Series, and Integral Representations

Let us now determine the main representations of the function \(\ln G(x)\).

  • Eulerian form and related identities. We have

    $$\displaystyle \begin{aligned} \ln G(x) ~=~ -\ln\Gamma(x)-\sum_{k=1}^{\infty}\left(\ln\Gamma(x+k)-\ln\Gamma(k)-x\ln k-{\textstyle{{{x}\choose{2}}}}\ln\left(1+\frac{1}{k}\right)\right), \end{aligned}$$
    $$\displaystyle \begin{aligned} G(x) ~=~ \frac{1}{\Gamma(x)}{\,} \prod_{k=1}^{\infty}\frac{\Gamma(k)}{\Gamma(x+k)}{\,}k^x(1+1/k)^{{x\choose 2}}. \end{aligned}$$

    Upon differentiation, we obtain

    $$\displaystyle \begin{aligned} x\,\psi(x) ~=~ x-\frac{1}{2}{\,}(1+\ln(2\pi))-\sum_{k=1}^{\infty}\left(\psi(x+k)-\ln k-\left(x-\frac{1}{2}\right)\ln\left(1+\frac{1}{k}\right)\right), \end{aligned}$$
    $$\displaystyle \begin{aligned} \psi(x)+x\,\psi_1(x) ~=~ 1-\sum_{k=1}^{\infty}\left(\psi_1(x+k)-\ln\left(1+\frac{1}{k}\right)\right), \end{aligned}$$
    $$\displaystyle \begin{aligned} (r+1)\,\psi_r(x)+x\,\psi_{r+1}(x) ~=~ -\sum_{k=1}^{\infty}\psi_{r+1}(x+k),\qquad r\in\mathbb{N}^*. \end{aligned}$$
  • Weierstrassian form and related identities. We have

    $$\displaystyle \begin{aligned}\displaystyle \ln G(x) ~=~ (-1-\gamma){\textstyle{{{x}\choose{2}}}}-\ln\Gamma(x)\\\displaystyle - \sum_{k=1}^{\infty}\left(\ln\Gamma(x+k)-\ln\Gamma(k)-x\ln k-{\textstyle{{{x}\choose{2}}}}\,\psi_1(k)\right), \end{aligned} $$
    $$\displaystyle \begin{aligned} G(x) ~=~ \frac{e^{(-\gamma -1){x\choose 2}}}{\Gamma(x)}{\,} \prod_{k=1}^{\infty}\frac{\Gamma(k)}{\Gamma(x+k)}{\,}k^xe^{\psi_{1}(k){\,}{x\choose 2}}{\,}, \end{aligned}$$

    Upon differentiation, we obtain

    $$\displaystyle \begin{aligned} x\,\psi(x) +\left(x-\frac{1}{2}\right)\gamma +\frac{1}{2}\ln(2\pi) ~=~ -\sum_{k=1}^{\infty}\left(\psi(x+k)-\left(x-\frac{1}{2}\right)\psi_1(k)-\ln k\right), \end{aligned}$$
    $$\displaystyle \begin{aligned} \psi(x)+x\,\psi_1(x)+\gamma ~=~ -\sum_{k=1}^{\infty}\left(\psi_1(x+k)-\psi_1(k)\right). \end{aligned}$$
  • Analogue of Gauss’ limit and related identities. The analogue of Gauss’ limit is

    $$\displaystyle \begin{aligned} \ln G(x) ~=~ \lim_{n\to\infty}\left(\sum_{k=1}^{n-1}\ln\Gamma(k)-\sum_{k=0}^{n-1}\ln\Gamma(x+k)+x\ln\Gamma(n)+{\textstyle{{{x}\choose{2}}}}\ln n\right), \end{aligned}$$
    $$\displaystyle \begin{aligned} G(x) ~=~ \lim_{n\to\infty}\frac{\Gamma(1)\Gamma(2){\,}\cdots{\,}\Gamma(n)}{\Gamma(x)\Gamma(x+1){\,}\cdots{\,}\Gamma(x+n)}{\,} n!^x{\,}n^{{x\choose 2}}{\,}. \end{aligned}$$

    Upon differentiation, we obtain

    $$\displaystyle \begin{aligned}\displaystyle (x-1)\,\psi(x)-x+\frac{1}{2}(1+\ln(2\pi))\\\displaystyle =~ \lim_{n\to\infty}\left(-\sum_{k=0}^{n-1}\psi(x+k)+\ln\Gamma(n)+\left(x-\frac{1}{2}\right)\ln n\right), \end{aligned} $$
    $$\displaystyle \begin{aligned} (x-1)\,\psi_1(x)+\psi(x)-1 ~=~ \lim_{n\to\infty}\left(\ln n-\sum_{k=0}^{n-1}\psi_1(x+t)\right). \end{aligned}$$
  • Integral representations. Using the elevator method on one and two levels, we obtain the following representations

    $$\displaystyle \begin{aligned} \ln G(x) ~=~ -\frac{1}{2}{\,}(x-1)(x-\ln(2\pi))+\int_1^x(t-1)\,\psi(t){\,}dt \end{aligned}$$

    and

    $$\displaystyle \begin{aligned} \ln G(x) ~=~ -\frac{1}{2}{\,}(x-1)(x-\ln(2\pi))+\int_1^x(x-t)(\psi(t)+(t-1)\,\psi_1(t)){\,}dt. \end{aligned}$$

    Each of these representations actually leads to identity (10.12).

  • Gregory’s formula-based series representation. For any x > 0 we have the series representation

    $$\displaystyle \begin{aligned} \begin{array}{rcl} \ln G(x) & =&\displaystyle \psi_{-2}(x)+\overline{\sigma}[g]-\frac{1}{2}\ln\Gamma(x)-\sum_{n=0}^{\infty}G_{n+2}\Delta^{n+1}g(x)\\ \\ & =&\displaystyle \psi_{-2}(x)+\overline{\sigma}[g]-\frac{1}{2}\ln\Gamma(x) -\sum_{n=0}^{\infty}|G_{n+2}|\,\sum_{k=0}^n(-1)^k{\textstyle{{{n}\choose{k}}}}{\,}\ln(x+k). \end{array} \end{aligned} $$

    Setting x = 1 in this identity yields the analogue of Fontana-Mascheroni series

    $$\displaystyle \begin{aligned} \overline{\sigma}[g] ~=~ -\frac{1}{2}\ln(2\pi)+\sum_{n=0}^{\infty}|G_{n+2}|\,\sum_{k=0}^n(-1)^k{\textstyle{{{n}\choose{k}}}}{\,}\ln(k+1). \end{aligned}$$

Note that the Eulerian and Weierstrassian forms above can also be integrated term by term on any bounded interval of [0, ). For instance, integrating on (1, x) provides series representations for the integral of \(\ln G(x)\) as defined in Project 10.5.

Analogue of Gauss’ Multiplication Formula

For any \(m\in \mathbb {N}^*\) and any x > 0, we have

$$\displaystyle \begin{aligned} \sum_{j=0}^{m-1}\ln G\left(\frac{x+j}{m}\right) ~=~ \sum_{j=1}^m\ln G\left(\frac{j}{m}\right)+\Sigma_x\ln\Gamma\left(\frac{x}{m}\right). \end{aligned}$$

For instance, setting m = 2 in this identity, we obtain

$$\displaystyle \begin{aligned} \ln G\left(\frac{x+1}{2}\right)+\ln G\left(\frac{x}{2}\right) ~=~ \ln G\left(\frac{1}{2}\right) + \Sigma_x\ln\Gamma\left(\frac{x}{2}\right). \end{aligned}$$

However, to make this multiplication formula interesting and usable, we need to find a simple expression for its right side. In particular, we need a closed-form expression for the function \(\Sigma _x\ln \Gamma (\frac {x}{m})\). Such a result would be most welcome.

We can nevertheless investigate the asymptotic behavior of the function

$$\displaystyle \begin{aligned} x ~\mapsto ~ \sum_{j=0}^{m-1}\ln G\left(\frac{x+j}{m}\right). \end{aligned}$$

In addition to the asymptotic expansion given in (10.14), Proposition 8.30 yields the following convergence result. We have

$$\displaystyle \begin{aligned}\displaystyle \sum_{j=0}^{m-1}\ln G\left(\frac{x+j}{m}\right)-m\,\psi_{-2}\left(\frac{x}{m}\right)+\frac{1}{2}\,\ln\Gamma\left(\frac{x}{m}\right)\\\displaystyle -\frac{1}{12}\left(\ln\Gamma\left(\frac{x+1}{m}\right)-\ln\Gamma\left(\frac{x}{m}\right)\right) ~\to ~ m\,\overline{\sigma}[g]\qquad \mbox{as }x\to\infty. \end{aligned} $$

Analogue of Wallis’s Product Formula

Using Legendre’s duplication formula for the gamma function, we obtain

$$\displaystyle \begin{aligned} \begin{array}{rcl} \Sigma_x\ln\Gamma(2x) & =&\displaystyle \textstyle{\ln G(x)+\ln G(x+\frac{1}{2})-\ln G(\frac{1}{2})}\\ & &\displaystyle +\textstyle{(x^2+1)\ln 2-\frac{x}{2}\ln(16\pi)}. \end{array} \end{aligned} $$

Using this identity with Proposition 8.49, we can derive the surprising analogue of Wallis’s formula

$$\displaystyle \begin{aligned} \lim_{n\to\infty}\frac{\Gamma(1)\Gamma(3){\,}\cdots{\,}\Gamma(2n-1)}{\Gamma(2)\Gamma(4){\,}\cdots{\,}\Gamma(2n)}\left(\frac{2n}{e}\right)^n ~=~ \frac{1}{\sqrt{2}}{\,}. \end{aligned}$$

Note that a shorter proof of this formula can be obtained using the second sequence described in Remark 8.53.

Project 10.8

Find the analogue of Wallis’s formula for the function \(g(x)=\ln G(x)\). After some algebra, we obtain

$$\displaystyle \begin{aligned} \lim_{n\to\infty}\frac{G(1)G(3){\,}\cdots{\,}G(2n-1)}{G(2)G(4){\,}\cdots{\,}G(2n)}\,\frac{n^{n^2-\frac{1}{2}n -\frac{1}{24}}{\,}2^{n^2-\frac{7}{24}}\,\pi^{\frac{1}{2}n}}{e^{\frac{3}{2}n^2-\frac{1}{2}n-\frac{1}{24}}} ~=~ A^{\frac{1}{2}}{\,}. \end{aligned}$$

This latter formula is a little harder to obtain than the former one. Using Proposition 8.49 requires the computation of both functions \(\Sigma \ln G(x)\) and \(2\,\Sigma _x\ln G(2x)\) using the elevator method (Corollary 7.20) with r = 1. That is,

$$\displaystyle \begin{aligned} \begin{array}{rcl} \Sigma\ln G(x) & =&\displaystyle -\frac{1}{8}{\,}x(x-1)(2x-5)+\frac{1}{4}{\,}x(x-3)\ln(2\pi)-x\ln A\\ & &\displaystyle + \frac{1}{2}{\,}(x-1)(x-2)\ln\Gamma(x)-\frac{1}{2}{\,}(2x-3)\,\psi_{-2}(x)+\psi_{-3}(x) \end{array} \end{aligned} $$

and

$$\displaystyle \begin{aligned} \begin{array}{rcl} 2\,\Sigma_x\ln G(2x) & =&\displaystyle -\frac{1}{4}{\,}x(2x-1)(4x-7)-2x\ln A\\ & &\displaystyle +\frac{1}{2}{\,}(2x^2-3x-1)\ln 2 +x(x-2)\ln\pi\\ & &\displaystyle +\frac{1}{2}\ln\Gamma(x)+\frac{1}{2}(2x-1)(2x-3)\ln\Gamma(2x)\\ & &\displaystyle -2(x-1)\,\psi_{-2}(2x)+\psi_{-3}(2x). \end{array} \end{aligned} $$

Here again, a shorter proof of the limit above can be obtained using the second sequence described in Remark 8.53. \(\lozenge \)

Restriction to the Natural Integers

For any \(n\in \mathbb {N}^*\) we have

$$\displaystyle \begin{aligned} G(n) ~=~ \prod_{k=0}^{n-2}k!{\,}. \end{aligned}$$

Generalized Webster’s Functional Equation

For any \(m\in \mathbb {N}^*\), there is a unique solution \(f\colon \mathbb {R}_+\to \mathbb {R}_+\) to the equation

$$\displaystyle \begin{aligned} \prod_{j=0}^{m-1}f\left(x+\frac{j}{m}\right) ~=~ \Gamma(x) \end{aligned}$$

such that \(\ln f\) lies in \(\mathcal {K}^1\), namely

$$\displaystyle \begin{aligned} f(x) ~=~ \frac{G(x+\frac{1}{m})}{G(x)}{\,}. \end{aligned}$$

Analogue of Euler’s Series Representation of γ

The Taylor series expansion of \(\ln G(x+1)\) about x = 0 is (see, e.g., [93, p. 311])

$$\displaystyle \begin{aligned} \ln G(x+1) ~=~\frac{1}{2}\left(\ln(2\pi)-1\right)x-\frac{\gamma +1}{2}{\,}x^2 -\sum_{k=2}^{\infty}\,\frac{\zeta(k)}{k+1}{\,}(-x)^{k+1}{\,},\qquad |x|<1. \end{aligned}$$

Integrating both sides of this equation on (0, 1), we obtain the identity

$$\displaystyle \begin{aligned} \sum_{k=2}^{\infty}(-1)^k\,\frac{\zeta(k)}{(k+1)(k+2)} ~=~ \frac{1}{2}+\frac{1}{6}{\,}\gamma -2\ln A. \end{aligned}$$

Also, the exponential generating function for the sequence nσ[g (n)] is

$$\displaystyle \begin{aligned} \mathrm{egf}_{\sigma}[g](x) ~=~ \ln G(x+1)-\psi_{-2}(x+1)+\frac{1}{4}\ln(2\pi)-\frac{1}{12}+2\ln A \end{aligned}$$

Integrating both sides of this equation on (0, 1) (i.e., we use (7.5)), after some algebra we obtain

$$\displaystyle \begin{aligned} \sum_{k=2}^{\infty}(-1)^k\,\frac{k-1}{k(k+1)(k+2)}\,\zeta(k) ~=~ \frac{5}{4}-3\ln A-\frac{1}{4}\ln(2\pi){\,}. \end{aligned}$$

Analogue of the Reflection Formula

A reflection formula for the Barnes G-function is given in (8.27); see, e.g., [93, p. 45].

10.6 The Hurwitz Zeta Function

For any x > 0, the Hurwitz zeta function sζ(s, x) is defined as an analytic continuation to \(\mathbb {C}\setminus \{1\}\) of the series (see, e.g., [93, p. 155])

$$\displaystyle \begin{aligned} \sum_{k=0}^{\infty}(x+k)^{-s} ~=~ \frac{1}{\Gamma(s)}\,\int_0^{\infty}\frac{t^{s-1}e^{-xt}}{1-e^{-t}}{\,}dt,\qquad \Re(s)>1. \end{aligned}$$

It is known (see, e.g., [93, p. 159–160]) that this function satisfies the identity

$$\displaystyle \begin{aligned} D^k_x\zeta(s,x) ~=~ (-s)^{\underline{k}}\,\zeta(s+k,x){\,},\qquad k\in\mathbb{N}, \end{aligned}$$

and the difference equation

$$\displaystyle \begin{aligned} \zeta(s,x+1)-\zeta(s,x) ~=~ -x^{-s}. \end{aligned} $$
(10.15)

For any fixed \(s\in \mathbb {R}\setminus \{1\}\), define the function \(g_s\colon \mathbb {R}_+\to \mathbb {R}\) by the equation

$$\displaystyle \begin{aligned} g_s(x) ~=~ -x^{-s}\qquad \mbox{for }x>0. \end{aligned}$$

We then have \(g_s\in \mathcal {C}^{\infty }\cap \mathcal {K}^{\infty }\). If s > 0 and s ≠ 1, then \(g_s\in \mathcal {D}^0_{\mathbb {N}}\). If s > 1, then \(g_s\in \mathcal {D}^{-1}_{\mathbb {N}}\). If − p < s < 1 for some \(p\in \mathbb {N}\), then \(g_s\in \mathcal {D}^p_{\mathbb {N}}\), and hence we can consider

$$\displaystyle \begin{aligned} p ~=~ 1+\deg g_s ~=~ \lfloor 1-s\rfloor. \end{aligned}$$

In all cases, we have

$$\displaystyle \begin{aligned} \Sigma g_s(x) ~=~ \zeta(s,x)-\zeta(s), \end{aligned}$$

where sζ(s) = ζ(s, 1) is the Riemann zeta function.

ID Card

The basic information about the Hurwitz zeta function is summarized in the following table.

g s(x)

Membership

deg g s

Σg s(x)

\(\mathstrut ^{\mathstrut }_{\mathstrut } -x^{-s}\)

\(\begin {array}{rl}\mathstrut ^{\mathstrut }_{\mathstrut }\mathcal {C}^{\infty }\cap \mathcal {D}^{-1}\cap \mathcal {K}^{\infty },&\mbox{if }s>1,\\ \mathcal {C}^{\infty }\cap \mathcal {D}^{\lfloor 1-s\rfloor }\cap \mathcal {K}^{\infty },&\mbox{if }s<1.\end {array}\)

− 1 + ⌊1 − s+

ζ(s, x) − ζ(s)

Project 10.9

Find a closed-form expression for Σg, where

$$\displaystyle \begin{aligned} g(x) ~=~ \frac{x^2}{\sqrt{x+1}}{\,}. \end{aligned}$$

Expanding x 2 = (x + 1 − 1)2, we obtain

$$\displaystyle \begin{aligned} g(x) ~=~ (x+1)^{\frac{3}{2}}-2(x+1)^{\frac{1}{2}}+(x+1)^{-\frac{1}{2}} \end{aligned}$$

and hence

$$\displaystyle \begin{aligned} \Sigma g(x) ~=~ \textstyle{c-\zeta(-\frac{3}{2},x+1)+2\zeta(-\frac{1}{2},x+1)-\zeta(\frac{1}{2},x+1)} \end{aligned}$$

for some \(c\in \mathbb {R}\). \(\lozenge \)

Analogue of Bohr-Mollerup’s Theorem

The function ζ(s, x) can be characterized as follows.

All solutions \(f_s\colon \mathbb {R}_+\to \mathbb {R}\) to the equation

$$\displaystyle \begin{aligned} f_s(x+1)-f_s(x) ~=~ -x^{-s} \end{aligned}$$

that lie in \(\mathcal {K}^{\lfloor 1-s\rfloor _+}\) are of the form f s(x) = c s + ζ(s, x), where \(c_s\in \mathbb {R}\).

Extended ID Card

The asymptotic constant σ[g s] satisfies the following identity

$$\displaystyle \begin{aligned} \sigma[g_s] ~=~ \int_0^1\zeta(s,t+1){\,}dt-\zeta(s) ~=~ \frac{1}{s-1}-\zeta(s). \end{aligned}$$

Hence we have the following values

\(\overline {\sigma }[g_s]\)

σ[g s]

γ[g s]

\(\mathstrut ^{\mathstrut }_{\mathstrut } \begin {array}{rl}\infty ,&\mbox{if }s>1,\\ -\zeta (s),&\mbox{if }s<1.\end {array}\)

\(\frac {1}{s-1}-\zeta (s)\)

\(\sigma [g_s]-\sum _{j=1}^{\lfloor 1-s\rfloor _+}G_j\,\Delta ^{j-1}g_s(1)\)

We also have the following identities.

  • Alternative representations of σ[g s]

    $$\displaystyle \begin{aligned} \begin{array}{rcl} \sigma[g_s] & =&\displaystyle \lim_{n\to\infty}\left(\frac{1-n^{1-s}}{s-1}-\sum_{k=1}^{n-1}k^{-s}+\sum_{j=1}^{\lfloor 1-s\rfloor_+}G_j\,\Delta^{j-1}g_s(n)\right), \\ \sigma[g_s] & =&\displaystyle \lim_{n\to\infty}\left(\frac{1}{s-1}-\sum_{k=1}^{n-1}k^{-s}+\frac{1}{1-s}\,\sum_{j=0}^{\lfloor 1-s\rfloor_+}{\textstyle{{{1-s}\choose{j}}}}{\,}\frac{B_j}{n^{s+j-1}}\right), \\ \sigma[g_s] & =&\displaystyle \sum_{j=1}^{\lfloor 1-s\rfloor_+}G_j\,\Delta^{j-1}g_s(1)\\ & &\displaystyle +\sum_{k=1}^{\infty}\left(\frac{k^{1-s}-(k+1)^{1-s}}{s-1}+\sum_{j=0}^{\lfloor 1-s\rfloor_+}G_j\,\Delta^jg_s(k)\right). \end{array} \end{aligned} $$

    If s > −1, then

    $$\displaystyle \begin{aligned} \sigma[g_s] ~=~ -\frac{1}{2}+s\int_1^{\infty}\frac{\{t\}-\frac{1}{2}}{t^{s+1}}{\,}dt. \end{aligned}$$

    If s ≤−1, then for any integer q ≥⌈(1 − s)∕2⌉,

    $$\displaystyle \begin{aligned} \sigma[g_s] ~=~ -\frac{1}{2}+\sum_{k=1}^q\frac{B_{2k}}{(2k)!}{\,}(-s)^{\underline{2k-1}} +\frac{(-s)^{\underline{2q}}}{(2q)!}\,\int_1^{\infty}\frac{B_{2q}(\{t\})}{t^{s+2q}}{\,}dt. \end{aligned}$$
  • Representations of γ[g s]

    $$\displaystyle \begin{aligned} \begin{array}{rcl} \gamma[g_s] & =&\displaystyle \sigma[g_s]-\sum_{j=1}^{\lfloor 1-s\rfloor_+}G_j\,\Delta^{j-1}g_s(1){\,},\\ \gamma[g_s] & =&\displaystyle \int_1^{\infty}\bigg(\sum_{j=0}^{\lfloor 1-s\rfloor_+}G_j\,\Delta^jg_s(\lfloor t\rfloor)-g_s(t)\bigg){\,}dt{\,},\\ \gamma[g_s] & =&\displaystyle \int_1^{\infty}\bigg(\sum_{j=0}^{\lfloor 1-s\rfloor_+}{\{t\}\choose j}\,\Delta^jg_s(\lfloor t\rfloor)-g_s(t)\bigg){\,}dt{\,}. \end{array} \end{aligned} $$
  • Generalized Binet’s function. For any \(q\in \mathbb {N}\) and any x > 0

    $$\displaystyle \begin{aligned} J^{q+1}[\Sigma g_s](x) ~=~ \zeta(s,x)-\frac{x^{1-s}}{s-1}+\sum_{j=1}^qG_j\,\Delta^{j-1}g_s(x). \end{aligned}$$
  • Analogue of Raabe’s formula

    $$\displaystyle \begin{aligned} \int_x^{x+1}\zeta(s,t){\,}dt ~=~ \frac{x^{1-s}}{s-1}{\,},\qquad x>0. \end{aligned}$$
  • Alternative characterization. The function f s(x) = ζ(s, x) is the unique solution lying in \(\mathcal {C}^0\cap \mathcal {K}^{\lfloor 1-s\rfloor _+}\) to the equation

    $$\displaystyle \begin{aligned} \int_x^{x+1}f_s(t){\,}dt ~=~ \frac{x^{1-s}}{s-1}{\,}, \qquad x>0. \end{aligned}$$

Inequalities

The following inequalities hold for any x > 0, any a > 0, and any \(n\in \mathbb {N}^*\).

  • Symmetrized generalized Wendel’s inequality (equality if a ∈{0, 1, …, ⌊1 − s+})

    $$\displaystyle \begin{aligned} &\left|\zeta(s,x+a)-\zeta(s,x)-\sum_{j=1}^{\lfloor 1-s\rfloor_+}{\textstyle{{{a}\choose{j}}}}\,\Delta^{j-1}g_s(x)\right| \\&\quad \leq ~ \lceil a\rceil\left|{\textstyle{{{a-1}\choose{\lfloor 1-s\rfloor_+}}}}\right|\left|\Delta^{\lfloor 1-s\rfloor_+}g_s(x)\right|. \end{aligned} $$

    If s ≤ 0, then

    $$\displaystyle \begin{aligned} \begin{array}{rcl} & &\displaystyle {\left|\zeta(s,x+a)-\zeta(s,x)-\sum_{j=1}^{\lfloor 1-s\rfloor}{\textstyle{{{a}\choose{j}}}}\,\Delta^{j-1}g_s(x)\right|}\\ & \leq &\displaystyle \left|{\textstyle{{{a-1}\choose{\lfloor 1-s\rfloor}}}}\right|\left|\Delta^{\lfloor -s\rfloor}g_s(x+a)-\Delta^{\lfloor -s\rfloor}g_s(x)\right|. \end{array} \end{aligned} $$
  • Symmetrized generalized Wendel’s inequality (discrete version)

    $$\displaystyle \begin{aligned} \left|\zeta(s,x)-\zeta(s)-f_n^{\lfloor 1-s\rfloor_+}[g_s](x)\right| ~\leq ~ \lceil x\rceil\left|{\textstyle{{{x-1}\choose{\lfloor 1-s\rfloor_+}}}}\right|\left|\Delta^{\lfloor 1-s\rfloor_+}g_s(n)\right|. \end{aligned}$$

    If s ≤ 0, then

    $$\displaystyle \begin{aligned} \left|\zeta(s,x)-\zeta(s)-f_n^{\lfloor 1-s\rfloor}[g_s](x)\right| ~\leq ~ \left|{\textstyle{{{x-1}\choose{\lfloor 1-s\rfloor}}}}\right|\left|\Delta^{\lfloor -s\rfloor}g_s(x+n)-\Delta^{\lfloor -s\rfloor}g_s(n)\right|. \end{aligned}$$

    Here

    $$\displaystyle \begin{aligned} f^{\lfloor 1-s\rfloor_+}_n[g_s](x) ~=~ \sum_{k=0}^{n-1}(x+k)^{-s}-\sum_{k=1}^{n-1}k^{-s} -\sum_{j=1}^{\lfloor 1-s\rfloor_+}{\textstyle{{{x}\choose{j}}}}\,\Delta_n^{j-1}n^{-s}. \end{aligned}$$
  • Symmetrized Stirling’s formula-based inequality

    $$\displaystyle \begin{aligned} \left|J^{\lfloor 1-s\rfloor_++1}[\Sigma g_s](x)\right| ~\leq ~ \overline{G}_{\lfloor 1-s\rfloor_+}\left|\Delta^{\lfloor 1-s\rfloor_+}g_s(x)\right|. \end{aligned}$$

    If s ≤ 0, then

    $$\displaystyle \begin{aligned} \left|J^{\lfloor 2-s\rfloor}[\Sigma g_s](x)\right| ~\leq ~ \int_0^1\left|{\textstyle{{{t-1}\choose{\lfloor 1-s\rfloor}}}}\right|\left|\Delta^{\lfloor -s\rfloor}g_s(x+t)-\Delta^{\lfloor -s\rfloor}g_s(x)\right|dt. \end{aligned}$$
  • Burnside’s formula-based inequality if s > −1

    $$\displaystyle \begin{aligned} \left|\zeta\left(s,x+\frac{1}{2}\right)-\frac{x^{1-s}}{s-1}\right| ~\leq ~ \left|J^{\lfloor 1-s\rfloor_++1}[\Sigma g_s](x)\right|. \end{aligned}$$
  • Additional inequality if s > 1.

    $$\displaystyle \begin{aligned} 0 ~\leq ~ \zeta(s,x+n) ~=~ \sum_{k=n}^{\infty}(x+k)^{-s} ~\leq ~ \zeta(s,n). \end{aligned}$$
  • Generalized Gautschi’s inequality

    If s ≥ 0, s ≠ 1,

    $$\displaystyle \begin{aligned} \begin{array}{rcl} (\lceil a\rceil -a)(x+\lceil a\rceil)^{-s} & \leq &\displaystyle s(\lceil a\rceil -a)\,\zeta(s+1,x+\lceil a\rceil)\\ & \leq &\displaystyle \zeta(s,x+a)-\zeta(s,x+\lceil a\rceil) ~\leq ~ (\lceil a\rceil -a)(x+\lfloor a\rfloor)^{-s}. \end{array} \end{aligned} $$

    If s ≤ 0, then these inequalities must be reversed and they are valid only if the Hurwitz zeta function is concave on [x + ⌊a⌋, ).

Generalized Stirling’s and Related Formulas

For any a ≥ 0, we have the following limits and asymptotic equivalences as x →,

$$\displaystyle \begin{aligned} \zeta(s,x+a)-\zeta(s,x)-\sum_{j=1}^{\lfloor 1-s\rfloor_+}{\textstyle{{{a}\choose{j}}}}\,\Delta^{j-1}g_s(x) ~\to ~0, \end{aligned}$$
$$\displaystyle \begin{aligned} \zeta(s,x)-\frac{x^{1-s}}{s-1}+\sum_{j=1}^{\lfloor 1-s\rfloor_+}G_j\,\Delta^{j-1}g_s(x) ~\to ~0, \end{aligned}$$
$$\displaystyle \begin{aligned} \zeta(s,x)+\frac{1}{1-s}\sum_{j=0}^{\lfloor 1-s\rfloor_+}{\textstyle{{{1-s}\choose{j}}}}\,\frac{B_j}{x^{s+j-1}} ~\to ~0, \end{aligned}$$
$$\displaystyle \begin{aligned} \zeta(s,x+a) ~\sim ~ \frac{x^{1-s}}{s-1}{\,}. \end{aligned}$$

In particular, if s > 1, then ζ(s, x) → 0 as x →.

For instance, setting \(s=-\frac {3}{2}\) in these latter two asymptotic formulas, we obtain

$$\displaystyle \begin{aligned} \begin{array}{rcl} \textstyle{\zeta\left(-\frac{3}{2},x\right)+\frac{2}{5}{\,}x^{5/2}-\frac{7}{12}{\,}x^{3/2} +\frac{1}{12}{\,}(x+1)^{3/2}} & \to &\displaystyle 0{\,},\\ \textstyle{\zeta\left(-\frac{3}{2},x\right)+\frac{2}{5}{\,}x^{5/2}-\frac{1}{2}{\,}x^{3/2}+\frac{1}{8}{\,}x^{1/2}} & \to &\displaystyle 0{\,}. \end{array} \end{aligned} $$

If s > −1, then we have the analogue of Burnside’s formula

$$\displaystyle \begin{aligned} \textstyle{\zeta(s,x)-\frac{1}{s-1}{\,}(x-\frac{1}{2})^{1-s}} ~\to ~ 0{\,},\qquad \mbox{as }x\to\infty, \end{aligned}$$

which provides a better approximation of ζ(s, x) than the generalized Stirling formula.

Asymptotic Expansions

For any \(m, q\in \mathbb {N}^*\) we have the following expansion as x →

$$\displaystyle \begin{aligned} \frac{1}{m}\sum_{j=0}^{m-1}\zeta\left(s,x+\frac{j}{m}\right) ~=~ \frac{1}{s-1}\,\sum_{k=0}^q{\textstyle{{{1-s}\choose{k}}}}\frac{B_k}{m^k{\,}x^{s+k-1}}+O\left(x^{-q-s}\right). \end{aligned}$$

Setting m = 1 in this formula, we obtain

$$\displaystyle \begin{aligned} \zeta(s,x) ~=~ \frac{1}{s-1}\,\sum_{k=0}^q{\textstyle{{{1-s}\choose{k}}}}\frac{B_k}{x^{s+k-1}}+O\left(x^{-q-s}\right). \end{aligned}$$

In particular, this clearly shows that ζ(s, x) is a (1 − s)-degree polynomial whenever 1 − s is a positive integer. More precisely, we have

$$\displaystyle \begin{aligned} \zeta(1-n,x) ~=~ -\frac{1}{n}{\,}\sum_{k=0}^n{\textstyle{{{n}\choose{k}}}}{\,}B_k{\,}x^{n-k},\qquad n\in\mathbb{N}^*, \end{aligned}$$

that is,

$$\displaystyle \begin{aligned} \zeta(1-n,x) ~=~ -\frac{1}{n}{\,}B_n(x),\qquad n\in\mathbb{N}^*. \end{aligned} $$
(10.16)

Generalized Liu’s Formula

We have the following formulas for x > 0.

  • If s > −1, then

    $$\displaystyle \begin{aligned} \zeta(s,x) ~=~ \frac{x^{1-s}}{s-1}+\frac{1}{2}{\,}x^{-s}-s{\,}\int_0^{\infty}\frac{\{t\}-\frac{1}{2}}{(x+t)^{s+1}}{\,}dt. \end{aligned}$$
  • If s ≤−1, then for any integer q ≥⌈(1 − s)∕2⌉,

    $$\displaystyle \begin{aligned} \zeta(s,x) ~=~ \frac{x^{1-s}}{s-1}+\frac{1}{2}{\,}x^{-s}-\sum_{k=1}^q\frac{B_{2k}}{(2k)!}{\,}\frac{(-s)^{\underline{2k-1}}}{x^{s+2k-1}} -\frac{(-s)^{\underline{2q}}}{(2q)!}\,\int_0^{\infty}\frac{B_{2q}(\{t\})}{(x+t)^{s+2q}}{\,}dt. \end{aligned}$$

Limit and Series Representations When s > 1

We simply have

$$\displaystyle \begin{aligned} \zeta(s,x) ~=~ \sum_{k=0}^{\infty}(x+k)^{-s} \end{aligned}$$

and this series converges uniformly on \(\mathbb {R}_+\). In particular, we retrieve the identity

$$\displaystyle \begin{aligned} \psi_{\nu}(x) ~=~ (-1)^{\nu +1}\nu!\,\zeta(\nu +1,x){\,},\qquad \nu\in\mathbb{N}^*. \end{aligned}$$

Limit and Series Representations When s < 1

We have the following Eulerian form

$$\displaystyle \begin{aligned} \begin{array}{rcl} \zeta(s,x)-\zeta(s) & =&\displaystyle -g_s(x)+\sum_{j=0}^{\lfloor -s\rfloor}{\textstyle{{{x}\choose{j+1}}}}\Delta^jg_s(1)\\ & &\displaystyle + \sum_{k=1}^{\infty}\left(-g_s(x+k)+\sum_{j=0}^{\lfloor 1-s\rfloor}{\textstyle{{{x}\choose{j}}}}\,\Delta^j g_s(k)\right), \end{array} \end{aligned} $$

and the Weierstrassian form can be obtained similarly. The associated series converge uniformly on any bounded subset of [0, ).

For instance, we have

$$\displaystyle \begin{aligned} \begin{array}{rcl} & &\displaystyle {\textstyle{\zeta\left(-\frac{3}{2},x\right)-\zeta\left(-\frac{3}{2}\right)}}\\& =&\displaystyle ~ \displaystyle{x^{\frac{3}{2}}+\lim_{n\to\infty}\left(\sum_{k=1}^{n-1}\left((x+k)^{\frac{3}{2}}-k^{\frac{3}{2}}\right) -x{\,}n^{\frac{3}{2}}-{\textstyle{{{x}\choose{2}}}}\Delta_nn^{\frac{3}{2}}\right)}\\ & =&\displaystyle x^{\frac{3}{2}}-x-(2\sqrt{2}-1){\textstyle{{{x}\choose{2}}}} + \sum_{k=1}^{\infty}\left((x+k)^{\frac{3}{2}}-k^{\frac{3}{2}} -x\Delta_kk^{\frac{3}{2}}-{\textstyle{{{x}\choose{2}}}}\Delta_k^2k^{\frac{3}{2}}\right)\\ & =&\displaystyle x^{\frac{3}{2}}-x+\textstyle{\frac{3}{4}{\,}\zeta\left(\frac{1}{2}\right)}{\textstyle{{{x}\choose{2}}}} + \displaystyle{\sum_{k=1}^{\infty}\left((x+k)^{\frac{3}{2}}-k^{\frac{3}{2}} -x\Delta_kk^{\frac{3}{2}}-\textstyle{\frac{3}{4}}{\textstyle{{{x}\choose{2}}}}k^{-\frac{1}{2}}\right)}. \end{array} \end{aligned} $$

The analogue of Gauss’ limit is

$$\displaystyle \begin{aligned} \zeta(s,x) ~=~ \zeta(s)+\lim_{n\to\infty}f^{\lfloor 1-s\rfloor}_n[g_s](x),\qquad x>0. \end{aligned}$$

where

$$\displaystyle \begin{aligned} f^{\lfloor 1-s\rfloor}_n[g_s](x) ~=~ \sum_{k=0}^{n-1}(x+k)^{-s}-\sum_{k=1}^{n-1}k^{-s} -\sum_{j=1}^{\lfloor 1-s\rfloor}{\textstyle{{{x}\choose{j}}}}\,\Delta_n^{j-1}n^{-s}. \end{aligned}$$

Gregory’s Formula-Based Series Representation

For any x > 0 we have

$$\displaystyle \begin{aligned} \begin{array}{rcl} \zeta(s,x) & =&\displaystyle \frac{x^{1-s}}{s-1}-\sum_{n=0}^{\infty}G_{n+1}\Delta^ng_s(x)\\ & =&\displaystyle \frac{x^{1-s}}{s-1}+\sum_{n=0}^{\infty}|G_{n+1}|\,\sum_{k=0}^n(-1)^k{\textstyle{{{n}\choose{k}}}}(x+k)^{-s}{\,}. \end{array} \end{aligned} $$

Setting x = 1 in this identity yields a known series expression for ζ(s) that is the analogue of Fontana-Mascheroni series

$$\displaystyle \begin{aligned} \zeta(s) ~=~ \frac{1}{s-1}+\sum_{n=0}^{\infty}|G_{n+1}|\,\sum_{k=0}^n(-1)^k{\textstyle{{{n}\choose{k}}}}(k+1)^{-s}{\,}. \end{aligned}$$

Analogue of Gauss’ Multiplication Formula

For any \(m\in \mathbb {N}^*\) and any x > 0, we have

$$\displaystyle \begin{aligned} \sum_{j=0}^{m-1}\zeta\left(s,\frac{x+j}{m}\right) ~=~ m^s\,\zeta(s,x). \end{aligned}$$

Corollary 8.33 provides the following limits for any x > 0

$$\displaystyle \begin{aligned} \begin{array}{rcl} \lim_{m\to\infty}m^{s-1}\zeta(s,mx) & =&\displaystyle \frac{x^{1-s}}{s-1}{\,},\qquad s<1,\\ \lim_{m\to\infty}m^{s-1}(\zeta(s,mx)-\zeta(s,m)) & =&\displaystyle \frac{x^{1-s}-1}{s-1}{\,},\qquad s\neq 1. \end{array} \end{aligned} $$

Analogue of Wallis’s Product Formula

If s > 1, then we have

$$\displaystyle \begin{aligned} \sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{k^s} ~=~ (1-2^{1-s})\,\zeta(s) ~=~ \eta(s), \end{aligned} $$
(10.17)

where sη(s) is Dirichlet’s eta function. When s < 1, the form of the formula strongly depends upon the value of s. When \(s=-\frac {3}{2}\) for instance, we obtain

$$\displaystyle \begin{aligned} \lim_{n\to\infty}\left(h(n)+\sum_{k=1}^{2n}(-1)^kk^{\frac{3}{2}}\right) ~=~ \textstyle{(4\sqrt{2}-1)\,\zeta(-\frac{3}{2})}. \end{aligned}$$

where \(h(n)=-\frac {8n+3}{4}\,\sqrt {\frac {n}{2}}\).

Restriction to the Natural Integers

For any \(n\in \mathbb {N}^*\) we have

$$\displaystyle \begin{aligned} \zeta(s,n)-\zeta(s) ~=~ -\sum_{k=1}^{n-1}k^{-s}\qquad \mbox{and}\qquad \zeta(s,n) ~=~ \sum_{k=n}^{\infty}k^{-s}. \end{aligned}$$

Gregory’s formula states that for any \(n\in \mathbb {N}^*\) and any \(q\in \mathbb {N}\) we have

$$\displaystyle \begin{aligned} \sum_{k=1}^{n-1}k^{-s} ~=~ \frac{1-n^{1-s}}{s-1}+\sum_{j=1}^qG_j\left(\Delta^{j-1}g_s(n)-\Delta^{j-1}g_s(1)\right)+R^q_{s,n}{\,}, \end{aligned}$$

with

$$\displaystyle \begin{aligned} |R^q_{s,n}| ~\leq ~\overline{G}_q{\,}|\Delta^qg_s(n)-\Delta^qg_s(1)|. \end{aligned}$$

Many other representations of this sum can be derived from, e.g., the limit and series representations of the Hurwitz zeta function.

Generalized Webster’s Functional Equation

For any \(m\in \mathbb {N}^*\) and any a > 0, there is a unique solution \(f_s\colon \mathbb {R}_+\to \mathbb {R}\) to the equation

$$\displaystyle \begin{aligned} \sum_{j=0}^{m-1}f_s\left(x+a{\,}j\right) ~=~ -x^{-s} \end{aligned}$$

that lies in \(\mathcal {K}^{\lfloor -s\rfloor _+}\), namely

$$\displaystyle \begin{aligned} f_s(x) ~=~ \frac{1}{(am)^s}\,\zeta\left(s,\frac{x+a}{am}\right)-\frac{1}{(am)^s}\,\zeta\left(s,\frac{x}{am}\right){\,}. \end{aligned}$$

Analogue of Euler’s Series Representation of γ

We have

$$\displaystyle \begin{aligned} (\Sigma g_s)^{(k)}(1) ~=~ (-s)^{\underline{k}}\,\zeta(s+k),\qquad k\in\mathbb{N}^*. \end{aligned}$$

Thus, the Taylor series expansion of ζ(s, x + 1) about x = 0 is

$$\displaystyle \begin{aligned} \zeta(s,x+1) ~=~ \sum_{k=0}^{\infty}{\textstyle{{{-s}\choose{k}}}}\,\zeta(s+k){\,}x^k{\,},\qquad |x|<1. \end{aligned}$$

Integrating both sides of this equation on (0, 1), we obtain the identity

$$\displaystyle \begin{aligned} \sum_{k=1}^{\infty}{\textstyle{{{1-s}\choose{k}}}}\,\zeta(s+k-1) ~=~ -1{\,},\qquad s<2,~s\notin\mathbb{Z}{\,}. \end{aligned}$$

(When s > 2, the summand in the series above does not approach zero as k increases.)

Analogue of the Reflection Formula

A reflection formula can be derived when s is an integer. Recall that we have the following special values for any \(n\in \mathbb {N}^*\)

$$\displaystyle \begin{aligned} \zeta(1+n,x) ~=~ (-1)^{n-1}\frac{1}{n!}\,\psi_n(x) \end{aligned}$$

and

$$\displaystyle \begin{aligned} \zeta(1-n,x) ~=~ -\frac{1}{n}{\,}B_n(x). \end{aligned}$$

It follows that for any \(x\in \mathbb {R}\setminus \mathbb {Z}\), we have

$$\displaystyle \begin{aligned} \zeta(s,x)+(-1)^s\,\zeta(s,1-x) ~=~ \begin{cases} \frac{(-1)^{s-1}}{(s-1)!}\,\pi{\,}D^{s-1}\cot{}(\pi x), & \mbox{if }s-1\in\mathbb{N}^*,\\ 0, & \mbox{if }-s\in\mathbb{N}. \end{cases} \end{aligned}$$

10.7 The Generalized Stieltjes Constants

Recall that the generalized Stieltjes constants are the numbers γ n(x) that occur in the Laurent series expansion of the Hurwitz zeta function

$$\displaystyle \begin{aligned} \zeta(s,x) ~=~ \frac{1}{s-1}+\sum_{n=0}^{\infty}\frac{(-1)^n}{n!}\,\gamma_n(x)(s-1)^n. \end{aligned} $$
(10.18)

Recall also that the numbers γ n = γ n(1), where \(n\in \mathbb {N}\), are called the Stieltjes constants. The Stieltjes constants and generalized Stieltjes constants are known to satisfy the relations

$$\displaystyle \begin{aligned} \gamma_0(x) ~=~ -\psi(x)\qquad \mbox{and}\qquad \gamma_0 ~=~ \gamma \end{aligned}$$

as well as the following identities for every \(q\in \mathbb {N}\)

$$\displaystyle \begin{aligned} \begin{array}{rcl} \gamma_q & =&\displaystyle \lim_{n\to\infty}\left(\sum_{k=1}^n\frac{(\ln k)^q}{k}-\frac{(\ln n)^{q+1}}{q+1}\right),\\ \gamma_q(x) & =&\displaystyle \lim_{n\to\infty}\left(\sum_{k=0}^n\frac{(\ln(x+k))^q}{x+k}-\frac{(\ln(x+n))^{q+1}}{q+1}\right). \end{array} \end{aligned} $$

For recent background on these constants, see, e.g., Blagouchine [19, 20] and Blagouchine and Coppo [22] (see also Nan-Yue and Williams [80]).

Here we naturally restrict the values of x to the set \(\mathbb {R}_+\). Interestingly, the generalized Stieltjes constants also satisfy the difference equation

$$\displaystyle \begin{aligned} \gamma_q(x+1)-\gamma_q(x) ~=~ g_q(x), \end{aligned}$$

where \(g_q\colon \mathbb {R}_+\to \mathbb {R}\) is the function defined by the equation

$$\displaystyle \begin{aligned} g_q(x) ~=~ -\frac{1}{x}(\ln x)^q\qquad \mbox{for }x>0. \end{aligned}$$

Thus, our theory is particularly suitable for the investigation of these constants. For any \(q\in \mathbb {N}\), the function g q lies in \(\mathcal {C}^{\infty }\cap \mathcal {D}^0\cap \mathcal {K}^{\infty }\) and is increasing on [e q, ). By uniqueness of Σg q, it follows that

$$\displaystyle \begin{aligned} \Sigma g_q(x) ~=~ \gamma_q(x)-\gamma_q. \end{aligned}$$

ID Card

The introduction above enables us to provide the following basic information about the generalized Stieltjes constants.

g q(x)

Membership

deg g q

Σg q(x)

\(\mathstrut ^{\mathstrut }_{\mathstrut } -\frac {1}{x}(\ln x)^q\)

\(\mathcal {C}^{\infty }\cap \mathcal {D}^0\cap \mathcal {K}^{\infty }\)

− 1

γ q(x) − γ q

Analogue of Bohr-Mollerup’s Theorem

The function γ q can be characterized as follows.

All eventually monotone solutions \(f_q\colon \mathbb {R}_+\to \mathbb {R}\) to the equation

$$\displaystyle \begin{aligned} f_q(x+1)-f_q(x) ~=~ -\frac{1}{x}(\ln x)^q \end{aligned}$$

are of the form f q(x) = c q + γ q(x), where \(c_q\in \mathbb {R}\).

Using Proposition 3.9, we can also derive the following alternative characterization of the function γ q.

All solutions \(f_q\colon \mathbb {R}_+\to \mathbb {R}\) to the equation

$$\displaystyle \begin{aligned} f_q(x+1)-f_q(x) ~=~ -\frac{1}{x}(\ln x)^q \end{aligned}$$

that satisfy the asymptotic condition that, for each x > 0,

$$\displaystyle \begin{aligned} f_q(x+n)-f_q(n) ~\to ~0\qquad \mbox{as }n\to_{\mathbb{N}}\infty \end{aligned}$$

are of the form f q(x) = c q + γ q(x), where \(c_q\in \mathbb {R}\).

Extended ID Card

Using identity (8.11), we can immediately make the remarkable observation that the asymptotic constant σ[g q] is exactly the opposite of the Stieltjes constant γ q. We then have the following values

  • Alternative representations of σ[g q] = γ[g q]

    $$\displaystyle \begin{aligned} \begin{array}{rcl} \gamma_q & =&\displaystyle \sum_{k=1}^{\infty}\left(\frac{(\ln k)^q}{k}-\frac{(\ln(k+1))^{q+1}-(\ln(k))^{q+1}}{q+1}\right),\\ \gamma_q & =&\displaystyle \int_1^{\infty}\frac{\{t\}-\frac{1}{2}}{t^2}{\,}(\ln t)^{q-1}(q-\ln t){\,}dt{\,}\qquad (q\geq 1),\\ \gamma_q & =&\displaystyle \int_1^{\infty}\left(\frac{(\ln\lfloor t\rfloor)^q}{\lfloor t\rfloor}-\frac{(\ln t)^q}{t}\right)dt{\,}. \end{array} \end{aligned} $$
  • Generalized Binet’s function. For any \(r\in \mathbb {N}\) and any x > 0

    $$\displaystyle \begin{aligned} J^{r+1}[\gamma_q](x) ~=~ \gamma_q(x)+\frac{(\ln x)^{q+1}}{q+1}+\sum_{j=1}^rG_j\,\Delta^{j-1}g_q(x). \end{aligned}$$
  • Analogue of Raabe’s formula

    $$\displaystyle \begin{aligned} \int_x^{x+1}\gamma_q(t){\,}dt ~=~ -\frac{(\ln x)^{q+1}}{q+1}{\,},\qquad x>0. \end{aligned} $$
    (10.19)
  • Alternative characterization. The function f(x) = γ q(x) is the unique solution lying in \(\mathcal {C}^0\cap \mathcal {K}^0\) to the equation

    $$\displaystyle \begin{aligned} \int_x^{x+1}f(t){\,}dt ~=~ -\frac{(\ln x)^{q+1}}{q+1}{\,}, \qquad x>0. \end{aligned}$$

\(\overline {\sigma }[g_q]\)

σ[g q]

γ[g q]

\(\mathstrut ^{\mathstrut }_{\mathstrut } \infty \)

− γ q

− γ q

Inequalities

The following inequalities hold for any x > 0, any a > 0, and any \(n\in \mathbb {N}\).

  • Symmetrized generalized Wendel’s inequality (equality if a ∈{0, 1})

    If x ≥ e q, we have

    $$\displaystyle \begin{aligned} \left|\gamma_q(x+a)-\gamma_q(x)\right| ~\leq ~ \lceil a\rceil\left|\frac{(\ln x)^q}{x}\right|. \end{aligned}$$
  • Symmetrized generalized Wendel’s inequality (discrete version)

    If n ≥ e q, we have

    $$\displaystyle \begin{aligned} \left|\gamma_q(x) - \gamma_q-\frac{(\ln x)^q}{x}-\sum_{k=1}^{n-1}\left(\frac{(\ln(x+k))^q}{x+k}-\frac{(\ln k)^q}{k}\right)\right| ~\leq ~ \lceil x\rceil\left|\frac{(\ln n)^q}{n}\right|. \end{aligned}$$
  • Symmetrized Stirling’s and Burnside’s formulas-based inequalities

    If x ≥ e q, we have

    $$\displaystyle \begin{aligned} \left|\gamma_q\left(x+\frac{1}{2}\right)+\frac{(\ln x)^{q+1}}{q+1}\right| ~\leq ~ \left|\gamma_q(x)+\frac{(\ln x)^{q+1}}{q+1}\right| ~\leq ~ \left|\frac{(\ln x)^q}{x}\right|{\,}. \end{aligned}$$
  • Further inequalities. For 0 < x ≤ 1, we use the following approximations (see Nan-Yue and Williams [80, p. 148])

    $$\displaystyle \begin{aligned} \left|\gamma_0(x)-\frac{1}{x}\right| ~\leq ~\gamma \end{aligned}$$

    and

    $$\displaystyle \begin{aligned} \left|\gamma_q(x)-\frac{(\ln x)^q}{x}\right| ~\leq ~ \frac{(3+(-1)^q)(2q)!}{q^{q+1}(2\pi)^q}{\,},\quad q\in\mathbb{N}^*. \end{aligned}$$

Generalized Stirling’s and Related Formulas

For any a ≥ 0, we have the following limits and asymptotic equivalence as x →,

$$\displaystyle \begin{aligned} \gamma_q(x+a)-\gamma_q(x) ~\to~ 0,\qquad \gamma_q(x)+\frac{(\ln x)^{q+1}}{q+1} ~\to ~ 0, \end{aligned}$$
$$\displaystyle \begin{aligned} \gamma_q(x+a) ~\sim ~ -\frac{(\ln x)^{q+1}}{q+1}{\,}. \end{aligned}$$

Burnside-like approximation (better than Stirling-like approximation)

$$\displaystyle \begin{aligned} \gamma_q(x)+\frac{1}{q+1}\left(\ln\left(x-\frac{1}{2}\right)\right)^{q+1} ~\to ~ 0. \end{aligned}$$

Further results (obtained by differentiation)

$$\displaystyle \begin{aligned} \gamma^{\prime}_q(x)+\frac{(\ln x)^q}{x} ~\to ~ 0,\qquad \gamma^{\prime}_q(x+a) ~\sim ~ -\frac{(\ln x)^q}{x}{\,}. \end{aligned}$$

For any \(r\in \mathbb {N}\),

$$\displaystyle \begin{aligned} \gamma^{(r)}_q(x+a)-\gamma^{(r)}_q(x) ~\to~ 0,\qquad D^r_x\left(\gamma_q(x)+\frac{(\ln x)^{q+1}}{q+1}\right) \to ~ 0. \end{aligned}$$
$$\displaystyle \begin{aligned} D^r_x\left(\gamma_q(x)+\frac{1}{q+1}\left(\ln\left(x-\frac{1}{2}\right)\right)^{q+1}\right) \to ~ 0. \end{aligned}$$

Asymptotic Expansions

For any \(m,r\in \mathbb {N}^*\) we have the following expansion as x →

$$\displaystyle \begin{aligned} \frac{1}{m}\,\sum_{j=0}^{m-1}\gamma_q\left(x+\frac{j}{m}\right) ~=~ -\frac{(\ln x)^{q+1}}{q+1}+\sum_{k=1}^r\frac{B_k}{m^kk!}{\,}g_q^{(k-1)}(x)+O\left(g_q^{(r)}(x)\right). \end{aligned}$$

Setting m = 1 in this latter formula, we obtain

$$\displaystyle \begin{aligned} \gamma_q(x) ~=~ -\frac{(\ln x)^{q+1}}{q+1}+\sum_{k=1}^r\frac{B_k}{k!}{\,}g_q^{(k-1)}(x)+O\left(g_q^{(r)}(x)\right). \end{aligned}$$

Let us detail this expansion when q = 1. We first observe that

$$\displaystyle \begin{aligned} g_1^{(k-1)}(x) ~=~ (-1)^k(k-1)!\,\frac{\ln x -H_{k-1}}{x^k}{\,},\qquad k\in\mathbb{N}^*. \end{aligned}$$

Using (10.4), we then obtain

$$\displaystyle \begin{aligned} \begin{array}{rcl} & &\displaystyle {\frac{1}{m}\sum_{j=0}^{m-1}\gamma_1\left(x+\frac{j}{m}\right) + (\ln x){\,}\frac{1}{m}\sum_{j=0}^{m-1}\psi\left(x+\frac{j}{m}\right)}\\ & =&\displaystyle \frac{(\ln x)^2}{2}+\sum_{k=1}^r\frac{(-1)^{k-1}{\,}B_k{\,}H_{k-1}}{k(mx)^k}+O\left(x^{-r-1}\right). \end{array} \end{aligned} $$

Setting m = 1 in this latter formula, we get

$$\displaystyle \begin{aligned} \gamma_1(x) ~=~ \frac{(\ln x)^2}{2}-\psi(x)\ln x+ \sum_{k=1}^r\frac{(-1)^{k-1}{\,}B_k{\,}H_{k-1}}{k{\,}x^k}+O\left(x^{-r-1}\right){\,}. \end{aligned}$$

Setting r = 5 for instance, we obtain

$$\displaystyle \begin{aligned} \gamma_1(x) ~=~ \frac{(\ln x)^2}{2}-\psi(x)\ln x-\frac{1}{12x^2}+\frac{11}{720x^4}+O\left(x^{-6}\right). \end{aligned}$$

Generalized Liu’s Formula

For any q ≥ 1 and any x > 0 we have

$$\displaystyle \begin{aligned} \gamma_q(x) ~=~ -\frac{(\ln x)^{q+1}}{q+1}+\frac{(\ln x)^q}{2x}+\int_0^{\infty}\frac{\{t\}-\frac{1}{2}}{(x+t)^2}{\,}(\ln(x+t))^{q-1}(q-\ln(x+t)){\,}dt. \end{aligned}$$

Series Representations

Since the function g q(x) lies in \(\mathcal {D}^{-1}_{\mathbb {N}}\), we only have the following series representations of γ q(x).

  • Eulerian and Weierstrassian forms. We have

    $$\displaystyle \begin{aligned} \gamma_q(x) ~=~ \gamma_q+\frac{(\ln x)^q}{x}+\sum_{k=1}^{\infty}\left(\frac{(\ln(x+k))^q}{x+k}-\frac{(\ln k)^q}{k}\right), \end{aligned}$$
    $$\displaystyle \begin{aligned} \gamma_q(x) ~=~ \frac{(\ln x)^q}{x}+\sum_{k=1}^{\infty}\left(\frac{(\ln(x+k))^q}{x+k}-\frac{(\ln(k+1))^{q+1}-(\ln k)^{q+1}}{q+1}\right). \end{aligned}$$

    The series can be differentiated term by term infinitely many times. For instance, we get

    $$\displaystyle \begin{aligned} \gamma^{\prime}_q(x) ~=~ \sum_{k=0}^{\infty}\frac{(\ln(x+k))^{q-1}}{(x+k)^2}{\,}(q-\ln(x+k)). \end{aligned}$$
  • The analogue of Gauss’ limit coincides with the Eulerian form.

  • Gregory’s formula-based series representation. For any x > 0 satisfying the assumptions of Proposition 8.11, we obtain

    $$\displaystyle \begin{aligned} \begin{array}{rcl} \gamma_q(x) + \frac{(\ln x)^{q+1}}{q+1} & =&\displaystyle \sum_{n=0}^{\infty}G_{n+1}\Delta_x^n\frac{(\ln x)^q}{x}\\ & =&\displaystyle \sum_{n=0}^{\infty}|G_{n+1}|\sum_{k=0}^n(-1)^k{\textstyle{{{n}\choose{k}}}}\frac{(\ln(x+k))^q}{x+k}{\,}. \end{array} \end{aligned} $$

    Setting x = 1 in this identity (provided that x = 1 satisfies the assumptions of Proposition 8.11), we obtain the Fontana-Mascheroni’s series expression for γ q

    $$\displaystyle \begin{aligned} \gamma_q ~=~ \sum_{n=0}^{\infty}|G_{n+1}|\sum_{k=0}^n(-1)^k{\textstyle{{{n}\choose{k}}}}\frac{(\ln(k+1))^q}{k+1}{\,}. \end{aligned}$$

    This latter expression can be found in Blagouchine [20, p. 383] and the references therein.

Analogue of Gauss’ Multiplication Formula

The following analogue of Gauss’ multiplication formula was previously known (see also Blagouchine [19, p. 542]) but it can be derived straightforwardly from our results.

For any \(m\in \mathbb {N}^*\) and any x > 0, we have

$$\displaystyle \begin{aligned} \sum_{j=0}^{m-1}\gamma_q\left(\frac{x+j}{m}\right) ~=~ -\frac{m}{q+1}\left(\ln\frac{1}{m}\right)^{q+1}+m\sum_{j=0}^q{\textstyle{{{q}\choose{j}}}}\left(\ln\frac{1}{m}\right)^j\gamma_{q-j}(x). \end{aligned}$$

In particular,

$$\displaystyle \begin{aligned} \sum_{j=1}^m\gamma_q\left(\frac{j}{m}\right) ~=~ -\frac{m}{q+1}\left(\ln\frac{1}{m}\right)^{q+1}+m\sum_{j=0}^q{\textstyle{{{q}\choose{j}}}}\left(\ln\frac{1}{m}\right)^j\gamma_{q-j}{\,}. \end{aligned}$$

Corollary 8.33 provides the following limits for x > 0

$$\displaystyle \begin{aligned} \begin{array}{rcl} \lim_{m\to\infty} \sum_{j=0}^q{\textstyle{{{q}\choose{j}}}}\left(\ln\frac{1}{m}\right)^j\left(\gamma_{q-j}(mx)-\gamma_{q-j}(m)\right) & =&\displaystyle -\frac{(\ln x)^{q+1}}{q+1}{\,},\\ \lim_{m\to\infty} \left(-\frac{1}{q+1}\left(\ln\frac{1}{m}\right)^{q+1}{\!} +\sum_{j=0}^q{\textstyle{{{q}\choose{j}}}}\left(\ln\frac{1}{m}\right)^j\gamma_{q-j}(mx)\right) & =&\displaystyle -\frac{(\ln x)^{q+1}}{q+1}{\,}. \end{array} \end{aligned} $$

For instance, setting q = 1 in these formulas yields

$$\displaystyle \begin{aligned} \begin{array}{rcl} \lim_{m\to\infty} \big(\gamma_1(mx)-\gamma_1(m)+(\ln m)(\psi(mx)-\psi(m))\big) & =&\displaystyle -\frac{1}{2}(\ln x)^2{\,},\\ \lim_{m\to\infty} \left(\gamma_1(mx)-\frac{1}{2}(\ln m)^2+\psi(mx)\ln m\right) & =&\displaystyle -\frac{1}{2}(\ln x)^2{\,}. \end{array} \end{aligned} $$

Now, setting m = 2 in the multiplication formula, we obtain the following analogue of Legendre’s duplication formula

$$\displaystyle \begin{aligned} \gamma_q\left(\frac{x}{2}\right)+\gamma_q\left(\frac{x+1}{2}\right) ~=~ -\frac{2}{q+1}\left(\ln\frac{1}{2}\right)^{q+1}+2\sum_{j=0}^q{\textstyle{{{q}\choose{j}}}}\left(\ln\frac{1}{2}\right)^j\gamma_{q-j}(x). \end{aligned}$$

When q = 0 and q = 1, the multiplication formula reduces to the known formulas

$$\displaystyle \begin{aligned} \begin{array}{rcl} \sum_{j=0}^{m-1}\psi\left(\frac{x+j}{m}\right) & =&\displaystyle m(\psi(x)-\ln m){\,},\\ \sum_{j=0}^{m-1}\gamma_1\left(\frac{x+j}{m}\right) & =&\displaystyle -\frac{m}{2}(\ln m)^2+m(\ln m)\,\psi(x)+m\,\gamma_1(x). \end{array} \end{aligned} $$

Analogue of Wallis’s Product Formula

The analogue of Wallis’s formula for the function g q(x) is

$$\displaystyle \begin{aligned} \sum_{k=1}^{\infty}(-1)^k\frac{(\ln k)^q}{k} ~=~ -\frac{(\ln 2)^{q+1}}{q+1}+\sum_{j=0}^{q-1}{\textstyle{{{q}\choose{j}}}}{\,}(\ln 2)^{q-j}\gamma_j{\,}. \end{aligned} $$
(10.20)

This formula was established by Briggs and Chowla [25, Eq. (8)]. For q = 1, it reduces to

$$\displaystyle \begin{aligned} \sum_{k=1}^{\infty}(-1)^k\,\frac{\ln k}{k} ~=~ -\frac{(\ln 2)^2}{2}+\gamma\ln 2{\,}. \end{aligned}$$

For q = 2, we obtain

$$\displaystyle \begin{aligned} \sum_{k=1}^{\infty}(-1)^k\,\frac{(\ln k)^2}{k} ~=~ -\frac{(\ln 2)^3}{3}+\gamma (\ln 2)^2+2\gamma_1 \ln 2{\,}. \end{aligned}$$

These latter two formulas were also established by Hardy [47].

As an aside, let us establish conversion formulas between the sequences qγ q and qη (q)(1), where η(s) is the Dirichlet eta function introduced in (10.17) and η (q)(1) stands for the limiting value of η (q)(s) as s → 1. To ease the computations, let us instead consider the conversion formulas between the sequences qγ q and qλ q, where

$$\displaystyle \begin{aligned} \lambda_q ~=~ \frac{1}{q+1}{\,}(\ln 2)^{q+1}+(-1)^{q+1}\,\eta^{(q)}(1){\,},\qquad q\in\mathbb{N}. \end{aligned}$$

Using (10.20), we can readily derive the following equations

$$\displaystyle \begin{aligned} \lambda_q ~=~ \sum_{k=0}^{q-1}{\textstyle{{{q}\choose{k}}}}{\,}(\ln 2)^{q-k}\,\gamma_k{\,},\qquad q\in\mathbb{N}. \end{aligned} $$
(10.21)

These equations actually consist of an infinite consistent triangular system. Solving this system provides the following conversion formula

$$\displaystyle \begin{aligned} \gamma_q ~=~ \sum_{k=0}^q{\textstyle{{{q}\choose{k}}}}\,\frac{B_{q-k}}{k+1}{\,}(\ln 2)^{q-k-1}\,\lambda_{k+1}{\,},\qquad q\in\mathbb{N}, \end{aligned} $$
(10.22)

that is,

$$\displaystyle \begin{aligned} \gamma_q ~=~ -\frac{B_{q+1}}{q+1}{\,}(\ln 2)^{q+1}+\sum_{k=0}^q(-1)^k\,{\textstyle{{{q}\choose{k}}}}\,\frac{B_{q-k}}{k+1}(\ln 2)^{q-k-1}\eta^{(k+1)}(1){\,},\quad q\in\mathbb{N}. \end{aligned}$$

Indeed, plugging (10.22) in the right side of (10.21) we obtain for any \(q\in \mathbb {N}\)

$$\displaystyle \begin{aligned} \begin{array}{rcl} \sum_{k=0}^{q-1}{\textstyle{{{q}\choose{k}}}}{\,}(\ln 2)^{q-k}\,\gamma_k & =&\displaystyle \sum_{k=0}^{q-1}{\textstyle{{{q}\choose{k}}}}{\,}(\ln 2)^{q-k}\,\sum_{j=0}^k{\textstyle{{{k}\choose{j}}}}\,\frac{B_{k-j}}{j+1}{\,}(\ln 2)^{k-j-1}\,\lambda_{j+1}\\ & =&\displaystyle \sum_{j=0}^{q-1}{\textstyle{{{q}\choose{j}}}}{\,}(\ln 2)^{q-j-1}\,\frac{\lambda_{j+1}}{j+1}\,\sum_{k=j}^{q-1}{\textstyle{{{q-j}\choose{k-j}}}}{\,}B_{k-j}{\,}, \end{array} \end{aligned} $$

where the inner sum reduces to 0qj−1. The latter quantity then reduces to λ q, as expected.

Remark 10.10

The conversion formulas (10.21) and (10.22) are not quite new. In essence, they were established by Liang and Todd [63, Eq. (3.6)] and Nan-Yue and Williams [80, Eqs. (1.9) and (7.1)]. \(\lozenge \)

Generalized Webster’s Functional Equation

For any \(m\in \mathbb {N}^*\) and any a > 0, there is a unique eventually monotone solution \(f\colon \mathbb {R}_+\to \mathbb {R}\) to the equation

$$\displaystyle \begin{aligned} \sum_{j=0}^{m-1}f\left(x+a{\,}j\right) ~=~ -\frac{1}{x}(\ln x)^q{\,}, \end{aligned}$$

namely

$$\displaystyle \begin{aligned} f(x) ~=~ S_{q,am}\left(\frac{x+a}{am}\right)-S_{q,am}\left(\frac{x}{am}\right){\,}, \end{aligned}$$

where

$$\displaystyle \begin{aligned} S_{q,am}(x) ~=~ \frac{1}{am}\,\sum_{j=0}^q{q\choose j}(\ln(am))^j\,\gamma_{q-j}(x). \end{aligned}$$

For instance, the unique eventually monotone solution \(f\colon \mathbb {R}_+\to \mathbb {R}\) to the equation

$$\displaystyle \begin{aligned} f(x)+f(x+1) ~=~ -\frac{1}{x}\ln x \end{aligned}$$

is

$$\displaystyle \begin{aligned} f(x) ~=~ \gamma_1(x)-\gamma_1\left(\frac{x}{2}\right)+(\ln 2)\,\psi(x)+\frac{1}{2}(\ln 2)^2. \end{aligned}$$

Rational Arguments Theorem

Let us apply Proposition 8.65 to the function g q(x). For any \(a,b\in \mathbb {N}^*\) with a < b and any j ∈{0, …, b − 1} we have

$$\displaystyle \begin{aligned} S_j^b[g_q] ~=~ b{\,}(-1)^{q+1}\sum_{i=0}^q{q\choose i}(\ln b)^{q-i}D^i_s\,\mathrm{Li}_s(z)\big|{}_{(s,z)=(1,\omega_b^j)}, \end{aligned}$$

where Lis(z) is the polylogarithm function. Hence, we have

$$\displaystyle \begin{aligned} \gamma_q\left(\frac{a}{b}\right)-\gamma_q ~=~ (-1)^{q+1}\sum_{i=0}^q{q\choose i}(\ln b)^{q-i}\sum_{j=0}^{b-1}(1-\omega_b^{-aj})D^i_s\,\mathrm{Li}_s(z)\big|{}_{(s,z)=(1,\omega_b^j)}{\,}. \end{aligned}$$

We note that a more practical formula was derived in the special case when q = 1 by Blagouchine [19] as a generalization of Gauss’ digamma theorem.

10.8 Higher Order Derivatives of the Hurwitz Zeta Function

Let \(s\in \mathbb {R}\setminus \{1\}\) and \(q\in \mathbb {N}\). Differentiating q times both sides of (10.15) we obtain

$$\displaystyle \begin{aligned} \zeta^{(q)}(s,x+1)-\zeta^{(q)}(s,x) ~=~ (-1)^{q+1}x^{-s}(\ln x)^q{\,},\qquad x>0, \end{aligned}$$

where ζ (q)(s, x) stands for \(D^q_s\,\zeta (s,x)\). This equation shows that the investigation of the higher order derivatives of the Hurwitz zeta function can be carried out using our results. To keep our presentation simple, we will focus on some selected results only.

The interested reader can find an earlier study of these functions in Ramanujan’s second notebook [18, p. 36 et seq.].

ID Card

The following basic information can be easily derived.

g s,q(x)

Membership

deg g s,q

Σg s,q(x)

\(\mathstrut ^{\mathstrut }_{\mathstrut } -x^{-s}(-\ln x)^q\)

\(\begin {array}{rl}\mathstrut ^{\mathstrut }_{\mathstrut }\mathcal {C}^{\infty }\cap \mathcal {D}^{-1}\cap \mathcal {K}^{\infty },&\mbox{if }s>1,\\ \mathcal {C}^{\infty }\cap \mathcal {D}^{\lfloor 1-s\rfloor }\cap \mathcal {K}^{\infty },&\mbox{if }s<1.\end {array}\)

\(\begin {array}{c} -1\\ +\lfloor 1-s\rfloor _+\end {array}\)

\(\begin {array}{c}\zeta ^{(q)}(s,x)\\ -\zeta ^{(q)}(s)\end {array}\)

We observe that this investigation can be regarded as a simultaneous generalization of the studies of the Hurwitz zeta function and the generalized Stieltjes constants. For the latter, we observe that

$$\displaystyle \begin{aligned} (-1)^q\,\lim_{s\to 1}g_{s,q}(x) ~=~ -\frac{1}{x}(\ln x)^q. \end{aligned}$$

Setting s = 0 in our results may also be very informative as it produces formulas involving the well-studied quantities ζ (q)(0) and ζ (q)(0, x) − ζ (q)(0) for any \(q\in \mathbb {N}\).

Project 10.11

Find a closed-form expression for the integral

$$\displaystyle \begin{aligned} \int_1^x\gamma_q(t){\,}dt. \end{aligned}$$

We apply Proposition 8.20 to \(g_q(x)=-\frac {1}{x}(\ln x)^q\). Using (10.19) we obtain

$$\displaystyle \begin{aligned} \begin{array}{rcl} \int_1^x\gamma_q(t){\,}dt & =&\displaystyle \Sigma_x\int_x^{x+1}\gamma_q(t){\,}dt ~=~ -\frac{1}{q+1}\,\Sigma(\ln x)^{q+1}\\ & =&\displaystyle \frac{(-1)^{q+1}}{q+1}\,\Sigma g_{0,q+1}(x){\,}, \end{array} \end{aligned} $$

that is,

$$\displaystyle \begin{aligned} \int_1^x\gamma_q(t){\,}dt ~=~ \frac{(-1)^{q+1}}{q+1}\left(\zeta^{(q+1)}(0,x)-\zeta^{(q+1)}(0)\right). \end{aligned}$$

In particular,

$$\displaystyle \begin{aligned} \gamma_q(x) ~=~ \frac{(-1)^{q+1}}{q+1}\, D_x\zeta^{(q+1)}(0,x){\,}. \end{aligned}$$

\(\lozenge \)

Analogue of Bohr-Mollerup’s Theorem

The function ζ (q)(s, x) can be characterized as follows.

All solutions \(f_{s,q}\colon \mathbb {R}_+\to \mathbb {R}\) to the equation

$$\displaystyle \begin{aligned} f_{s,q}(x+1)-f_{s,q}(x) ~=~ g_{s,q}(x) \end{aligned}$$

that lie in \(\mathcal {K}^{\lfloor 1-s\rfloor _+}\) are of the form f s,q(x) = c s,q + ζ (q)(s, x), where \(c_{s,q}\in \mathbb {R}\).

Extended ID Card

The asymptotic constant σ[g s,q] satisfies the identity

$$\displaystyle \begin{aligned} \sigma[g_{s,q}] ~=~ \int_0^1\zeta^{(q)}(s,t+1){\,}dt - \zeta^{(q)}(s) ~=~ \frac{-q!}{(1-s)^{q+1}}-\zeta^{(q)}(s). \end{aligned}$$

Hence we have the following values

  • Alternative representations of σ[g s,q]

    $$\displaystyle \begin{aligned} \begin{array}{rcl} \sigma[g_{s,q}] & =&\displaystyle \lim_{n\to\infty}\left(\sum_{k=1}^{n-1}g_{s,q}(k)-\int_1^ng_{s,q}(t){\,}dt +\sum_{j=1}^{\lfloor 1-s\rfloor_+}G_j\Delta^{j-1}g_{s,q}(n)\right),\\ \sigma[g_{s,q}] & =&\displaystyle \sum_{j=1}^{\lfloor 1-s\rfloor_+}G_j\Delta^{j-1}g_{s,q}(1)\\ & &\displaystyle -\sum_{k=1}^{\infty}\left(\int_k^{k+1}g_{s,q}(t){\,}dt-\sum_{j=0}^{\lfloor 1-s\rfloor_+}G_j\Delta^jg_{s,q}(k)\right). \end{array} \end{aligned} $$

    Setting s = 0 in the previous formulas, we obtain

    $$\displaystyle \begin{aligned} \begin{array}{rcl} (-1)^q(q!+\zeta^{(q)}(0)) & =&\displaystyle \lim_{n\to\infty}\left(\sum_{k=1}^n(\ln k)^q-\int_1^n(\ln t)^q{\,}dt-\frac{1}{2}(\ln n)^q\right)\\ & =&\displaystyle \sum_{k=1}^{\infty}\left(\frac{1}{2}(\ln k)^q-\int_k^{k+1}(\ln t)^q{\,}dt\right). \end{array} \end{aligned} $$

    The left-hand quantity can actually be related to the Stieltjes constants in a very simple way. Indeed, on differentiating both sides of (10.18), we obtain the following surprising identity

    $$\displaystyle \begin{aligned} (-1)^q(q!+\zeta^{(q)}(0)) ~=~ \sum_{n=0}^{\infty}\frac{\gamma_{n+q}}{n!}{\,}. \end{aligned}$$
  • Generalized Binet’s function. For any \(r\in \mathbb {N}\) and any x > 0

    $$\displaystyle \begin{aligned} J^{r+1}[\Sigma g_{s,q}](x) ~=~ \zeta^{(q)}(s,x)-\int_x^{x+1}\zeta^{(q)}(s,t){\,}dt+\sum_{j=1}^rG_j\,\Delta^{j-1}g_{s,q}(x). \end{aligned}$$
  • Analogue of Raabe’s formula. We have

    $$\displaystyle \begin{aligned} \int_1^x g_{s,q}(t){\,}dt ~=~ \frac{q!-\Gamma(q+1,(s-1)\ln x)}{(1-s)^{q+1}}{\,},\quad x>0, \end{aligned}$$

    and hence the analogue of Raabe’s formula is

    $$\displaystyle \begin{aligned} \begin{array}{rcl} \int_x^{x+1}\zeta^{(q)}(s,t){\,}dt & =&\displaystyle -\frac{\Gamma(q+1,(s-1)\ln x)}{(1-s)^{q+1}}\\ & =&\displaystyle -q!{\,}\frac{x^{1-s}}{(1-s)^{q+1}}{\,}\sum_{j=0}^q\frac{((s-1)\ln x)^j}{j!}{\,},\quad x>0. \end{array} \end{aligned} $$

\(\overline {\sigma }[g_{s,q}]\)

σ[g s,q]

γ[g s,q]

\(\mathstrut ^{\mathstrut }_{\mathstrut } \begin {array}{rl}\infty ,&\mbox{if }s>1,\\ -\zeta ^{(q)}(s),&\mbox{if }s<1.\end {array}\)

\(\frac {-q!}{(1-s)^{q+1}}-\zeta ^{(q)}(s)\)

\(\sigma [g_{s,q}]-\sum _{j=1}^{\lfloor 1-s\rfloor _+}G_j\Delta ^{j-1}g_{s,q}(1)\)

Generalized Stirling’s and Related Formulas

For any a ≥ 0 we have

$$\displaystyle \begin{aligned} \zeta^{(q)}(s,x+a)-\zeta^{(q)}(s,x)-\sum_{j=1}^{\lfloor 1-s\rfloor_+}{\textstyle{{{a}\choose{j}}}}\,\Delta^{j-1}g_{s,q}(x) ~\to ~0 \qquad \mbox{as }x\to\infty, \end{aligned}$$

with equality if a ∈{0, 1, …, ⌊1 − s+}. Also, we have the following analogue of Stirling’s formula

$$\displaystyle \begin{aligned} \zeta^{(q)}(s,x)-\int_x^{x+1}\zeta^{(q)}(s,t){\,}dt+\sum_{j=1}^{\lfloor 1-s\rfloor_+}G_j\,\Delta^{j-1}g_{s,q}(x) ~\to ~0\qquad \mbox{as }x\to\infty. \end{aligned}$$

Setting s = 0 in this latter formula and then simplifying the resulting expression, we obtain

$$\displaystyle \begin{aligned} \zeta^{(q)}(0,x)+\Gamma(q+1,-\ln x)+\frac{1}{2}(-1)^{q+1}(\ln x)^q ~\to ~0\qquad \mbox{as }x\to\infty. \end{aligned}$$

We also have

$$\displaystyle \begin{aligned} \zeta^{(q)}(s,x+a) ~\sim ~ \int_x^{x+1}\zeta^{(q)}(s,t){\,}dt\qquad \mbox{as }x\to\infty. \end{aligned}$$

Finally, if s > −1, then we have the following analogue of Burnside’s formula

$$\displaystyle \begin{aligned} \zeta^{(q)}(s,x)-\int_{x-\frac{1}{2}}^{x+\frac{1}{2}}\zeta^{(q)}(s,t){\,}dt ~\to ~ 0{\,},\qquad \mbox{as }x\to\infty, \end{aligned}$$

which provides a better approximation of ζ q(s, x) than the analogue of Stirling’s formula.

Eulerian and Weierstrassian Forms

If s > 1, then for any x > 0, we simply have

$$\displaystyle \begin{aligned} \zeta^{(q)}(s,x) ~=~ -\sum_{k=0}^{\infty}g_{s,q}(x+k) \end{aligned}$$

and this series converges uniformly on \(\mathbb {R}_+\) and can be integrated and differentiated term by term. If s < 1, then for any x > 0, we obtain the following Eulerian form

$$\displaystyle \begin{aligned} \begin{array}{rcl} \zeta^{(q)}(s,x)-\zeta^{(q)}(s) & =&\displaystyle -g_{s,q}(x)+\sum_{j=0}^{\lfloor -s\rfloor}{\textstyle{{{x}\choose{j+1}}}}\Delta^jg_{s,q}(1)\\ & &\displaystyle + \sum_{k=1}^{\infty}\left(-g_{s,q}(x+k)+\sum_{j=0}^{\lfloor 1-s\rfloor}{\textstyle{{{x}\choose{j}}}}\,\Delta^j g_{s,q}(k)\right) \end{array} \end{aligned} $$

and the Weierstrassian form can be obtained similarly. Both associated series converge uniformly on any bounded subset of [0, ) and can be integrated and differentiated term by term. Note that the case where (s, q) = (0, 2) can be found in Ramanujan’s second notebook [18, p. 26–27].

Gregory’s Formula-Based Series Representation

For any x > 0 satisfying the assumptions of Proposition 8.11, we have

$$\displaystyle \begin{aligned} \begin{array}{rcl} \zeta^{(q)}(s,x) & =&\displaystyle \int_x^{x+1}\zeta^{(q)}(s,t){\,}dt-\sum_{n=0}^{\infty}G_{n+1}\Delta^ng_{s,q}(x)\\ & =&\displaystyle \int_x^{x+1}\zeta^{(q)}(s,t){\,}dt-\sum_{n=0}^{\infty}|G_{n+1}|\,\sum_{k=0}^n(-1)^k{\textstyle{{{n}\choose{k}}}}{\,}g_{s,q}(x+k){\,}. \end{array} \end{aligned} $$

Setting x = 1 in this identity (provided that x = 1 satisfies the assumptions of Proposition 8.11) yields a series expression for ζ (q)(s) that is the analogue of Fontana-Mascheroni series

$$\displaystyle \begin{aligned} \zeta^{(q)}(s) ~=~ \frac{-q!}{(1-s)^{q+1}}-\sum_{n=0}^{\infty}|G_{n+1}|\,\sum_{k=0}^n(-1)^k{\textstyle{{{n}\choose{k}}}}{\,}g_{s,q}(k+1){\,}, \end{aligned}$$

which can also be obtained differentiating the analogue of Fontana-Mascheroni series for the Hurwitz zeta function. For instance, we have

$$\displaystyle \begin{aligned} \zeta''(0) ~=~ -2+\sum_{n=0}^{\infty}|G_{n+1}|\,\sum_{k=0}^n(-1)^k{\textstyle{{{n}\choose{k}}}}{\,}(\ln(k+1))^2 \end{aligned}$$

and this latter value is also known to be (see, e.g., Berndt [18, p. 25])

$$\displaystyle \begin{aligned} \frac{1}{2}\,\gamma^2-\frac{\pi^2}{24}-\frac{1}{2}(\ln(2\pi))^2+\gamma_1{\,}. \end{aligned}$$

Analogue of Gauss’ Multiplication Formula

Upon differentiating the analogue of Gauss’ multiplication formula for the Hurwitz zeta function, we immediately obtain the following multiplication formula. For any \(m\in \mathbb {N}^*\) and any x > 0, we have

$$\displaystyle \begin{aligned} \sum_{j=0}^{m-1}\zeta^{(q)}\left(s,\frac{x+j}{m}\right) ~=~ m^s\,\sum_{j=0}^q{\textstyle{{{q}\choose{j}}}}(\ln m)^{q-j}\,\zeta^{(j)}(s,x). \end{aligned}$$

Moreover, Corollary 8.33 provides the following limit for any x > 0 and any s < 1

$$\displaystyle \begin{aligned} \lim_{m\to\infty} \sum_{j=0}^q{\textstyle{{{q}\choose{j}}}}(\ln m)^{q-j}\,\frac{\zeta^{(j)}(s,mx)}{m^{1-s}} ~=~ -\frac{\Gamma(q+1,(s-1)\ln x)}{(1-s)^{q+1}}{\,}. \end{aligned}$$

Also, for any s ≠ 1, we have

$$\displaystyle \begin{aligned} \lim_{m\to\infty} \sum_{j=0}^q{\textstyle{{{q}\choose{j}}}}(\ln m)^{q-j}\,\frac{\zeta^{(j)}(s,mx)-\zeta^{(j)}(s,m)}{m^{1-s}} ~=~ \frac{q!-\Gamma(q+1,(s-1)\ln x)}{(1-s)^{q+1}}{\,}. \end{aligned}$$

Analogue of Wallis’s Product Formula

When s < 1, the form of the analogue of Wallis’s product formula strongly depends upon the value of s. If s > 1, then we have

$$\displaystyle \begin{aligned} \begin{array}{rcl} \eta^{(q)}(s) & =&\displaystyle \sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{k^s}{\,}(-\ln k)^q\\ & =&\displaystyle \zeta^{(q)}(s)-2^{1-s}\sum_{j=0}^q{\textstyle{{{q}\choose{j}}}}\left(\ln\frac{1}{2}\right)^{q-j}\zeta^{(j)}(s), \end{array} \end{aligned} $$

where sη(s) is Dirichlet’s eta function. Just as we did for the formulas (10.21) and (10.22), we can easily establish the following conversion formulas for s > 1

$$\displaystyle \begin{aligned} \begin{array}{rcl} \mu_q(s) & =&\displaystyle \sum_{k=0}^{q-1}{\textstyle{{{q}\choose{k}}}}\left(\ln\frac{1}{2}\right)^{q-k}\zeta^{(k)}(s){\,},\qquad q\in\mathbb{N}{\,},\\ \zeta^{(q)}(s) & =&\displaystyle \sum_{k=0}^q{\textstyle{{{q}\choose{k}}}}\,\frac{B_{q-k}}{k+1}\left(\ln\frac{1}{2}\right)^{q-k-1}\mu_{k+1}(s){\,},\qquad q\in\mathbb{N}{\,}, \end{array} \end{aligned} $$

where

$$\displaystyle \begin{aligned} \mu_q(s) ~=~ 2^{s-1}(\zeta^{(q)}(s)-\eta^{(q)}(s))-\zeta^{(q)}(s){\,},\qquad q\in\mathbb{N}. \end{aligned}$$

10.9 The Catalan Number Function

The Catalan number function is the restriction to \(\mathbb {R}_+\) of the map xC x defined on \((-\frac {1}{2},\infty )\) by

$$\displaystyle \begin{aligned} C_x ~=~ \frac{1}{x+1}{2x\choose x}. \end{aligned}$$

This function satisfies the equation

$$\displaystyle \begin{aligned} C_{x+1} ~=~ \left(4-\frac{6}{x+2}\right) C_x{\,}. \end{aligned}$$

The additive version of this equation reads Δf = g, where the function g is the logarithm of a rational function. We observe that such equations have been thoroughly investigated by Anastassiadis [7, p. 41] (see also Kuczma [57]).

The equation above shows that the Catalan number function can be investigated using our results. Let us briefly study this function.

ID Card

The function C x is clearly a Γ-type function and we immediately derive the following basic information.

g(x)

Membership

deg g

Σg(x)

\(\mathstrut ^{\mathstrut }_{\mathstrut } \ln \left (4-\frac {6}{x+2}\right )\)

\(\mathcal {C}^{\infty }\cap \mathcal {D}^1\cap \mathcal {K}^{\infty }\)

0

\(\ln C_x\)

Analogue of Bohr-Mollerup’s Theorem

The function C x can be characterized as follows.

All solutions \(f\colon \mathbb {R}_+\to \mathbb {R}_+\) to the equation

$$\displaystyle \begin{aligned} (x+2){\,}f(x+1) ~=~ (4x+2){\,}f(x) \end{aligned}$$

for which \(\ln f\) lies in \(\mathcal {K}^1\) are of the form f(x) = c C x, where c > 0.

Extended ID Card

We have the following values:

\(\overline {\sigma }[g]\)

σ[g]

γ[g]

\(\mathstrut ^{\mathstrut }_{\mathstrut } \frac {1}{2}\left (3+\ln \frac {1}{8\pi }\right )\)

\(\frac {1}{2}\left (3+\ln \frac {8}{27\pi }\right )\)

\(\frac {1}{2}\left (3+\ln \frac {4}{27\pi }\right )\)

We also have the inequality

$$\displaystyle \begin{aligned} |\gamma[g]| ~\leq ~ \frac{25}{8}\ln 5+\frac{39}{8}\ln 3-16\ln 2+\frac{3}{4} ~\approx ~ 0.04 \end{aligned}$$

and the following representations

$$\displaystyle \begin{aligned} \begin{array}{rcl} \gamma[g] & =&\displaystyle \int_1^{\infty}\frac{3(\{t\}-\frac{1}{2})}{(t+2)(2t+1)}{\,}dt{\,},\\ \sigma[g] & =&\displaystyle \int_0^1\ln C_{t+1}{\,}dt. \end{array} \end{aligned} $$

Moreover, the analogue of Raabe’s formula is

$$\displaystyle \begin{aligned} \int_x^{x+1}\ln C_t{\,}dt ~=~ \ln\left(\frac{e^{\frac{3}{2}}(4x+2)^{x+\frac{1}{2}}}{\sqrt{\pi}{\,}(x+2)^{x+2}}\right),\qquad x>0. \end{aligned}$$

Generalized Stirling’s and Related Formulas

For any a ≥ 0, we have

$$\displaystyle \begin{aligned} \frac{C_{x+a}}{C_x} ~\sim ~ 4^a\qquad \mbox{and}\qquad C_x ~\sim ~ \frac{4^x}{x^{3/2}\,\sqrt{\pi}}\qquad \mbox{as }x\to\infty. \end{aligned}$$

Also, the analogue of Burnside’s formula gives

$$\displaystyle \begin{aligned} \ln C_x - \ln\left(\frac{e^{\frac{3}{2}}(4x)^x}{\sqrt{\pi}{\,}(x+\frac{3}{2})^{x+\frac{3}{2}}}\right) ~\to ~0 \qquad \mbox{as }x\to\infty. \end{aligned}$$

Restriction to the Natural Integers

For any \(n\in \mathbb {N}^*\) we have

$$\displaystyle \begin{aligned} C_n ~=~ \frac{1}{n+1}{\,}{\textstyle{{{2n}\choose{n}}}}. \end{aligned}$$

Eulerian and Weierstrassian Forms

For any x > 0, we have

$$\displaystyle \begin{aligned} C_x ~=~ \frac{x+2}{4x+2}{\,}2^x\, \prod_{k=1}^{\infty}\frac{\left(2-\frac{3}{k+3}\right)^x}{\left(2-\frac{3}{k+2}\right)^{x-1}\left(2-\frac{3}{x+k+2}\right)} \end{aligned}$$

and

$$\displaystyle \begin{aligned} C_x ~=~ \frac{x+2}{4x+2}{\,}e^{-\frac{x}{2}}\, \prod_{k=1}^{\infty}\frac{1+\frac{x}{k+2}}{1+\frac{2x}{2k+1}}{\,}e^{\frac{3x}{(k+2)(2k+1)}}. \end{aligned}$$