Abstract
In this paper, we present the best possible parameter \(a\in(1/15, \infty)\) such that the functions \(\psi^{\prime}(x+1)-\mathcal{L}_{x}(x, a)\) and \(\psi^{\prime\prime }(x+1)-\mathcal{L}_{xx}(x, a)\) are strictly increasing or decreasing with respect to \(x\in(0, \infty)\), where \(\mathcal{L}(x,a)=\frac {1}{90a^{2}+2}\log (x^{2}+x+\frac{3a+1}{3} )+ \frac {45a^{2}}{90a^{2}+2}\log (x^{2}+x+\frac{15a-1}{45a} )\) and \(\psi(x)\) is the classical psi function. As applications, we get several new sharp bounds for the psi function and its derivatives.
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1 Introduction
For real and positive values of x, Euler’s gamma function Γ and its logarithmic derivative ψ, the so-called psi function, are defined by
respectively. For extensions of these functions to complex variables and for basic properties see [1]. Recently, the gamma function Γ and psi function ψ have been the subject of intensive research. In particular, many remarkable inequalities and monotonicity properties for these functions can be found in the literature [2–18].
Recently, Yang [19] introduced the function
and proved that the double inequality
holds for all \(x>0\) if and only if \(a\leq a_{0}=0.5129\ldots\) and \(b\geq(40+3\sqrt{205})/105=0.7900\ldots\) if \(a\in(1/15, \infty)\) and \(b\in(4/15, \infty)\), where \(a_{0}\) is the unique solution of the equation \(\mathcal{L}(0, a)=\psi(1)\).
Partial derivative computations give
It is not difficult to verify that
by use of the L’Hôspital’s rule and the formula
given in [20].
The main purpose of this paper is to present the best possible parameter \(a\in(1/15, \infty)\) such that the functions \(\psi^{\prime}(x+1)-\mathcal{L}_{x}(x, a)\) and \(\psi^{\prime\prime }(x+1)-\mathcal{L}_{xx}(x, a)\) are strictly increasing or decreasing with respect to \(x\in(0, \infty)\), and establish several new sharp bounds for the psi function and its derivatives. All numerical computations are carried out using the MATHEMATICA software.
2 Lemmas
In order to prove our main results we need several lemmas, which we present in this section.
Lemma 2.1
(see [19])
Let \(\mathcal{L}(x,a)\) be defined on \((0, \infty)\times (1/15, \infty)\) by (1.1). Then the following statements are true:
-
(i)
the functions \(a\mapsto\partial\mathcal{L}(x,a)/\partial x\) is strictly decreasing, \(a\mapsto\partial^{2}\mathcal{L}(x,a)/\partial x^{2}\) is strictly increasing and \(a\mapsto\partial^{3}\mathcal{L}(x,a)/\partial x^{3}\) is strictly decreasing on \((1/15,\infty)\);
-
(ii)
the function \(a \mapsto\mathcal{L}_{xx}(x,a)-\mathcal {L}_{xx}(y,a)\) is strictly decreasing on \((1/15,\infty)\) if \(x>y>0\);
-
(iii)
the inequalities \(\psi^{\prime}(x+1)-\mathcal{L}_{x}[x, (40+3\sqrt{205})/105]>0\), \(\psi^{\prime\prime}(x+1)-\mathcal{L}_{xx}[x, (40+3\sqrt{205})/105]<0\), and \(\psi^{\prime\prime\prime}(x+1)-\mathcal {L}_{xxx}[x, (40+3\sqrt{205})/105]>0\) hold for all \(x>0\).
Lemma 2.2
(see [20])
The identity
holds for all \(x>0\) and \(n\in\mathbb{N}\).
Lemma 2.3
(see [21])
Let \(\lambda\in\mathbb{R}\) and f be a real-valued function defined on the interval \(I=(\lambda,\infty)\) with \(\lim_{x\rightarrow\infty}f(x)=0\). Then \(f(x)<0\) if \(f(x+1)-f(x)>0\) for all \(x\in I\), and \(f(x)>0\) if \(f(x+1)-f(x)<0\) for all \(x\in I\).
3 Main results
Theorem 3.1
Let \(\mathcal{L}(x,a)\) be defined on \((0, \infty)\times(1/15, \infty)\) by (1.1) and \(F_{a}(x)=\psi(x+1)-\mathcal{L}(x,a)\). Then the following statements are true:
-
(i)
\(F_{a}(x)\) is strictly increasing with respect to x on \((0, \infty)\) if and only if \(a\geq a_{1}=(40+3\sqrt{205})/105=0.7900\ldots\) ;
-
(ii)
\(F_{a}(x)\) is strictly decreasing with respect to x on \((0, \infty)\) if and only if \(a\leq a_{2}=(45-4\pi^{2}+3\sqrt{4\pi^{4}-80\pi ^{2}+405})/[30(\pi^{2}-9)]=0.4705\ldots\) .
Proof
(i) If \(F_{a}(x)\) is strictly increasing with respect to x on \((0, \infty)\), then \(\lim_{x\rightarrow\infty} [x^{7}F^{\prime}_{a}(x) ]\geq0\). Making use of L’Hôspital’s rule and (1.5) we get
Therefore, \(a\geq a_{1}=(40+3\sqrt{205})/105\) follows easily from (3.1) and \(a\in(1/15, \infty)\).
If \(a\geq a_{1}=(40+3\sqrt{205})/105\), then Lemma 2.1(i) and (iii) lead to
for all \(x\in(0, \infty)\). Therefore, \(F_{a}(x)\) is strictly increasing with respect to x on \((0, \infty)\).
(ii) If \(F_{a}(x)\) is strictly decreasing with respect to x on \((0, \infty)\), then
It follows from (1.2) and \(\psi^{\prime}(1)=\pi^{2}/6\) that
Therefore, \(a\leq a_{2}=(45-4\pi^{2}+3\sqrt{4\pi^{4}-80\pi ^{2}+405})/[30(\pi^{2}-9)]\) follows from (3.2) and (3.3) together with \(a\in(1/15, \infty)\).
Next, we prove that \(F_{a}(x)\) is strictly decreasing with respect to x on \((0, \infty)\) if \(a\leq a_{2}\). From Lemma 2.1(i) we clearly see that it is enough to prove that \(F^{\prime}_{a_{2}}(x)<\psi^{\prime }(x+1)-\mathcal{L}_{x}(x, a_{2})<0\) for all \(x\in(0, \infty)\).
Let \(x>0\) and \(a>1/15\). Then it follows from (1.2), (1.6), and Lemma 2.2 that
where
and
We divide the proof into two cases.
Case 1. \(x\in(1/20, \infty)\). Then from (3.6) and (3.8) together with \(a_{2}<48/100\) we get
Equation (3.5) and inequalities (3.7) and (3.9) lead to
Therefore, \(F_{a_{2}}^{\prime}(x)<0\) follows from Lemma 2.3 and (3.4) together with (3.10).
Case 2. \(x\in(0, 1/20]\). Then Lemma 2.1(i) and \(a_{2}>9/20\) lead to
It follows from (1.3) and Lemma 2.1(iii) together with (3.11) that
where
and
From (3.14) we clearly see that
for \(x\in(0, 1/20]\) since \(P(x)\) is strictly increasing on \((0, 1/20]\) and
Equation (3.12) and inequalities (3.13) and (3.15) lead to the conclusion that \(F_{a_{2}}^{\prime}(x)\) is strictly decreasing on \((0, 1/20]\). Therefore, \(F_{a_{2}}^{\prime}(x)< F_{a_{2}}^{\prime}(0)=0\). □
Theorem 3.2
Let \(\mathcal{L}(x,a)\) be defined on \((0, \infty)\times(1/15, \infty)\) by (1.1), \(F_{a}(x)=\psi(x+1)-\mathcal{L}(x,a)\) and \(a_{3}=0.4321\ldots\) is the unique solution of the equation \(\mathcal{L}_{xx}(0,a)=\psi^{\prime\prime}(1)\). Then the following statements are true:
-
(i)
\(F^{\prime}_{a}(x)\) is strictly decreasing with respect to x on \((0, \infty)\) if and only if \(a\geq a_{1}=(40+3\sqrt {205})/105=0.7900\ldots\) ;
-
(ii)
\(F^{\prime}_{a}(x)\) is strictly increasing with respect to x on \((0, \infty)\) if and only if \(a\leq a_{3}\).
Proof
(i) If \(F^{\prime}_{a}(x)\) is strictly decreasing with respect to x on \((0, \infty)\), then \(\lim_{x\rightarrow\infty} [x^{8}F^{\prime\prime}_{a}(x) ]\leq0\). Making use of L’Hôspital’s rule and (1.5) we get
Therefore, \(a\geq a_{1}=(40+3\sqrt{205})/105\) follows easily from (3.16) and \(a\in(1/15, \infty)\).
If \(a\geq a_{1}=(40+3\sqrt{205})/105\), then
follows easily from Lemma 2.1(i) and (iii).
(ii) If \(F^{\prime}_{a}(x)\) is strictly increasing with respect to x on \((0, \infty)\), then
It follows from Lemma 2.1(i) that the function \(a\mapsto F^{\prime\prime}_{a}(0)\) is strictly decreasing on \((1/15, \infty)\). Note that
Therefore, \(a\leq a_{3}\) follows from (3.17)-(3.20) and the monotonicity of the function \(a\mapsto F^{\prime\prime}_{a}(0)\).
If \(a\leq a_{3}\), then we only need to prove that \(F^{\prime\prime }_{a_{3}}(x)>0\) for all \(x\in(0, \infty)\) by Lemma 2.1(i).
We divide the proof into two cases.
Case 1. \(x\in(3/50, \infty)\). Then from (1.3), (1.6), Lemma 2.1(ii), Lemma 2.2 and \(a_{3}<9/20\) we get
where
and
We clearly see that
for \(x\in(3/50, \infty)\) since \(r(x)\) is strictly increasing on \((3/50, \infty)\) and
Therefore, \(F^{\prime\prime}_{a_{3}}(x)>0\) for \(x\in(3/50, \infty)\) follows from (3.21)-(3.24) and Lemma 2.3.
Case 2. \(x\in(0, 3/50]\). Then from \(F^{\prime\prime }_{a_{3}}(0)=\psi^{\prime\prime}(1)-\mathcal{L}_{xx}(0, a_{3})=0\) we know that it is enough to prove that \(F^{\prime\prime\prime}_{a_{3}}(x)>0\).
It follows from (1.4) and Lemma 2.1(i) and (iii) together with \(a_{3}>21/50\) that
where
and
It follows from
and
that
Therefore, \(F^{\prime\prime\prime}_{a_{3}}(x)>0\) follows from (3.25)-(3.27). □
Let \(a_{1}=(40+3\sqrt{205})/105\), \(a_{2}=(45-4\pi^{2}+3\sqrt{4\pi ^{4}-80\pi^{2}+405})/[30(\pi^{2}-9)]\), and \(a_{3}=0.4321\ldots\) be the unique solution of the equation \(\mathcal{L}_{xx}(0,a)=\psi^{\prime \prime}(1)\). Then (1.2) and (1.3) lead to
From Lemma 2.1(i), Theorems 3.1 and 3.2, (3.28)-(3.31), and \(a_{3}>1/3\) we have the following.
Corollary 3.3
-
(i)
The double inequalities
$$ \mathcal{L}_{x}(x,a_{1})< \psi^{\prime}(x+1)< \mathcal{L}_{x}(x,a_{2}) $$and
$$ \mathcal{L}_{xx}(x,a_{3})< \psi^{\prime\prime}(x+1)< \mathcal{L}_{xx}(x,a_{1}) $$hold for all \(x>0\) with the best possible constants \(a_{1}\), \(a_{2}\), and \(a_{3}\).
-
(ii)
The double inequalities
$$\begin{aligned}& \biggl(x+\frac{1}{2} \biggr)\frac{x+x^{2}+\frac {23}{21}}{x^{4}+2x^{3}+\frac{17x^{2}}{7}+\frac{10x}{7}+\frac {12}{35}} \\& \quad < \psi^{\prime}(x+1) \\& \quad < \biggl(x+\frac{1}{2} \biggr) \frac{x^{2}+x+\frac{\pi^{2}}{15(\pi ^{2}-9)}}{x^{4}+2x^{3}+\frac{7\pi^{2}-60}{5(\pi^{2}-9)}x^{2}+\frac{2\pi ^{2}-15}{5(\pi^{2} -9)}x+\frac{1}{5(\pi^{2}-9)}} \\& \qquad {}-\frac{9}{2}\frac{450x^{6}+1\text{,}350x^{5}+1\text{,}965x^{4}+1\text{,}680x^{3}+897x^{2}+282x+38}{ (3x^{2}+3x+2 )^{2} (15x^{2}+15x+4 )^{2}} \\& \quad < \psi ^{\prime\prime}(x+1) \\& \quad < -\frac{5}{6}\frac{ (1\text{,}470x^{6}+4\text{,}410x^{5}+7\text{,}875x^{4}+8\text{,}400x^{3}+5\text{,}863x^{2}+2\text{,}398x+346 ) }{ ( 35x^{4}+70x^{3}+85x^{2}+50x+12 )^{2}} \end{aligned}$$hold for all \(x>0\).
Let \(a\in (\frac{1}{15}, \frac{45-4\pi^{2}+3\sqrt{4\pi^{4}-80\pi ^{2}+405}}{30(\pi^{2}-9)} ]\) and \(\gamma=0.577215\ldots\) be the Euler-Mascheroni constant. Then from Lemma 2.1(ii) and the fact that \(F_{a}(0)=-\gamma-\mathcal{L}(0, a)\) and \(\lim_{x\rightarrow\infty }F_{a}(x)=0\) we get Corollary 3.4 immediately.
Corollary 3.4
The double inequality
holds for all \(x>0\) and \(a\in (\frac{1}{15}, \frac{45-4\pi ^{2}+3\sqrt{4\pi^{4}-80\pi^{2}+405}}{30(\pi^{2}-9)} ]\) with the best possible constant \(-\gamma-\mathcal{L}(0,a)\).
In particular, taking \(a=1/3, 4/15, \sqrt{5}/15, 1/15\) and using (1.1) one has
and
for \(x>0\).
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Acknowledgements
The authors wish to thank the anonymous referees for their careful reading of the manuscript and their fruitful comments and suggestions. The research was supported by the Natural Science Foundation of China under Grants 11301127, 61374086 and 11171307, and the Natural Science Foundation of Zhejiang Province under Grant LY13A010004.
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Zhao, TH., Yang, ZH. & Chu, YM. Monotonicity properties of a function involving the psi function with applications. J Inequal Appl 2015, 193 (2015). https://doi.org/10.1186/s13660-015-0724-2
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DOI: https://doi.org/10.1186/s13660-015-0724-2