Monotonicity properties of a function involving the psi function with applications

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.

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

$$ \Gamma(x)=\int_{0}^{+\infty}t^{x-1}e^{-t} \, dt,\qquad \psi(x)=\frac{\Gamma '(x)}{\Gamma(x)}, $$

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

$$\begin{aligned} \mathcal{L}(x,a) =&\frac{1}{90a^{2}+2}\log \biggl(x^{2}+x+ \frac {3a+1}{3} \biggr) \\ &{}+\frac{45a^{2}}{90a^{2}+2}\log \biggl(x^{2}+x+\frac{15a-1}{45a} \biggr) \end{aligned}$$

(1.1)

and proved that the double inequality

$$ \mathcal{L}(x, a)< \psi(x+1)< \mathcal{L}(x-1, b)+\frac{1}{x} $$

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

$$\begin{aligned}& \mathcal{L}_{x}(x, a)=\frac{1}{90a^{2}+2}\frac{2x+1}{x^{2}+x+a+\frac {1}{3}}+ \frac{45a^{2}}{90a^{2}+2}\frac{2x+1}{ x^{2}+x+\frac{15a-1}{45a}}, \end{aligned}$$

(1.2)

$$\begin{aligned}& \mathcal{L}_{xx}(x,a)=-\frac{1}{45a^{2}+1}\frac{x^{2}+x-a+\frac {1}{6}}{ (x^{2}+x+a+\frac{1}{3} )^{2}} - \frac{45a^{2}}{45a^{2}+1}\frac{x^{2}+x+\frac{1}{45a}+\frac{1}{6}}{ (x^{2}+x+\frac{15a-1}{45a} )^{2}}, \end{aligned}$$

(1.3)

$$\begin{aligned}& \mathcal{L}_{xxx}(x,a)=\frac{1}{45a^{2}+1}\frac{(2x+1) (x^{2}+x-3a )}{ (x^{2}+x+a+\frac{1}{3} )^{3}} + \frac{45a^{2}}{45a^{2}+1}\frac{(2x+1) (x^{2}+x+\frac{1}{15a} )}{ (x^{2}+x+\frac{15a-1}{45a} )^{3}}. \end{aligned}$$

(1.4)

It is not difficult to verify that

$$ \lim_{x\rightarrow\infty}\frac{\psi( x+1)-\mathcal{L}(x,a)}{x^{-6}} =-\frac{ (a-\frac{40+3\sqrt{205}}{105} ) (a-\frac{40-3\sqrt{205}}{105} )}{85\text{,}050a} $$

(1.5)

by use of the L’Hôspital’s rule and the formula

$$ \psi^{\prime}(x)\sim\frac{1}{x}+\frac{1}{2x^{2}}+ \frac{1}{6x^{3}}-\frac {1}{30x^{5}}+\frac{1}{42x^{7}}-\frac{1}{30x^{9}}+\cdots \quad (x\rightarrow \infty) $$

(1.6)

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

$$ \psi^{n}(x+1)-\psi^{n}(x)=\frac{(-1)^{n}n!}{x^{n+1}} $$

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

$$\begin{aligned} \lim_{x\rightarrow\infty}\frac{\psi( x+1)-\mathcal {L}(x,a)}{x^{-6}} =&-;\frac{1}{6}\lim _{x\rightarrow\infty} \bigl[x^{7}F^{\prime}_{a}(x) \bigr] \\ =&-;\frac{ (a-\frac{40+3\sqrt{205}}{105} ) (a-\frac{40-3\sqrt{205}}{105} )}{85\text{,}050a}\leq0. \end{aligned}$$

(3.1)

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

$$ F^{\prime}_{a}(x)=\psi^{\prime}(x+1)-\mathcal{L}_{x}(x,a) \geq\psi ^{\prime}(x+1)-\mathcal{L}_{x}(x,a_{1})>0 $$

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

$$ F^{\prime}_{a}(0)=\psi^{\prime}(1)- \mathcal{L}_{x}(0,a)\leq0. $$

(3.2)

It follows from (1.2) and $\psi^{\prime}(1)=\pi^{2}/6$ that

$$ \psi^{\prime}(1)-\mathcal{L}_{x}(0,a)= \frac{45(\pi^{2}-9)a^{2}-3(45-4\pi ^{2})a-\pi^{2}+9}{6(3a+1)(15a-1)}. $$

(3.3)

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

$$\begin{aligned}& \lim_{x\rightarrow\infty}F^{\prime}_{a}(x)=\lim _{x\rightarrow\infty }\bigl[\psi^{\prime}(x+1)-\mathcal{L}_{x}(x, a)\bigr]=0, \end{aligned}$$

(3.4)

$$\begin{aligned}& F_{a}^{\prime}(x+1)-F_{a}^{\prime}(x) \\& \quad =\psi^{\prime}(x+2)-\psi^{\prime}(x+1)-\mathcal{L}_{x}(x+1,a)+ \mathcal {L}_{x}(x,a) \\& \quad =-\frac{2( x+1)+1}{ (90a^{2}+2 ) [(x+1)^{2}+(x+1) +\frac{3a+1}{3} ]} \\& \qquad {}-\frac{45a^{2}[2(x+1) +1]}{ (90a^{2}+2 ) [(x+1)^{2}+(x+1) +\frac{15a-1}{45a} ]} \\& \qquad {}+\frac{2x+1}{ (90a^{2}+2 ) (x^{2}+x+\frac{3a+1}{3} )}+\frac{45a^{2}(2x+1)}{ (90a^{2}+2 ) (x^{2}+x+\frac{15a-1}{45a} )}-\frac{1}{(x+1)^{2}} \\& \quad =\frac{q(x,a)}{p(x,a)}, \end{aligned}$$

(3.5)

where

$$\begin{aligned}& q(x,a)=\frac{7 (a+\frac{3\sqrt{205}-40}{105} )(\frac{40+3\sqrt {205}}{105}-a)}{45a}(x+1)^{2} -\frac{ (a+\frac{1}{3} )^{2} (a-\frac{1}{15} )^{2}}{9a^{2}}, \end{aligned}$$

(3.6)

$$\begin{aligned}& p(x,a)=(x+1)^{2} \biggl(x^{2}+3x+a+\frac{7}{3} \biggr) \biggl(x^{2}+x+a+\frac{1}{3} \biggr) \\& \hphantom{p(x,a)=}{}\times \biggl(x^{2}+x+\frac{1}{3}- \frac{1}{45a} \biggr) \biggl(x^{2}+3x+\frac{7}{3}- \frac{1}{45a} \biggr)>0 \end{aligned}$$

(3.7)

and

$$ \frac{\partial q(x,a)}{\partial a}=-\frac {7(45a^{2}+1)}{2\text{,}025a^{2}}(x+1)^{2}- \frac{2 (a+\frac{1}{3} ) (a-\frac{1}{15} )(2\text{,}025a^{2}+45)}{18\text{,}225a^{3}}< 0. $$

(3.8)

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

$$ q(x, a_{2})>q \biggl(x, \frac{48}{100} \biggr)>q \biggl(\frac{1}{20}, \frac {48}{100} \biggr)=\frac{2\text{,}341\text{,}501}{1\text{,}312\text{,}200\text{,}000}>0. $$

(3.9)

Equation (3.5) and inequalities (3.7) and (3.9) lead to

$$ F_{a_{2}}^{\prime}(x+1)-F_{a_{2}}^{\prime}(x)>0. $$

(3.10)

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

$$ \mathcal{L}_{xx}(x, a_{2})> \mathcal{L}_{xx} \biggl(x, \frac{9}{20} \biggr). $$

(3.11)

It follows from (1.3) and Lemma 2.1(iii) together with (3.11) that

$$ F_{a_{2}}^{\prime\prime}(x)=\psi^{\prime\prime}(x+1)- \mathcal {L}_{xx}(x, a_{2})< \mathcal{L}_{xx}(x, a_{1})-\mathcal{L}_{xx}(x, a_{2})= \frac{P(x)}{6Q(x)}, $$

(3.12)

where

$$\begin{aligned} Q(x) =& \bigl(60x^{2}+60x+47 \bigr)^{2} \bigl(81x^{2}+81x+23 \bigr)^{2} \\ &{}\times \bigl(570x+505x^{2}+210x^{3}+35x^{4}+252 \bigr)^{2}(x+1)^{3}>0 \end{aligned}$$

(3.13)

and

$$\begin{aligned} P(x) =&9;\text{,}756\text{,}595\text{,}800x^{11}+146 \text{,}348\text{,}937\text{,}000x^{10}+1\text{,}005\text{,}597 \text{,}383\text{,}250x^{9} \\ &{}+3\text{,}954\text{,}619\text{,}691\text{,}700x^{8}+9\text{,}800 \text{,}346\text{,}642\text{,}855x^{7}+16\text{,}058\text{,}808 \text{,}560\text{,}085x^{6} \\ &{}+17\text{,}731\text{,}092\text{,}059\text{,}926x^{5}+13 \text{,}107\text{,}900\text{,}251\text{,}862x^{4}+6\text{,}210 \text{,}045\text{,}031\text{,}977x^{3} \\ &{}+1\text{,}655\text{,}666\text{,}210\text{,}995x^{2}+153\text{,}061 \text{,}816\text{,}584x-15\text{,}463\text{,}394\text{,}658. \end{aligned}$$

(3.14)

From (3.14) we clearly see that

$$ P(x)< 0 $$

(3.15)

for $x\in(0, 1/20]$ since $P(x)$ is strictly increasing on $(0, 1/20]$ and

$$ P \biggl(\frac{1}{20} \biggr)=-\frac{2\text{,}874\text{,}530\text{,}403\text{,}954\text{,}909\text{,}124\text{,}821}{1\text{,}024\text{,}000\text{,}000\text{,}000}< 0. $$

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

$$\begin{aligned} \lim_{x\rightarrow\infty}\frac{\psi( x+1)-\mathcal {L}(x,a)}{x^{-6}} =& \frac{1}{42}\lim_{x\rightarrow\infty} \bigl[x^{8}F^{\prime\prime}_{a}(x) \bigr] \\ =&-;\frac{ (a-\frac{40+3\sqrt{205}}{105} ) (a-\frac{40-3\sqrt{205}}{105} )}{85\text{,}050a}\leq0. \end{aligned}$$

(3.16)

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

$$ F^{\prime\prime}_{a}(x)=\psi^{\prime\prime}(x+1)-\mathcal {L}_{xx}(x,a)\leq\psi^{\prime\prime}(x+1)-\mathcal{L}_{xx}(x,a_{1})< 0 $$

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

$$ F^{\prime\prime}_{a}(0)=\psi^{\prime\prime}(1)-\mathcal {L}_{xx}(0,a)\geq0. $$

(3.17)

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

$$\begin{aligned}& F^{\prime\prime}_{a_{3}}(0)=0,\quad a_{3}\in(1/3, 1/2), \end{aligned}$$

(3.18)

$$\begin{aligned}& F^{\prime\prime}_{1/3}(0)=\frac{171}{64}-2\sum _{n=1}^{\infty}\frac {1}{n^{3}}=0.2677 \ldots>0, \end{aligned}$$

(3.19)

$$\begin{aligned}& F^{\prime\prime}_{1/2}(0)=\frac{19\text{,}299}{8\text{,}450}-2\sum _{n=1}^{\infty}\frac {1}{n^{3}}=-0.1202 \ldots< 0. \end{aligned}$$

(3.20)

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

$$\begin{aligned}& \lim_{x\rightarrow\infty}F^{\prime\prime}_{a_{3}}(x)=\lim _{x\rightarrow\infty}\bigl[\psi^{\prime\prime}(x+1)-\mathcal{L}_{xx}(x,a_{3}) \bigr]=0, \end{aligned}$$

(3.21)

$$\begin{aligned}& F^{\prime\prime}_{a_{3}}(x+1)-F^{\prime\prime}_{a_{3}}(x) \\& \quad =\psi^{\prime\prime}(x+2)-\psi^{\prime\prime}(x+1)-\bigl[\mathcal {L}_{xx}(x+1, a_{3})-\mathcal{L}_{xx}(x, a_{3})\bigr] \\& \quad < \frac{2}{(x+1)^{3}}- \biggl[\mathcal{L}_{xx} \biggl(x+1, \frac{9}{20} \biggr)-\mathcal{L}_{xx} \biggl(x, \frac{9}{20} \biggr) \biggr]=-\frac{2r(x)}{s(x)}, \end{aligned}$$

(3.22)

where

$$\begin{aligned} s(x) =& \bigl(60x+60x^{2}+47 \bigr)^{2} \bigl(81x+81x^{2}+23 \bigr)^{2} \\ &{}\times \bigl(180x+60x^{2}+167 \bigr)^{2} \bigl(243x+81x^{2}+185 \bigr)^{2}(x+1)^{3}>0 \end{aligned}$$

(3.23)

and

$$\begin{aligned} r(x) =&125;\text{,}413\text{,}273\text{,}555\text{,}200x^{10}+1 \text{,}254\text{,}132\text{,}735\text{,}552\text{,}000x^{9} \\ &{}+5\text{,}518\text{,}250\text{,}043\text{,}762\text{,}960x^{8}+14 \text{,}046\text{,}814\text{,}696\text{,}855\text{,}680x^{7} \\ &{}+22 \text{,}840\text{,}386\text{,}490\text{,}946\text{,}664x^{6}+24\text{,}664\text{,}633\text{,}018\text{,}794\text{,}864x^{5} \\ &{}+17 \text{,}718\text{,}225\text{,}566\text{,}437\text{,}953x^{4}+8 \text{,}120\text{,}232\text{,}997\text{,}769\text{,}412x^{3} \\ &{}+2\text{,}081\text{,}281\text{,}129\text{,}927\text{,}908x^{2}+179 \text{,}154\text{,}971\text{,}702\text{,}976x \\ &{}-19\text{,}953\text{,}618\text{,}766 \text{,}474. \end{aligned}$$

We clearly see that

$$ r(x)>0 $$

(3.24)

for $x\in(3/50, \infty)$ since $r(x)$ is strictly increasing on $(3/50, \infty)$ and

$$ r \biggl(\frac{3}{50} \biggr)=\frac{1\text{,}114\text{,}560\text{,}148\text{,}894\text{,}087\text{,}067\text{,}992\text{,}508}{3\text{,}814\text{,}697\text{,}265\text{,}625}>0. $$

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

$$\begin{aligned} \begin{aligned}[b] F^{\prime\prime\prime}_{a_{3}}(x)&=\psi^{\prime\prime\prime }(x+1)- \mathcal{L}_{xxx}(x, a_{3}) \\ &>\mathcal{L}_{xxx}(x, a_{1})-\mathcal{L}_{xxx} \biggl(x, \frac {21}{50} \biggr)=-\frac{R(x)}{3S(x)}, \end{aligned} \end{aligned}$$

(3.25)

where

$$\begin{aligned} S(x) =&(x+1)^{4} \bigl(150x^{2}+150x+113 \bigr)^{3} \bigl(189x^{2}+189x+53 \bigr)^{3} \\ &{}\times \bigl(35x^{4}+210x^{3}+505x^{2}+570x+252 \bigr)^{3}>0 \end{aligned}$$

(3.26)

and

$$\begin{aligned} R(x) =&1;\text{,}439\text{,}970\text{,}288\text{,}529\text{,}500 \text{,}000x^{19} \\ &{}+33\text{,}839\text{,}301\text{,}780\text{,}443 \text{,}250\text{,}000x^{18} \\ &{}+377\text{,}685\text{,}219\text{,}317\text{,}959\text{,}507 \text{,}500x^{17} \\ &{}+2\text{,}619\text{,}038\text{,}198\text{,}507 \text{,}995\text{,}293\text{,}750x^{16} \\ &{}+12\text{,}578\text{,}516\text{,}662\text{,}166\text{,}748\text{,}200 \text{,}250x^{15} \\ &{}+44\text{,}394\text{,}499\text{,}254\text{,}715 \text{,}419\text{,}844\text{,}125x^{14} \\ &{}+119\text{,}436\text{,}801\text{,}689\text{,}614\text{,}664\text{,}479 \text{,}875x^{13} \\ &{}+250\text{,}817\text{,}342\text{,}412\text{,}016 \text{,}626\text{,}059\text{,}625x^{12} \\ &{}+417\text{,}457\text{,}335\text{,}039\text{,}758\text{,}233\text{,}395 \text{,}000x^{11} \\ &{}+555\text{,}642\text{,}395\text{,}442\text{,}917 \text{,}892\text{,}895\text{,}800x^{10} \\ &{}+593\text{,}602\text{,}907\text{,}219\text{,}352\text{,}981\text{,}396 \text{,}390x^{9} \\ &{}+508\text{,}233\text{,}654\text{,}389\text{,}427 \text{,}279\text{,}197\text{,}745x^{8} \\ &{}+346\text{,}198\text{,}219\text{,}129\text{,}731\text{,}218\text{,}829 \text{,}124x^{7} \\ &{}+184\text{,}849\text{,}155\text{,}080\text{,}550 \text{,}188\text{,}733\text{,}310x^{6} \\ &{}+75\text{,}353\text{,}569\text{,}007\text{,}634\text{,}565\text{,}613 \text{,}769x^{5} \\ &{}+22\text{,}380\text{,}430\text{,}314\text{,}381 \text{,}942\text{,}509\text{,}812x^{4} \\ &{}+4\text{,}414\text{,}609\text{,}088\text{,}286\text{,}249\text{,}144 \text{,}994x^{3} \\ &{}+450\text{,}421\text{,}073\text{,}304\text{,}504 \text{,}390\text{,}873x^{2} \\ &{}-4\text{,}721\text{,}565\text{,}008\text{,}851\text{,}422\text{,}102x \\ &{}-4 \text{,}420\text{,}688\text{,}040\text{,}144\text{,}642\text{,}816. \end{aligned}$$

It follows from

$$ R^{\prime\prime}(x)>0,\qquad R(0)< 0, $$

and

$$ R \biggl(\frac{3}{50} \biggr)=-\frac {337\text{,}711\text{,}343\text{,}455\text{,}989\text{,}855\text{,}048\text{,}292\text{,}675\text{,}691\text{,}209\text{,}992\text{,}531\text{,}618\text{,}111}{190\text{,}734\text{,}863\text{,}281\text{,}250\text{,}000\text{,}000\text{,}000\text{,}000}< 0 $$

that

$$ R(x)< \frac{3/50-x}{3/50}R(0)+\frac{x}{3/50}R(3/50)< 0. $$

(3.27)

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

$$\begin{aligned}& \mathcal{L}_{x}(x,a_{1})= \biggl( x+\frac{1}{2} \biggr)\frac {x+x^{2}+23/21}{x^{4}+2x^{3}+17x^{2}/7+10x/7+12/35}, \end{aligned}$$

(3.28)

$$\begin{aligned}& \mathcal{L}_{x}(x,a_{2})= \biggl(x+\frac{1}{2} \biggr)\frac{x^{2}+x+\frac {\pi^{2}}{15 (\pi^{2}-9 )}}{x^{4}+2x^{3} +\frac{7\pi^{2}-60}{5 (\pi^{2}-9 )}x^{2}+\frac{2\pi ^{2}-15}{5 (\pi^{2}-9 )}x+\frac{1}{5 (\pi^{2}-9 )}}, \end{aligned}$$

(3.29)

$$\begin{aligned}& \mathcal{L}_{xx}(x,a_{1}) \\& \quad =-\frac{5}{6} \frac{ (1\text{,}470x^{6}+4\text{,}410x^{5}+7\text{,}875x^{4}+8\text{,}400x^{3}+5\text{,}863x^{2}+2\text{,}398x+346 ) }{ ( 35x^{4}+70x^{3}+85x^{2}+50x+12 )^{2}}, \end{aligned}$$

(3.30)

$$\begin{aligned}& \mathcal{L}_{xx} \biggl(x,\frac{1}{3} \biggr)=- \frac{9}{2}\frac {450x^{6}+1\text{,}350x^{5}+1\text{,}965x^{4}+1\text{,}680x^{3}+897x^{2}+282x+38}{ (3x^{2}+3x+2 )^{2} (15x^{2}+15x+4 )^{2}}. \end{aligned}$$

(3.31)

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

$$ \mathcal{L}(x,a)< \psi(x+1)< \mathcal{L}(x,a)-\gamma-\mathcal{L}(0,a) $$

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

$$\begin{aligned}& \frac{1}{12}\log \biggl(x^{2}+x+\frac{2}{3} \biggr)+ \frac{5}{12}\log \biggl(x^{2}+x+\frac{4}{15} \biggr) \\& \quad < \psi(x+1)< \frac{1}{12}\log \biggl(x^{2}+x+ \frac{2}{3} \biggr)+\frac{5}{12}\log \biggl(x^{2}+x+ \frac{4}{15} \biggr) +\frac{1}{12}\log\frac{3}{2}+ \frac{5}{12}\log\frac{15}{4}-\gamma, \\& \frac{16}{21}\log \biggl(x+\frac{1}{2} \biggr)+\frac{5}{42} \log \biggl(x^{2}+x+\frac{3}{5} \biggr) \\& \quad < \psi(x+1)< \frac{16}{21}\log \biggl(x+\frac{1}{2} \biggr)+ \frac{5}{42}\log \biggl(x^{2}+x+\frac{3}{5} \biggr)+ \frac{16}{21}\log2+\frac{5}{42}\log\frac {5}{3}-\gamma, \\& \frac{1}{4}\log \biggl[ \biggl(x^{2}+x+\frac{1}{3} \biggr)^{2}-\frac {1}{45} \biggr]< \psi(x+1) < \frac{1}{4}\log \biggl[ \biggl(x^{2}+x+\frac{1}{3} \biggr)^{2}- \frac {1}{45} \biggr]+\frac{1}{4}\log\frac{45}{4}-\gamma \end{aligned}$$

and

$$ \psi(x+1)>\frac{1}{12}\log\bigl(x^{2}+x\bigr)+\frac{5}{12} \log \biggl(x^{2}+x+\frac {2}{5} \biggr) $$

for $x>0$.