Some New Inequalities for the Gamma and Polygamma Functions

Abstract

In this paper, we present some new symmetric bounds for the gamma and polygamma functions. For this goal, we present two functions involving gamma and polygamma functions and we investigate their complete monotonicity. Also, we investigate their completely monotonic degrees. This concept gives more accuracy in measuring the complete monotonicity property. These new bounds are better than some of the recently published results.

1. Introduction

The gamma function was presented by Euler [1] as:

$$\Gamma \left(l\right)=\underset{n\to \infty }{lim}{\displaystyle \frac{n^l}{l}}\prod _{r=1}^n\left({\displaystyle \frac{r}{l+r}}\right),l\in {\mathbb{R}}^{+}$$

and this leads to

$${\displaystyle \frac{\Gamma (n+l)}{\Gamma \left(l\right)}}=\prod _{r=0}^{n-1}(r+l),l\in {\mathbb{R}}^{+}.$$

(1)

$ln\Gamma \left(l\right)$ has the asymptotic expansion [2]:

$$ln\Gamma \left(l\right)\sim (l-1/2)lnl+{\displaystyle \frac{1}{2}}ln\left(2\pi \right)-l+\sum _{r=1}^{\infty }{\displaystyle \frac{B_{2r}}{\left(2r\right)(2r-1)l^{2r-1}}},l⟶\infty$$

(2)

where $B_{2r}$ denotes the Bernoulli numbers. The Psi and polygamma functions are presented by [2]:

$$\psi \left(l\right)={\displaystyle \frac{-\gamma l-1}{l}}+\sum _{r=1}^{\infty }{\displaystyle \frac{l}{r(r+l)}}={\int }_0^{\infty }\left({\left(t{e}^t\right)}^{-1}-{\displaystyle \frac{e^{(-l+1)t}}{-1+e^t}}\right)dt,l\in {\mathbb{R}}^{+}$$

(3)

where the Euler–Mascheroni’s constant $\gamma \simeq 0.5772156649$, and

$${\psi }^{\left(s\right)}\left(l\right)=\sum _{r=0}^{\infty }{\displaystyle \frac{{(-1)}^{s+1}s!}{{(r+l)}^{s+1}}}={\int }_0^{\infty }\left({\displaystyle \frac{{(-1)}^{s+1}{t}^s{e}^{(1-l)t}}{-1+e^t}}\right)dt,l\in {\mathbb{R}}^{+},s\in \mathbb{N}$$

(4)

and they have the relation:

$${\psi }^{\left(s\right)}(1+l)={\displaystyle \frac{{(-1)}^{s}s!}{l^{1+s}}}+{\psi }^{\left(s\right)}\left(l\right),s=0,1,2,\cdots .$$

(5)

Also, they have the asymptotic expansions [2]:

$$\psi \left(l\right)\sim -{\displaystyle \frac{1}{2l}}+lnl-\sum _{r=1}^{\infty }{\displaystyle \frac{B_{2r}}{\left(2r\right)l^{2r}}},l⟶\infty$$

(6)

and

$${\psi }^{\left(s\right)}\left(l\right)\sim {\displaystyle \frac{{(-1)}^{s-1}s!}{2l^{s+1}}}+{\displaystyle \frac{{(-1)}^{s-1}(s-1)!}{l^s}}+{(-1)}^{s-1}\sum _{r=1}^{\infty }{\displaystyle \frac{(s+2r-1)!B_{2r}}{\left(2r\right)!l^{2r+s}}},s\in \mathbb{N},l⟶\infty .$$

(7)

For extra information about the gamma and polygamma functions, see [3,4,5,6]. A function S defined on an interval J is completely monotonic (CM) if it satisfies that

$${(-1)}^s{S}^{\left(s\right)}\left(l\right)\ge 0l\in J;s=0,1,2,....$$

The necessary and sufficient condition for the function $S\left(l\right)$ to be CM for $l\in {\mathbb{R}}^{+}$ is that [7]:

$$S\left(l\right)={\int }_0^{\infty }{e}^{-lt}d\vartheta \left(t\right),$$

where $\vartheta \left(t\right)$ is a non-negative measure such that the integral converges for $l\in {\mathbb{R}}^{+}$. Let $S\left(l\right)$ be a CM function for $l\in {\mathbb{R}}^{+}$ and suppose the notation $S(\infty )=\underset{l\to \infty }{lim}S\left(l\right)$. If $l^{\rho }[S\left(l\right)-S(\infty )]$ is a CM function for $l\in {\mathbb{R}}^{+}$ if and only if $\rho \in \left[0,\sigma \right],$ then the number $\sigma \in {\mathbb{R}}^{+}$ is called the CM degree of $S\left(l\right)$ for $l\in {\mathbb{R}}^{+}$ and denoted by ${deg}_{CM}^l\left[S\left(l\right)\right]=\sigma .$ For extra information about this topic, see [8,9,10].

In 2007, Alzer and Batir [11] introduced some CM functions involving the gamma the digamma functions and the following symmetric inequalities are deduced:

$$exp\left(-l-{\displaystyle \frac{1}{2}}\psi (l+1/3)\right)<{\displaystyle \frac{\Gamma \left(l\right)}{\sqrt{2\pi }{l}^l}}<exp\left(-l-{\displaystyle \frac{1}{2}}\psi \left(l\right)\right),l\in {\mathbb{R}}^{+}$$

(8)

$${\displaystyle \frac{1}{2}}{\psi }^{\prime }(l+1/3)<lnl-\psi \left(l\right)<{\displaystyle \frac{1}{2}}{\psi }^{\prime }\left(l\right),l\in {\mathbb{R}}^{+}$$

(9)

and for $s=2,3,\cdots ,$

$${\displaystyle \frac{{(-1)}^{s+1}}{2}}{\psi }^{\left(s\right)}(l+1/3)<{(-1)}^s{\psi }^{(s-1)}\left(l\right)-{\displaystyle \frac{(s-2)!}{l^{s-1}}}<{\displaystyle \frac{{(-1)}^{s+1}}{2}}{\psi }^{\left(s\right)}\left(l\right),l\in {\mathbb{R}}^{+}.$$

(10)

Batir [12] improved (8) by:

$$exp\left(-l-{\displaystyle \frac{1}{2}}\psi \left(l\right)-{\displaystyle \frac{1}{6(l-\frac{1}{4})}}\right)<{\displaystyle \frac{\Gamma \left(l\right)}{\sqrt{2\pi }{l}^l}}<exp\left(-l-{\displaystyle \frac{1}{2}}\psi \left(l\right)-{\displaystyle \frac{1}{6l}}\right),l\in {\mathbb{R}}^{+}.$$

(11)

Next, Şevli and Batir [13] investigated some CM functions involving the $\Gamma$ and ${\psi }^{\prime }$ functions and the following symmetric inequalities are deduced:

$${\displaystyle \frac{1}{\sqrt{l}}}exp\left(-l+{\displaystyle \frac{1}{12}}{\psi }^{\prime }(l+1/2)\right)<{\displaystyle \frac{\Gamma \left(l\right)}{\sqrt{2\pi }{l}^l}}<{\displaystyle \frac{1}{\sqrt{l}}}exp\left(-l+{\displaystyle \frac{1}{12}}{\psi }^{\prime }\left(l\right)\right),l\in {\mathbb{R}}^{+}$$

(12)

$${\displaystyle \frac{1}{2l}}-{\displaystyle \frac{1}{12}}{\psi }^{″}(l+1/2)<lnl-\psi \left(l\right)<{\displaystyle \frac{1}{2l}}-{\displaystyle \frac{1}{12}}{\psi }^{″}\left(l\right),l\in {\mathbb{R}}^{+}$$

(13)

and for $s=2,3,\cdots$ and $l\in {\mathbb{R}}^{+},$

$${\displaystyle \frac{(s-1)!}{2l^s}}+{\displaystyle \frac{{(-1)}^s}{12}}{\psi }^{(s+1)}(l+1/2)<{(-1)}^s{\psi }^{(s-1)}\left(l\right)-{\displaystyle \frac{(s-2)!}{l^{s-1}}}<{\displaystyle \frac{{(-1)}^s}{12}}{\psi }^{(s+1)}\left(l\right)+{\displaystyle \frac{(s-1)!}{2l^s}}.$$

(14)

In 2024, Moustafa [14] refined the upper bounds of (8) and (11) for all $l\in {\mathbb{R}}^{+}$ and $l>{\displaystyle \frac{1}{8}}$ consecutively by:

$$exp\left(-l-{\displaystyle \frac{1}{2}}\psi \left(l\right)-{\displaystyle \frac{1}{6}}{\psi }^{\prime }\left(l\right)\right)<{\displaystyle \frac{\Gamma \left(l\right)}{\sqrt{2\pi }{l}^l}}<exp\left(-l-{\displaystyle \frac{1}{2}}\psi \left(l\right)-{\displaystyle \frac{1}{6}}{\psi }^{\prime }(l+1/4)\right),l\in {\mathbb{R}}^{+}$$

(15)

and he improved the upper bounds of (9) and (13) for all $l\in {\mathbb{R}}^{+}$ by:

$${\displaystyle \frac{1}{2}}{\psi }^{\prime }\left(l\right)+{\displaystyle \frac{1}{6}}{\psi }^{″}\left(l\right)<lnl-\psi \left(l\right)<{\displaystyle \frac{1}{2}}{\psi }^{\prime }\left(l\right)+{\displaystyle \frac{1}{6}}{\psi }^{″}(l+1/4),l\in {\mathbb{R}}^{+}$$

(16)

and for $l\in {\mathbb{R}}^{+}$ and $s=2,3,\cdots ,$ he improved the upper bounds of (10) and (14) by:

$${\displaystyle \frac{{(-1)}^{s+1}}{2}}{\psi }^{\left(s\right)}\left(l\right)+{\displaystyle \frac{{(-1)}^{s+1}}{6}}{\psi }^{(s+1)}\left(l\right)<{(-1)}^s{\psi }^{(s-1)}\left(l\right)-{\displaystyle \frac{(s-2)!}{l^{s-1}}}$$

$$<{\displaystyle \frac{{(-1)}^{s+1}}{2}}{\psi }^{\left(s\right)}\left(l\right)+{\displaystyle \frac{{(-1)}^{s+1}}{6}}{\psi }^{(s+1)}(l+1/4)$$

(17)

And finally, he improved the lower bound of (8) for all $l\ge 2.2$ by:

$$exp\left(-l-{\displaystyle \frac{1}{2}}\psi \left(l\right)-{\displaystyle \frac{1}{6}}{\psi }^{\prime }\left(l\right)+{\displaystyle \frac{1}{24l^2}}\right)<{\displaystyle \frac{\Gamma \left(l\right)}{\sqrt{2\pi }{l}^l}}<exp\left(-l-{\displaystyle \frac{1}{2}}\psi \left(l\right)-{\displaystyle \frac{1}{6}}{\psi }^{\prime }\left(l\right)+{\displaystyle \frac{1}{24{\left(l-\frac{3}{10}\right)}^2}}\right).$$

(18)

Most of the last results were generalized to the generalized gamma and polygamma functions in [15,16,17].

The researchers studying mathematics and applied sciences benefit from the new inequalities involving the gamma and polygamma functions, which gives better approximations and a better understanding of the behavior of special functions. Our paper aims to present new inequalities involving $\Gamma$ and ${\psi }^{\left(s\right)}$ ($s\in \mathbb{N}\cup \left\{0\right\}$). These new results are better than the previous inequalities presented by Alzer and Batir, Batir, Şevli and Batir and Moustafa. For this goal, we will study some CM functions containing $\Gamma ,{\psi }^{\prime }$ and ${\psi }^{″}$ and we will present some new symmetric bounds for $\Gamma$ and ${\psi }^{\left(s\right)}$ ($s\in \mathbb{N}\cup \left\{0\right\}$), which improve the results (8)–(18).

2. Auxiliary Results

In [10], (Lemma 1), the next corollary was introduced:

Corollary 1.

Suppose that the real-valued function $T\left(l\right)$ is defined on $(l_0,\infty )$ and $\underset{l\to \infty }{lim}T\left(l\right)=0$. Then, for $\upsilon \in {\mathbb{R}}^{+}$, $T\left(l\right)>0$, if $T(l+\upsilon )<T\left(l\right)$ for every $l\in (l_0,\infty )$ and $T\left(l\right)<0$, if $T(l+\upsilon )>T\left(l\right)$ for every $l\in (l_0,\infty )$.

In the following Lemma, we present some inequalities involving the digamma and polygamma functions, which will be useful in Section 4, in proving that the new bounds for $\Gamma$ and ${\psi }^s$ are better than the old ones.

Lemma 1. 

1. 

For $l\in [0.5,\infty ),$ we have

$${\psi }^{\prime }\left(l\right)>{\displaystyle \frac{1}{l}}-{\displaystyle \frac{1}{6}}{\psi }^{″}\left(l\right)-{\displaystyle \frac{1}{12}}{\psi }^{\prime \prime \prime }(l+3/10),$$

(19)

$${\psi }^{″}\left(l\right)<{\displaystyle \frac{-1}{l^2}}-{\displaystyle \frac{1}{6}}{\psi }^{\prime \prime \prime }\left(l\right)-{\displaystyle \frac{1}{12}}{\psi }^{\left(4\right)}(l+3/10).$$

(20)

$$lnl-\psi \left(l\right)>{\displaystyle \frac{1}{6}}{\psi }^{\prime }\left(l\right)+{\displaystyle \frac{1}{12}}{\psi }^{″}\left(l+3/10\right)+{\displaystyle \frac{1}{3l}},$$

(21)

and

$$lnl-\psi \left(l\right)>{\displaystyle \frac{1}{6}}{\psi }^{\prime }\left(l\right)+{\displaystyle \frac{1}{12}}{\psi }^{″}(l+3/10).$$

(22)

2. 

For all $l\ge 5.1,$ we have

$$lnl-{\displaystyle \frac{1}{6}}{\psi }^{\prime }\left(l\right)-{\displaystyle \frac{1}{12}}{\psi }^{″}\left(l\right)<\psi (l+1/3),$$

(23)

$${\displaystyle \frac{1}{l}}-{\displaystyle \frac{1}{6}}{\psi }^{″}\left(l\right)-{\displaystyle \frac{1}{12}}{\psi }^{\prime \prime \prime }\left(l\right)>{\psi }^{\prime }(l+1/3)$$

(24)

and

$${\displaystyle \frac{-1}{l^2}}-{\displaystyle \frac{1}{6}}{\psi }^{\prime \prime \prime }\left(l\right)-{\displaystyle \frac{1}{12}}{\psi }^{\left(4\right)}\left(l\right)<{\psi }^{″}(l+1/3).$$

(25)

3. 

For all $l\ge {\displaystyle \frac{13}{10}},$ we have

$$lnl-\psi \left(l\right)>{\displaystyle \frac{1}{6}}{\psi }^{\prime }\left(l\right)+{\displaystyle \frac{1}{12}}{\psi }^{″}(l+3/10)+{\displaystyle \frac{1}{3}}{\psi }^{\prime }(l+1/4),$$

(26)

$${\displaystyle \frac{1}{l}}-{\displaystyle \frac{1}{6}}{\psi }^{″}\left(l\right)-{\displaystyle \frac{1}{12}}{\psi }^{\prime \prime \prime }(l+3/10)<{\psi }^{\prime }\left(l\right)+{\displaystyle \frac{1}{3}}{\psi }^{″}(l+1/4),$$

(27)

$${\psi }^{″}\left(l\right)<{\displaystyle \frac{-1}{l^2}}-{\displaystyle \frac{1}{6}}{\psi }^{\prime \prime \prime }\left(l\right)-{\displaystyle \frac{1}{12}}{\psi }^{\left(4\right)}(l+3/10)-{\displaystyle \frac{1}{3}}{\psi }^{″}(l+1/4),$$

(28)

and

$$lnl-\psi \left(l\right)<{\displaystyle \frac{1}{2}}{\psi }^{\prime }\left(l\right)+{\displaystyle \frac{1}{12}}{\psi }^{″}\left(l\right)+{\displaystyle \frac{1}{20l^3}}-{\displaystyle \frac{1}{12l^2}}.$$

(29)

Proof. 

First, let ${\Omega }_1\left(l\right)={\displaystyle \frac{1}{l}}-{\psi }^{\prime }\left(l\right)-{\displaystyle \frac{1}{6}}{\psi }^{″}\left(l\right)-{\displaystyle \frac{1}{12}}{\psi }^{\prime \prime \prime }(l+3/10).$ Then ${\Omega }_1^{\prime }\left(l\right)={\displaystyle \frac{-1}{l^2}}-{\psi }^{″}\left(l\right)-{\displaystyle \frac{1}{6}}{\psi }^{\prime \prime \prime }\left(l\right)-{\displaystyle \frac{1}{12}}{\psi }^{\left(4\right)}(l+3/10)$ and by using (5), we obtain

$$-{\Omega }_1^{\prime }\left(l\right)+{\Omega }_1^{\prime }(1+l)={\displaystyle \frac{\frac{-h\left(l\right)}{{(1+l)}^2}}{l^4{(3+10l)}^5}}<0,l\ge {\displaystyle \frac{1}{2}}$$

where

$$h\left(l\right)=11741+225636\left(l-{\displaystyle \frac{1}{2}}\right)+1206636{\left(l-{\displaystyle \frac{1}{2}}\right)}^2+3013600{\left(l-{\displaystyle \frac{1}{2}}\right)}^3$$

$$+4054000{\left(l-{\displaystyle \frac{1}{2}}\right)}^4+3030000{\left(l-{\displaystyle \frac{1}{2}}\right)}^5+1200000{\left(l-{\displaystyle \frac{1}{2}}\right)}^6+200000{\left(l-{\displaystyle \frac{1}{2}}\right)}^7.$$

Using the asymptotic Formula (7), we obtain $\underset{l\to \infty }{lim}{\Omega }_1^{\prime }\left(l\right)=0$ and by Corollary 1, we obtain ${\Omega }_1^{\prime }\left(l\right)$ is positive for $l\ge 0.5$ and hence we obtain (20). Then, ${\Omega }_1\left(l\right)$ is increasing on $[0.5,\infty )$ with $\underset{l\to \infty }{lim}{\Omega }_1\left(l\right)=0$, and consequently, ${\Omega }_1\left(l\right)$ is negative for $l\ge 0.5$ and this proves (19). Next, we assume that ${\Omega }_2\left(l\right)=lnl-\psi \left(l\right)-{\displaystyle \frac{1}{6}}{\psi }^{\prime }\left(l\right)-{\displaystyle \frac{1}{12}}{\psi }^{″}(l+3/10)-{\displaystyle \frac{1}{3l}}$ and then ${\Omega }_2^{\prime }\left(l\right)={\displaystyle \frac{1}{l}}-{\psi }^{\prime }\left(l\right)-{\displaystyle \frac{1}{6}}{\psi }^{″}\left(l\right)-{\displaystyle \frac{1}{12}}{\psi }^{\prime \prime \prime }(l+3/10)+{\displaystyle \frac{1}{3l^2}}$ and by using (5), we have

$$-{\Omega }_2^{\prime }\left(l\right)+{\Omega }_2^{\prime }(1+l)={\displaystyle \frac{\frac{1}{4}}{3l^3{(3+10l)}^4{(1+l)}^2}}[491+41830\left(l-{\displaystyle \frac{1}{2}}\right)+191400{\left(l-{\displaystyle \frac{1}{2}}\right)}^2$$

$$+322000{\left(l-{\displaystyle \frac{1}{2}}\right)}^3+230000{\left(l-{\displaystyle \frac{1}{2}}\right)}^4+60000{\left(l-{\displaystyle \frac{1}{2}}\right)}^5]>0,l\ge {\displaystyle \frac{1}{2}}.$$

Similarly, as before, we obtain ${\Omega }_2^{\prime }\left(l\right)<0$ for all $l\ge {\displaystyle \frac{1}{2}}$ and then ${\Omega }_2\left(l\right)>0$ for $l\ge 0.5$ and this proves (21), and consequently, we have (22). Now, we assume that ${\Omega }_3\left(l\right)=\psi (l+1/3)-lnl+{\displaystyle \frac{1}{6}}{\psi }^{\prime }\left(l\right)+{\displaystyle \frac{1}{12}}{\psi }^{″}\left(l\right).$ Then ${\Omega }_3^{\prime }\left(l\right)={\psi }^{\prime }(l+1/3)-{\displaystyle \frac{1}{l}}+{\displaystyle \frac{1}{6}}{\psi }^{″}\left(l\right)+{\displaystyle \frac{1}{12}}{\psi }^{\prime \prime \prime }\left(l\right)$ and ${\Omega }_3^{″}\left(l\right)={\psi }^{″}(l+1/3)+{\displaystyle \frac{1}{l^2}}+{\displaystyle \frac{1}{6}}{\psi }^{\prime \prime \prime }\left(x\right)+{\displaystyle \frac{1}{12}}{\psi }^{\left(4\right)}\left(x\right)$ and hence

$$-{\Omega }_3^{″}\left(l\right)+{\Omega }_3^{″}(l+1)={\displaystyle \frac{\frac{-t\left(l\right)}{{(1+l)}^2}}{l^5{(1+3l)}^3}}<0,l\ge 5.1$$

where

$$25000t\left(l\right)=24458152+432369375\left(l-{\displaystyle \frac{51}{10}}\right)+306040500{\left(l-{\displaystyle \frac{51}{10}}\right)}^2+82610000{\left(l-{\displaystyle \frac{51}{10}}\right)}^3$$

$$+9950000{\left(l-{\displaystyle \frac{51}{10}}\right)}^4+450000{\left(l-{\displaystyle \frac{51}{10}}\right)}^5.$$

By the same way, we obtain ${\Omega }_3^{″}\left(l\right)>0$ for all $l\ge 5.1$ and then we have (25). Then, ${\Omega }_3^{\prime }\left(l\right)$ is increasing on $[5.1,\infty )$ with $\underset{l\to \infty }{lim}{\Omega }_3^{\prime }\left(l\right)=0$ and then ${\Omega }_3^{\prime }\left(l\right)<0$ for all $l\ge 5.1$ and this proves (24). Also, ${\Omega }_3\left(l\right)$ is decreasing on $[5.1,\infty )$ with $\underset{l\to \infty }{lim}{\Omega }_3\left(l\right)=0$ and then we obtain (23). After that, we assume that ${\Omega }_4\left(l\right)=lnl-\psi \left(l\right)-{\displaystyle \frac{1}{6}}{\psi }^{\prime }\left(l\right)-{\displaystyle \frac{1}{12}}{\psi }^{″}(l+3/10)-{\displaystyle \frac{1}{3}}{\psi }^{\prime }(l+1/4).$ Then ${\Omega }_4^{\prime }\left(l\right)={\displaystyle \frac{1}{l}}-{\psi }^{\prime }\left(l\right)-{\displaystyle \frac{1}{6}}{\psi }^{″}\left(l\right)-{\displaystyle \frac{1}{12}}{\psi }^{\prime \prime \prime }(l+3/10)-{\displaystyle \frac{1}{3}}{\psi }^{″}(l+1/4)$ and ${\Omega }_4^{″}\left(l\right)={\displaystyle \frac{-1}{l^2}}-{\psi }^{″}\left(l\right)-{\displaystyle \frac{1}{6}}{\psi }^{\prime \prime \prime }\left(l\right)-{\displaystyle \frac{1}{12}}{\psi }^{\left(4\right)}(l+3/10)-{\displaystyle \frac{1}{3}}{\psi }^{\prime \prime \prime }(l+1/4)$ and similarly, we have

$$-{\Omega }_4^{″}\left(l\right)+{\Omega }_4^{″}(l+1)={\displaystyle \frac{\frac{-\eta \left(l\right)}{{(1+4l)}^4}}{l^4{(1+l)}^2{(3+10l)}^5}}<0,l\ge 1.3$$

where

$$3125\eta \left(l\right)=128702632481+2752894096900\left(l-{\displaystyle \frac{13}{10}}\right)+12291217933500{\left(l-{\displaystyle \frac{13}{10}}\right)}^2$$

$$+26291877140000{\left(l-{\displaystyle \frac{13}{10}}\right)}^3+32922564950000{\left(l-{\displaystyle \frac{13}{10}}\right)}^4+25998075950000{\left(l-{\displaystyle \frac{13}{10}}\right)}^5$$

$$+1318515\left({10}^7\right){\left(l-{\displaystyle \frac{13}{10}}\right)}^6+4183225\left({10}^6\right){\left(l-{\displaystyle \frac{13}{10}}\right)}^7+758\left({10}^9\right){\left(l-{\displaystyle \frac{13}{10}}\right)}^8+6\left({10}^{10}\right){\left(l-{\displaystyle \frac{13}{10}}\right)}^9.$$

Then, ${\Omega }_4^{″}\left(l\right)>0$ for all $l\ge 1.3$ and then we have (28). Also, ${\Omega }_4^{\prime }\left(l\right)$ is increasing on $[1.3,\infty )$ with $\underset{l\to \infty }{lim}{\Omega }_4^{\prime }\left(l\right)=0$ and then we obtain (27). Then, ${\Omega }_4\left(l\right)$ is decreasing on $[1.3,\infty )$ with $\underset{l\to \infty }{lim}{\Omega }_4\left(l\right)=0$ and then we obtain (26). Finally, we set ${\Omega }_5\left(l\right)=lnl-\psi \left(l\right)-{\displaystyle \frac{1}{2}}{\psi }^{\prime }\left(l\right)-{\displaystyle \frac{1}{12}}{\psi }^{″}\left(l\right)-{\displaystyle \frac{1}{20l^3}}+{\displaystyle \frac{1}{12l^2}}$ and then

$${\Omega }_5^{\prime }\left(l\right)={\displaystyle \frac{1}{l}}-{\psi }^{\prime }\left(l\right)-{\displaystyle \frac{1}{2}}{\psi }^{″}\left(l\right)-{\displaystyle \frac{1}{12}}{\psi }^{\prime \prime \prime }\left(l\right)+{\displaystyle \frac{3}{20l^4}}-{\displaystyle \frac{1}{6l^3}}$$

and we obtain, for $l\ge {\displaystyle \frac{13}{10}},$

$${\Omega }_5^{\prime }(l+1)-{\Omega }_5^{\prime }\left(l\right)={\displaystyle \frac{\frac{-1}{125}\left[4694+23115\left(l-\frac{13}{10}\right)+19300{\left(l-\frac{13}{10}\right)}^2+4500{\left(l-\frac{13}{10}\right)}^3\right]}{60l^4{(1+l)}^4}}<0.$$

By the same way, we obtain ${\Omega }_5^{\prime }\left(l\right)>0$ for all $l\ge {\displaystyle \frac{13}{10}}$ and then ${\Omega }_5\left(l\right)$ is increasing on $[1.3,\infty )$ with $\underset{l\to \infty }{lim}{\Omega }_5\left(l\right)=0$ and then we obtain (29). □

The following lemma is deduced from (3) and (4):

Lemma 2.

The following limits are valid for $l\in {\mathbb{R}}^{+}$:

$$\underset{l\to 0}{lim}{l}^{s+1}{\psi }^{\left(s\right)}\left(l\right)={(-1)}^{s+1}s!,s\in \mathbb{N}\cup \left\{0\right\},$$

(30)

$$\underset{l\to 0}{lim}{l}^s{\psi }^{\left(s\right)}(l+\tau )=0,s\in \mathbb{N},\tau >0$$

(31)

and

$$\underset{l\to 0}{lim}{l}^{s+3}{\psi }^{\left(s\right)}\left(l\right)=0,s=0,1,2,\cdots .$$

(32)

3. CM Degrees of Some Functions Containing $\Gamma ,{\mathit{\psi }}^{\prime }$ and ${\mathit{\psi }}^{″}$

In this section, we present two completely monotonic functions involving the gamma and polygamma functions, which will be useful in Section 4 in deducing some new bounds for the gamma and polygamma functions. Also, we investigate their completely monotonic degrees.

Theorem 1.

Suppose that $l\in {\mathbb{R}}^{+}.$ Then, the function

$$T_{\varrho }\left(l\right)=(l-1/2)lnl+ln\sqrt{2\pi }-l-ln\Gamma \left(l\right)+{\displaystyle \frac{1}{12}}{\psi }^{\prime }\left(l\right)+{\displaystyle \frac{1}{24}}{\psi }^{″}(l+\varrho ),\varrho \ge 0$$

is CM on ${\mathbb{R}}^{+}$ if and only if $\varrho \ge {\displaystyle \frac{3}{10}}$ with $1\le {deg}_{CM}^l\left[T_{\frac{3}{10}}\left(l\right)\right]<2.$ Also, the function $-T_{\varrho }\left(l\right)$ is CM on ${\mathbb{R}}^{+}$ if and only if $\varrho =0$ with $2\le {deg}_{CM}^l\left[-T_0\left(l\right)\right]<3.$

Proof. 

We use the Binet first Formula [18]:

$$ln\Gamma \left(l\right)=(l-1/2)lnl+ln\sqrt{2\pi }-l+{\int }_0^{\infty }\left(1/2+{\displaystyle \frac{1}{e^t-1}}-{\displaystyle \frac{1}{t}}\right){\displaystyle \frac{e^{-lt}}{t}}dt,l\in {\mathbb{R}}^{+}$$

(33)

we obtain

$$T_{\varrho }\left(l\right)={\int }_0^{\infty }\left[-1/2-{\displaystyle \frac{1}{e^t-1}}+{\displaystyle \frac{1}{t}}\right]{\displaystyle \frac{e^{-lt}}{t}}dt+{\displaystyle \frac{1}{12}}{\psi }^{\prime }\left(l\right)+{\displaystyle \frac{1}{24}}{\psi }^{″}(l+\varrho )$$

and by using (4), we have $T_{\varrho }\left(l\right)={\int }_0^{\infty }{\displaystyle \frac{e^{-lt}}{t^2(e^t-1)}}{\chi }_{\varrho }\left(t\right)dt,$ where

$${\chi }_{\varrho }\left(t\right)=-1+e^t-{\displaystyle \frac{t(e^t+1)}{2}}+{\displaystyle \frac{t^3{e}^t}{12}}-{\displaystyle \frac{t^4{e}^{(1-\varrho )t}}{24}}.$$

Let $\varrho \ge {\displaystyle \frac{3}{10}},$ then we obtain

$${\chi }_{\varrho }\left(t\right)\ge -1+e^t-{\displaystyle \frac{t(e^t+1)}{2}}+{\displaystyle \frac{t^3{e}^t}{12}}-{\displaystyle \frac{t^4}{24}}{e}^{\frac{7}{10}t}={\displaystyle \frac{13t^6}{14400}}+\sum _{p=3}^{\infty }{\displaystyle \frac{a_p}{24(p+4)!}}{t}^{p+4}>0,$$

where

$$\begin{array}{ccc}\hfill {\displaystyle \frac{\frac{{10}^p}{(p+1)}{a}_p}{(p+2)}}& =& 2(6+p){10}^p-7^p(4+p)(3+p)\hfill \\ & =& (12+2p)\sum _{s=3}^p\left( _s^p\right)7^{p-s}{3}^s+p(p-1)(9p+47)7^{p-2}>0,p\ge 3\hfill \end{array}$$

and consequently, $T_{\varrho }\left(l\right)$ is CM on ${\mathbb{R}}^{+}$ for $\varrho \ge {\displaystyle \frac{3}{10}}.$ On the other side, if $T_{\varrho }\left(l\right)$ is CM on ${\mathbb{R}}^{+}$, then by using the asymptotic Formulas (2) and (7), we obtain for $l>0$:

$$l^3{T}_{\varrho }\left(l\right)=l^3\left[{\displaystyle \frac{1}{24l^2}}-{\displaystyle \frac{1}{24{(l+\varrho )}^2}}+{\displaystyle \frac{1}{60l^3}}-{\displaystyle \frac{1}{24{(l+\varrho )}^3}}+O\left({\displaystyle \frac{1}{l^4}}\right)\right]>0.$$

Then, $\underset{l\to \infty }{lim}{l}^3{T}_{\varrho }\left(l\right)={\displaystyle \frac{\varrho }{12}}-{\displaystyle \frac{1}{40}}\ge 0$ and then $\varrho \ge {\displaystyle \frac{3}{10}}.$ Next, letting $\varrho =0,$ we obtain

$$T_0\left(l\right)={\int }_0^{\infty }\left(\sum _{p=1}^{\infty }{\displaystyle \frac{(p+5)(p+2)(p+1)p{t}^{p+4}}{24(p+4)!}}\right){\displaystyle \frac{e^{-lt}}{t^2(1-e^t)}}dt$$

and then the function $-T_0\left(l\right)$ is CM on ${\mathbb{R}}^{+}$. On the other side, assuming that $-T_{\varrho }\left(l\right)$ is CM on ${\mathbb{R}}^{+}$ with $\varrho >0,$ and then

$$l^2{T}_{\varrho }\left(l\right)<0,l>0,\varrho >0.$$

(34)

Using (1), (30), and (31), we have $\underset{l\to 0}{lim}{l}^2{T}_{\varrho }\left(l\right)={\displaystyle \frac{1}{12}}>0,$ which is a contradiction with (34). Then, $\varrho =0.$ By using the asymptotic Formulas (2) and (7), we obtain $T_{\frac{3}{10}}(\infty )=\underset{l\to \infty }{lim}{T}_{\frac{3}{10}}\left(l\right)=0.$

• Next,

$$l{T}_{\frac{3}{10}}\left(l\right)={\int }_0^{\infty }{\displaystyle \frac{\frac{{\Delta }_1\left(t\right)}{{\left(1+e^{\frac{t}{5}}+e^{\frac{2t}{5}}-e^{\frac{3t}{10}}-e^{\frac{t}{10}}\right)}^2}}{240t^3{(e^{\frac{t}{10}}-1)}^2{(e^{\frac{t}{10}}+1)}^2{\left(1+e^{\frac{t}{5}}+e^{\frac{2t}{5}}+e^{\frac{t}{10}}+e^{\frac{3t}{10}}\right)}^2}}{e}^{-lt}dt,l\in {\mathbb{R}}^{+}$$

where

$$\begin{array}{ccc}\hfill {\Delta }_1\left(t\right)& =& 480\left(-1+2e^t-e^{2t}\right)+120t\left(e^{2t}-1\right)+240t^2{e}^t\hfill \\ & & +20t^3\left(e^{2t}-e^t\right)-20t^4\left(-e^{\frac{7t}{10}}+e^t+e^{\frac{17t}{10}}\right)+t^5\left(7e^{\frac{7t}{10}}+3e^{\frac{17t}{10}}\right)\hfill \\ & =& {\displaystyle \frac{13t^7}{20}}+{\displaystyle \frac{551t^8}{700}}+{\displaystyle \frac{896094t^9}{5(9!)}}+{\displaystyle \frac{96987924t^{10}}{125(10!)}}+{\displaystyle \frac{3575122551t^{11}}{1250(11!)}}\hfill \\ & & +{\displaystyle \frac{29569195002t^{12}}{3125(12!)}}+{\displaystyle \frac{7239762907611t^{13}}{250000(13!)}}+{\displaystyle \frac{10445121705591t^{14}}{125000(14!)}}\hfill \\ & & +{\displaystyle \frac{5755638359336499t^{15}}{25000000(15!)}}+\sum _{p=11}^{\infty }{\displaystyle \frac{b_p}{(p+5)!}}{t}^{p+5}>0,p\ge 11\hfill \end{array}$$

with

$$\begin{array}{ccc}\hfill b_p& =& 960\left(1+(p-3)2^{p+1}\right)+20(p+5)(p+4)\left(12+(p+3)\left(2^{p+2}-1\right)\right)\hfill \\ & & +(p+5)(p+4)(p+3)(p+2)\left(7(p+3){\left({\displaystyle \frac{7}{10}}\right)}^p+(3p-31){\left({\displaystyle \frac{17}{10}}\right)}^p-20\right)\hfill \\ & =& 960\left(1+(p-3)2^{p+1}\right)+20(p+5)(p+4)\left(12+(p+3)\left(2^{p+2}-1\right)\right)\hfill \\ & & +(p+5)(p+4)(p+3)(p+2)(7(p+3){\left({\displaystyle \frac{7}{10}}\right)}^p+\left(3(p-11)+2\right)\sum _{s=3}^p\left( _s^p\right){\left({\displaystyle \frac{7}{10}}\right)}^s\hfill \\ & & +{\displaystyle \frac{1}{200}}\left(10260+23728(p-11)+3605{(p-11)}^2+147{(p-11)}^3\right))>0,p\ge 11.\hfill \end{array}$$

Hence, $1\le {deg}_{CM}^l\left[T_{\frac{3}{10}}\left(l\right)\right].$ But,

$$l^2{T}_{\frac{3}{10}}\left(l\right)=\frac{\frac{{\Delta }_2\left(t\right)}{2400t^4{(e^{\frac{t}{10}}-1)}^3{(e^{\frac{t}{10}}+1)}^3}}{{\left(1-e^{\frac{t}{10}}+e^{\frac{t}{5}}-e^{\frac{3t}{10}}+e^{\frac{2t}{5}}\right)}^3{\left(1+e^{\frac{t}{10}}+e^{\frac{t}{5}}+e^{\frac{3t}{10}}+e^{\frac{2t}{5}}\right)}^3}{e}^{-lt}dt,l\in {\mathbb{R}}^{+}$$

where

$$\begin{array}{ccc}\hfill {\Delta }_2\left(t\right)& =& -2400e^{3t}\left(-6+t\right)-2400\left(6+t\right)-e^{\frac{27t}{10}}\left(200-120t+9t^2\right)t^4\hfill \\ & & -2e^{\frac{17t}{10}}{t}^4\left(-200-80t+71t^2\right)+200e^{2t}\left(-216+12t-24t^2-12t^3-2t^4+t^5\right)\hfill \\ & & +200e^t\left(216+12t+24t^2-12t^3+2t^4+t^5\right)-e^{\frac{7t}{10}}{t}^4\left(200+280t+49t^2\right)\hfill \end{array}$$

with ${\Delta }_2\left(11.1\right)=1.59388\left({10}^{17}\right)$ and ${\Delta }_2\left(11.2\right)=-1.57813\left({10}^{18}\right)$. Hence, $l^2{T}_{\frac{3}{10}}\left(l\right)$ is not CM on ${\mathbb{R}}^{+}$ and then ${deg}_{CM}^l\left[T_{\frac{3}{10}}\left(l\right)\right]<2.$ Now, we have $T_0(\infty )=-\underset{l\to \infty }{lim}{T}_0\left(l\right)=0$ and

$$-l^2{T}_0\left(l\right)={\int }_0^{\infty }{\displaystyle \frac{{\Delta }_3\left(t\right)}{24t^4{(e^t-1)}^3}}{e}^{-lt}dt,l\in {\mathbb{R}}^{+}$$

where

$$\begin{array}{ccc}\hfill {\Delta }_3\left(t\right)& =& 144\left(1-3e^t+3e^{2t}-e^{3t}\right)+24t\left(1-e^t-e^{2t}+e^{3t}\right)+48t^2\left(e^{2t}-e^t\right)\hfill \\ & & +24t^3\left(e^t+e^{2t}\right)+2t^4\left(-e^t+e^{3t}\right)+2t^5\left(e^t-3e^{2t}\right)+t^6\left(e^t+e^{2t}\right)\hfill \\ & =& {\displaystyle \frac{3t^7}{5}}+{\displaystyle \frac{7t^8}{5}}+{\displaystyle \frac{54t^9}{35}}+{\displaystyle \frac{4033440t^{10}}{10!}}+{\displaystyle \frac{23923152t^{11}}{11!}}+{\displaystyle \frac{124711488t^{12}}{12!}}\hfill \\ & & +{\displaystyle \frac{593986224t^{13}}{13!}}+{\displaystyle \frac{2649720528t^{14}}{14!}}+{\displaystyle \frac{11256805296t^{15}}{15!}}+\sum _{p=10}^{\infty }{\displaystyle \frac{c_p}{(p+6)!}}{t}^{p+6}>0\hfill \end{array}$$

with

$$\begin{array}{ccc}\hfill c_p& =& 144\left(-3+3\left(2^{p+6}\right)-3^{p+6}\right)+24(p+6)\left(-1-2^{p+5}+3^{p+5}\right)\hfill \\ & & +48(p+6)(p+5)\left(2^{p+4}-1\right)+24(p+6)(p+5)(p+4)\left(1+2^{p+3}\right)\hfill \\ & & +2(p+6)(p+5)(p+4)(p+3)\left(3^{p+2}-1\right)\hfill \\ & & +2(p+6)(p+5)(p+4)(p+3)(p+2)\left(1-3\left(2^{p+1}\right)\right)\hfill \\ & & +(p+6)(p+5)(p+4)(p+3)(p+2)(p+1)\left(1+2^p\right)\hfill \\ & =& (320640+487464(p-10)+171322{(p-10)}^2+25467{(p-10)}^3+1885{(p-10)}^4\hfill \\ & & +69{(p-10)}^5+{(p-10)}^6)2^p+2304+4392p+2858p^2+1033p^3+213p^4+23p^5+p^6\hfill \\ & & +18\left(108+1974(p-3)+335{(p-3)}^2+30{(p-3)}^3+{(p-3)}^4\right)3^p>0,p\ge 10.\hfill \end{array}$$

Hence, $2\le {deg}_{CM}^l\left[-T_0\left(l\right)\right].$ But,

$$-l^3{T}_0\left(l\right)={\int }_0^{\infty }{\displaystyle \frac{{\Delta }_4\left(t\right)}{24t^5{(e^t-1)}^4}}{e}^{-lt}dt,l>0$$

where

$$\begin{array}{ccc}\hfill {\Delta }_4\left(t\right)& =& -72e^{4t}(-8+t)+72(8+t)-4e^{2t}\left(-864-72t^2+24t^4-3t^5-2t^6+t^7\right)\hfill \\ & & -e^{3t}\left(2304-144t+144t^2+72t^3+24t^4+12t^5-8t^6+t^7\right)\hfill \\ & & -e^t\left(2304+144t+144t^2-72t^3+24t^4+4t^6+t^7\right)\hfill \end{array}$$

with ${\Delta }_4\left(6\right)\simeq 1.91564\left({10}^{11}\right)$ and ${\Delta }_4\left(6.1\right)\simeq -6.76109\left({10}^{10}\right)$. Hence, $-l^3{T}_0\left(l\right)$ is not CM on ${\mathbb{R}}^{+}$ and then ${deg}_{CM}^l\left[T_0\left(l\right)\right]<3.$ □

Theorem 2.

For $l\in {\mathbb{R}}^{+}$ and $\delta \ge 0,$ the function

$$M_{\delta }\left(l\right)=(l+\delta -1/2)ln(l+\delta )+ln\sqrt{2\pi }+{\displaystyle \frac{1}{12}}{\psi }^{\prime }(l+\delta )+{\displaystyle \frac{1}{24}}{\psi }^{″}(l+\delta )+{\displaystyle \frac{1}{40l^3}}-ln\Gamma (l+\delta )-(l+\delta )$$

is CM on ${\mathbb{R}}^{+}$ if and only if $\delta \ge {\displaystyle \frac{5}{18}}$ with $2\le {deg}_{CM}^l\left[M_{\frac{5}{18}}\left(l\right)\right]<3$ and the function $-M_{\delta }\left(l\right)$ is CM on ${\mathbb{R}}^{+}$ if and only if $\delta =0$ with $2\le {deg}_{CM}^l\left[-M_0\left(l\right)\right]<3.$

Proof. 

We use the identity ${\displaystyle \frac{1}{l^p}}={\displaystyle \frac{1}{(p-1)!}}{\int }_0^{\infty }{t}^{p-1}{e}^{-lt}dt$ for $l\in {\mathbb{R}}^{+},$ (see [2]) and (4), (33), we obtain $M_{\delta }\left(l\right)={\int }_0^{\infty }{\displaystyle \frac{e^{-(l+\delta )t}}{t^2(e^t-1)}}{{\rm Y}}_{\delta }\left(t\right)dt,$ where

$${{\rm Y}}_{\delta }\left(t\right)=-1+e^t-{\displaystyle \frac{t(e^t+1)}{2}}+{\displaystyle \frac{t^3{e}^t}{12}}-{\displaystyle \frac{t^4{e}^t}{24}}+{\displaystyle \frac{t^4}{80}}(e^t-1)e^{\delta t}.$$

Letting $\delta \ge {\displaystyle \frac{5}{18}},$ we obtain ${{\rm Y}}_{\delta }\left(t\right)\ge -1+e^t-{\displaystyle \frac{t(e^t+1)}{2}}+{\displaystyle \frac{t^3{e}^t}{12}}-{\displaystyle \frac{t^4{e}^t}{24}}+{\displaystyle \frac{t^4}{80}}\left(e^{\frac{23t}{18}}-e^{\frac{5t}{18}}\right).$ Then,

$${{\rm Y}}_{\delta }\left(t\right)\ge {\displaystyle \frac{121t^7}{362880}}+{\displaystyle \frac{1319t^8}{4898880}}+\sum _{p=5}^{\infty }{\displaystyle \frac{d_p}{80(p+4)!}}{t}^{p+4}>0,$$

where

$$\begin{array}{ccc}\hfill {\displaystyle \frac{\frac{d_p}{(p+2)}}{(p+1)}}& =& -{\displaystyle \frac{10p(p+5)}{3}}+(p+4)(p+3)\left({\left({\displaystyle \frac{23}{18}}\right)}^p-{\left({\displaystyle \frac{5}{18}}\right)}^p\right)\hfill \\ & =& {\displaystyle \frac{(p-2)}{18}}\left(32+53(p-5)+5{(p-5)}^2\right)+(p+4)(p+3)\sum _{s=2}^{p-1}\left( _s^p\right){\left({\displaystyle \frac{5}{18}}\right)}^s>0,p\ge 5.\hfill \end{array}$$

Hence $M_{\delta }\left(l\right)$ is CM on ${\mathbb{R}}^{+}$ for $\delta \ge {\displaystyle \frac{5}{18}}.$ On the other hand, assuming that $M_{\delta }\left(l\right)$ is CM on ${\mathbb{R}}^{+}$, then we obtain $M_{\delta }\left(l\right)>0$ for $l>0$ and we use (2) and (7), we obtain

$$\begin{array}{ccc}\hfill \underset{l\to \infty }{lim}{l}^4{M}_{\delta }\left(l\right)& =& \underset{l\to \infty }{lim}{l}^4\left({\displaystyle \frac{1}{40l^3}}-{\displaystyle \frac{1}{40{(l+\delta )}^3}}-{\displaystyle \frac{1}{48{(l+\beta )}^4}}+O\left({\displaystyle \frac{1}{l^5}}\right)\right)={\displaystyle \frac{3\delta }{40}}-{\displaystyle \frac{1}{48}}\ge 0\hfill \end{array}$$

and then $\delta \ge {\displaystyle \frac{5}{18}}$. Next letting $\delta =0,$ we obtain

$$M_0\left(l\right)={\int }_0^{\infty }\left(\sum _{p=2}^{\infty }{\displaystyle \frac{(-1+p)(1+p)(2+p)(36+7p)t^{p+4}}{240(p+4)!}}\right){\displaystyle \frac{e^{-lt}}{t^2(1-e^t)}}dt$$

and then $-M_0\left(l\right)$ is CM on ${\mathbb{R}}^{+}.$ On the other side, assuming that $-M_{\delta }\left(l\right)$ is CM on ${\mathbb{R}}^{+}$ with $\delta \in {\mathbb{R}}^{+},$ and then

$$l^3{M}_{\delta }\left(l\right)<0,l\in {\mathbb{R}}^{+},\delta \in {\mathbb{R}}^{+}.$$

(35)

We use (4), we have $\underset{l\to 0}{lim}{l}^3{M}_{\delta }\left(l\right)={\displaystyle \frac{1}{40}}>0$ which has a contradiction with (35) and then $\delta =0.$ Using the asymptotic Formulas (2) and (7), we obtain

$$M_{\frac{5}{18}}(\infty )=\underset{l\to \infty }{lim}{M}_{\frac{5}{18}}\left(l\right)=0.$$

Next,

$$l^2{M}_{\frac{5}{18}}\left(l\right)={\int }_0^{\infty }\frac{\frac{{\Delta }_5\left(t\right)}{38880t^4{e}^{\frac{5t}{18}}{(e^{\frac{t}{18}}-1)}^3{(e^{\frac{t}{18}}+1)}^3{\left(1-e^{\frac{t}{18}}+e^{\frac{t}{9}}\right)}^3}}{{\left(1+e^{\frac{t}{18}}+e^{\frac{t}{9}}\right)}^3{\left(1-e^{\frac{t}{6}}+e^{\frac{t}{3}}\right)}^3{\left(1+e^{\frac{t}{6}}+e^{\frac{t}{3}}\right)}^3}{e}^{-lt}dt,l\in {\mathbb{R}}^{+}$$

where

$$\begin{array}{ccc}\hfill {\Delta }_5\left(t\right)& =& 233280\left(e^{3t}-3e^{2t}+3e^t-1\right)+4320t\left(-19+39e^t-21e^{2t}+e^{3t}\right)\hfill \\ & & -120t^2\left(115-813e^t+633e^{2t}+65e^{3t}\right)-60t^3\left(25+263e^t+983e^{2t}+25e^{3t}\right)\hfill \\ & & +36t^4\left(-27e^{\frac{5t}{18}}+40e^t+81e^{\frac{23t}{18}}+100e^{2t}-81e^{\frac{41t}{18}}-140e^{3t}+27e^{\frac{59t}{18}}\right)\hfill \\ & & +10t^5\left(-299e^t+742e^{2t}+205e^{3t}\right)-5t^6\left(169e^t+454e^{2t}+25e^{3t}\right)\hfill \\ & =& u\left(t\right)+\sum _{p=20}^{\infty }{\displaystyle \frac{A_p}{(p+6)!}}{t}^{p+6}>\sum _{p=20}^{\infty }{\displaystyle \frac{A_p}{(p+6)!}}{t}^{p+6},\hfill \end{array}$$

where $u\left(t\right)$ is a polynomial of 25-th degree of t with positive coefficients (by using Wolfram Mathematica 10), and

$$\begin{array}{ccc}\hfill A_p& =& 233280\left(3^{p+6}-3\left(2^{p+6}\right)+3\right)+4320(p+6)\left(39-21\left(2^{p+5}\right)+3^{p+5}\right)\hfill \\ & & -120(p+6)(p+5)\left(-813+633\left(2^{p+4}\right)+65\left(3^{p+4}\right)\right)\hfill \\ & & -60(p+6)(p+5)(p+4)\left(263+983\left(2^{p+3}\right)+25\left(3^{p+3}\right)\right)\hfill \\ & & +36(p+6)(p+5)(p+4)(p+3)\left(40+81{\left({\displaystyle \frac{23}{18}}\right)}^{p+2}+100\left(2^{p+2}\right)+7{\left({\displaystyle \frac{59}{18}}\right)}^{p+2}+B_p\right)\hfill \\ & & +10(p+6)(p+5)(p+4)(p+3)(p+2)\left(-299+742\left(2^{p+1}\right)+205\left(3^{p+1}\right)\right)\hfill \\ & & -5(p+6)(p+5)(p+4)(p+3)(p+2)(p+1)\left(169+454\left(2^p\right)+25\left(3^p\right)\right),\hfill \end{array}$$

where

$${\displaystyle \frac{B_p}{3^{p+2}}}=-27{\left({\displaystyle \frac{5}{54}}\right)}^{p+2}-81{\left({\displaystyle \frac{41}{54}}\right)}^{p+2}-140+20{\left({\displaystyle \frac{59}{54}}\right)}^{p+2}>0.128139>0,p\ge 20$$

as ${\displaystyle \frac{B_p}{3^{p+2}}}$ is the sum of some increasing sequences, we have ${\displaystyle \frac{B_p}{3^{p+2}}}>0.128139>0$ for $p\ge 20$. By using the binomial theorem, we have

$$\begin{array}{ccc}\hfill A_p& >& 233280\left(\sum _{s=1}^{p-5}\left( _s^{p+6}\right)2^s+\sum _{s=p-4}^{p+6}\left( _s^{p+6}\right)2^s-3\left(2^{p+6}\right)+4\right)\hfill \\ & & +4320(p+6)\left(\sum _{s=1}^{p-5}\left( _s^{p+5}\right)2^s+\sum _{s=p-4}^{p+5}\left( _s^{p+5}\right)2^s-21\left(2^{p+5}\right)+40\right)\hfill \\ & & -120\prod _{i=5}^6(p+i)\left(65\sum _{s=1}^{p-5}\left( _s^{p+4}\right)2^s+65\sum _{s=p-4}^{p+4}\left( _s^{p+4}\right)2^s+633\left(2^{p+4}\right)-748\right)\hfill \\ & & -60\prod _{i=4}^6(p+i)\left(25\sum _{s=1}^{p-5}\left( _s^{p+3}\right)2^s+25\sum _{s=p-4}^{p+3}\left( _s^{p+3}\right)2^s+983\left(2^{p+3}\right)+288\right)\hfill \\ & & +36\prod _{i=3}^6(p+i)(7\sum _{s=1}^{p-5}\left( _s^{p+2}\right)2^s{\left({\displaystyle \frac{23}{18}}\right)}^{p+2-s}+7\sum _{s=p-4}^{p+2}\left( _s^{p+2}\right)2^s{\left({\displaystyle \frac{23}{18}}\right)}^{p+2-s}\hfill \\ & & +100\left(2^{p+2}\right)+88{\left({\displaystyle \frac{23}{18}}\right)}^{p+2}+40)\hfill \\ & & +10\prod _{i=2}^6(p+i)\left(205\sum _{s=1}^{p-5}\left( _s^{p+1}\right)2^s+205\sum _{s=p-4}^{p+1}\left( _s^{p+1}\right)2^s+742\left(2^{p+1}\right)-94\right)\hfill \\ & & -5\prod _{i=1}^6(p+i)\left(25\sum _{s=1}^{p-5}\left( _s^p\right)2^s+25\sum _{s=p-4}^p\left( _s^p\right)2^s+454\left(2^p\right)+194\right).\hfill \\ & =& {\displaystyle \frac{2^{p-9}}{295245\left(504\right)}}(1280188765694361600+2098647666504997920p\hfill \\ & & +1003140178986264408p^2+302033387875264572p^3+57665513357826134p^4\hfill \\ & & +9907268194068903p^5+1659133062425685p^6+214596730342998p^7\hfill \\ & & +21045208371756p^8+1583093175207p^9+61823537617p^{10})\hfill \\ & & +C_p+3168\prod _{i=3}^6(p+i)\sum _{n=7}^{p+2}\left( _n^{p+2}\right){\left({\displaystyle \frac{5}{18}}\right)}^n+D_p>0,p\ge 20.\hfill \end{array}$$

where

$$\begin{array}{ccc}\hfill C_p& =& {\displaystyle \frac{1}{17006112}}(183149035104407040+320624709301629504(p-19)\hfill \\ & & +93286455487888896{(p-19)}^2+12866690543164964{(p-19)}^3\hfill \\ & & +1053248892332396{(p-19)}^4+56077574544255{(p-19)}^5\hfill \\ & & +2020495004535{(p-19)}^6+49561734750{(p-19)}^7\hfill \\ & & +802353750{(p-19)}^8+7789375{(p-19)}^9+34375{(p-19)}^{10})\hfill \\ & >& 0,p\ge 20\hfill \end{array}$$

and

$$\begin{array}{ccc}\hfill D_p& =& 233280\sum _{s=1}^{p-5}\left( _s^{p+6}\right)2^s+4320(p+6)\sum _{s=1}^{p-5}\left( _s^{p+5}\right)2^s-7800(p+6)(p+5)\sum _{s=1}^{p-5}\left( _s^{p+4}\right)2^s\hfill \\ & & -1500(p+6)(p+5)(p+4)\sum _{s=1}^{p-5}\left( _s^{p+3}\right)2^s+252\prod _{i=3}^6\left(p+i\right)\sum _{s=1}^{p-5}\left( _s^{p+2}\right)2^s{\left({\displaystyle \frac{23}{18}}\right)}^{p+2-s}\hfill \\ & & +2050\prod _{i=2}^6\left(p+i\right)\sum _{s=1}^{p-5}\left( _s^{p+1}\right)2^s-125\prod _{i=1}^6\left(p+i\right)\sum _{s=1}^{p-5}\left( _s^p\right)2^s=\prod _{i=1}^6\left(p+i\right)\sum _{s=1}^{p-5}\left( _s^p\right)2^s{m}_s\hfill \end{array}$$

with

$$\begin{array}{ccc}\hfill \prod _{i=1}^6\left(p-s+i\right)m_s& =& -5(-246240-345444(p-s)-191140{(p-s)}^2-44875{(p-s)}^3\hfill \\ & & -3825{(p-s)}^4+115{(p-s)}^5+25{(p-s)}^6)\hfill \\ & & +252\prod _{i=3}^6\left(p-s+i\right){\left({\displaystyle \frac{23}{18}}\right)}^{p-s+2}\hfill \\ & =& {\displaystyle \frac{1}{136048896}}(11034829531434240+5577195933147744(p-s-5)\hfill \\ & & +1027054854920088{(p-s-5)}^2+75913157989196{(p-s-5)}^3\hfill \\ & & +1090067398942{(p-s-5)}^4+23389059495{(p-s-5)}^5\hfill \\ & & +24455541975{(p-s-5)}^6+2126055750{(p-s-5)}^7\hfill \\ & & +78960000{(p-s-5)}^8+1894375{(p-s-5)}^9+21875{(p-s-5)}^{10})\hfill \\ & & +252\prod _{i=3}^6\left(p-s+i\right)\sum _{n=7}^{p-s+2}\left( _n^{p-s+2}\right){\left({\displaystyle \frac{5}{18}}\right)}^n\hfill \\ & >& 0,5\le p-s\le p-1.\hfill \end{array}$$

Then, $2\le {deg}_{CM}^l\left[M_{\frac{5}{18}}\left(l\right)\right].$ But,

$$l^3{M}_{\frac{5}{18}}\left(l\right)={\int }_0^{\infty }\frac{\frac{{\Delta }_6\left(t\right)}{139968t^5{e}^{\frac{5t}{18}}{(e^{\frac{t}{18}}-1)}^4{(e^{\frac{t}{18}}+1)}^4{\left(1-e^{\frac{t}{18}}+e^{\frac{t}{9}}\right)}^4}}{{\left(1+e^{\frac{t}{18}}+e^{\frac{t}{9}}\right)}^4{\left(1-e^{\frac{t}{6}}+e^{\frac{t}{3}}\right)}^4{\left(1+e^{\frac{t}{6}}+e^{\frac{t}{3}}\right)}^4}{e}^{-lt}dt,l\in {\mathbb{R}}^{+}$$

where

$$\begin{array}{ccc}\hfill {\Delta }_6\left(t\right)& =& -3359232{\left(-1+e^t\right)}^4-279936\left(-4+e^t\right){\left(-1+e^t\right)}^{3}t\hfill \\ & & +5184{\left(-1+e^t\right)}^2\left(-35+187e^t+10e^{2t}\right)t^2\hfill \\ & & +48\left(-1+e^t\right)\left(400+3363e^t+13458e^{2t}+275e^{3t}\right)t^3\hfill \\ & & +12\left(-125+4894e^t+41256e^{2t}+23834e^{3t}+125e^{4t}\right)t^4\hfill \\ & & +108e^t\left(-1+e^t\right)\left(-65+598e^t+115e^{2t}\right)t^5\hfill \\ & & -2e^t\left(-6929+5613e^t+34833e^{2t}+1475e^{3t}\right)t^6\hfill \\ & & +e^t\left(2197+21003e^t+11667e^{2t}+125e^{3t}\right)t^7\hfill \end{array}$$

with ${\Delta }_6\left(5.8\right)=2.51761\left({10}^{15}\right)$ and ${\Delta }_6\left(5.9\right)=-1.40575\left({10}^{16}\right)$. Hence, $l^3{M}_{\frac{5}{18}}\left(l\right)$ is not CM on ${\mathbb{R}}^{+}$ and then ${deg}_{CM}^l\left[M_{\frac{3}{10}}\left(l\right)\right]<3.$ We also have $M_0(\infty )=\underset{l\to \infty }{lim}{M}_0\left(l\right)=0$ and

$$-l^2{M}_0\left(l\right)={\int }_0^{\infty }{\displaystyle \frac{{\Delta }_7\left(t\right)}{120t^4{(e^t-1)}^3}}{e}^{-lt}dt,l\in {\mathbb{R}}^{+}$$

where

$$\begin{array}{ccc}\hfill {\Delta }_7\left(t\right)& =& -720\left(e^{3t}-3e^{2t}+3e^t-1\right)+120t\left(e^{3t}-e^{2t}-e^t+1\right)+240t^2\left(e^{2t}-e^t\right)\hfill \\ & & +120t^3\left(e^{2t}+e^t\right)+t^4\left(3-19e^t+9e^{2t}+7e^{3t}\right)-10t^5\left(3e^{2t}-e^t\right)+5t^6\left(e^{2t}+e^t\right)\hfill \\ & =& {\displaystyle \frac{5t^8}{2}}+{\displaystyle \frac{111t^9}{28}}+{\displaystyle \frac{1667t^{10}}{504}}+{\displaystyle \frac{1937t^{11}}{1008}}+{\displaystyle \frac{195t^{12}}{224}}+{\displaystyle \frac{326213t^{13}}{997920}}+\sum _{p=8}^{\infty }{\displaystyle \frac{{\theta }_p}{(p+6)!}}{t}^{p+6}\hfill \end{array}$$

with

$$\begin{array}{ccc}\hfill {\theta }_p& =& 8280+18882p+13219p^2+5003p^3+1056p^4+115p^5+5p^6\hfill \\ & & +\left(9360+20138(p-4)+3017{(p-4)}^2+238{(p-4)}^3+7{(p-4)}^4\right)3^{p+2}\hfill \\ & & +(141264+542940(p-8)+326060{(p-8)}^2+66735{(p-8)}^3+6311{(p-8)}^4\hfill \\ & & +285{(p-8)}^5+5{(p-8)}^6)2^p>0,p\ge 8.\hfill \end{array}$$

Hence, $2\le {deg}_{CM}^l\left[-M_0\left(l\right)\right].$ But,

$$-l^3{M}_0\left(l\right)={\int }_0^{\infty }{\displaystyle \frac{{\Delta }_8\left(t\right)}{24t^5{(e^t-1)}^4}}{e}^{-lt}dt,l\in {\mathbb{R}}^{+}$$

where

$$\begin{array}{ccc}\hfill {\Delta }_8\left(t\right)& =& -72e^{4t}(-8+t)+72(8+t)-4e^{2t}\left(-864-72t^2+24t^4-3t^5-2t^6+t^7\right)\hfill \\ & & -e^{3t}\left(2304-144t+144t^2+72t^3+24t^4+12t^5-8t^6+t^7\right)\hfill \\ & & -e^t\left(2304+144t+144t^2-72t^3+24t^4+4t^6+t^7\right)\hfill \end{array}$$

with ${\Delta }_8\left(6\right)\simeq 1.91564\left({10}^{11}\right)$ and ${\Delta }_8\left(6.1\right)\simeq -6.76109\left({10}^{10}\right)$. Hence, $-l^3{M}_0\left(l\right)$ is not CM on ${\mathbb{R}}^{+}$ and then ${deg}_{CM}^l\left[-M_0\left(l\right)\right]<3.$ □

4. Some New Symmetric Bounds for $\Gamma$ and ${\mathit{\psi }}^{\mathit{s}}$

In this section, we present some new bounds for $\Gamma$ and ${\psi }^s,$ which are better than some of the recently published results. The following results are deduced from Theorem 1 and 2.

Corollary 2.

Let $0\le \omega ,\tau <\infty$ and $l\in {\mathbb{R}}^{+}.$ Then,

$${\displaystyle \frac{1}{\sqrt{l}}}exp\left[-l+{\displaystyle \frac{{\psi }^{\prime }\left(l\right)}{12}}+{\displaystyle \frac{{\psi }^{″}(l+\tau )}{24}}\right]<{\displaystyle \frac{\Gamma \left(l\right)}{\sqrt{2\pi }{l}^l}}<{\displaystyle \frac{1}{\sqrt{l}}}exp\left[-l+{\displaystyle \frac{{\psi }^{\prime }\left(l\right)}{12}}+{\displaystyle \frac{{\psi }^{″}(l+\omega )}{24}}\right],$$

(36)

where $\omega ={\displaystyle \frac{3}{10}}$ and $\tau =0$ being the best.

Proof. 

The right-hand side (R.H.S) of (36) is deduced from $l^3{T}_{\omega }\left(l\right)>0$ which gives $\omega \ge {\displaystyle \frac{3}{10}}$ as we mentioned in the proof of Theorem 1. Since ${\psi }^{″}\left(l\right)$ is increasing on ${\mathbb{R}}^{+},$ we obtain ${\psi }^{″}(l+{\displaystyle \frac{3}{10}})\le {\psi }^{″}(l+\omega )$ for $\omega \ge {\displaystyle \frac{3}{10}}.$ Hence, $\omega ={\displaystyle \frac{3}{10}}$ is the best in (36). The other side of (36) for $\tau =0$ is deduced from Theorem 1. We assume that (36) is true for $\tau >0$ and $l\in {\mathbb{R}}^{+},$ then we obtain

$$\underset{l\to 0}{lim}{l}^{2}ln\Gamma \left(l\right)>\underset{l\to 0}{lim}\left[l^2\left(l-{\displaystyle \frac{1}{2}}\right)lnl-l^3+l^{2}ln\sqrt{2\pi }+{\displaystyle \frac{l^2}{12}}{\psi }^{\prime }\left(l\right)+{\displaystyle \frac{l^2}{24}}{\psi }^{″}(l+\tau )\right]$$

and we use (30) and (31), we obtain $\underset{l\to 0}{lim}{l}^{2}ln\Gamma \left(l\right)>{\displaystyle \frac{1}{12}}$ and this contradicts that $\underset{l\to 0}{lim}{l}^{2}ln\Gamma \left(l\right)=0.$ Then, $\tau =0$ is the best in (36). □

Remark 1.

• Since $-{\psi }^{″}\left(l\right)$ is CM on ${\mathbb{R}}^{+},$ we have that the upper bound of (36) improves its counterpart of (12) for all $l\in {\mathbb{R}}^{+}.$
• By using the relations (22) and (23), we obtain that the upper and lower bounds of (36) at $\omega ={\displaystyle \frac{3}{10}}$ and $\tau =0$, respectively, improve their counterparts of (8) for all $l\ge 0.5$ and $l\ge 5.1$, respectively.
• Using (21), we have that the upper bound of (36) at $\omega ={\displaystyle \frac{3}{10}}$, improves its counterpart of (11) for all $l\ge 0.5.$
• Using (26), we obtain that the upper bound of (36) at $\omega ={\displaystyle \frac{3}{10}}$, improves its counterpart of (15) for all $l\ge 1.3.$

Corollary 3.

Let $l\in {\mathbb{R}}^{+}$ and $0\le \omega ,\tau <\infty$. Then,

$${\displaystyle \frac{1}{2l}}-{\displaystyle \frac{{\psi }^{″}\left(l\right)}{12}}-{\displaystyle \frac{{\psi }^{\prime \prime \prime }(l+\tau )}{24}}<lnl-\psi \left(l\right)<{\displaystyle \frac{1}{2l}}-{\displaystyle \frac{{\psi }^{″}\left(l\right)}{12}}-{\displaystyle \frac{{\psi }^{\prime \prime \prime }(l+\omega )}{24}},$$

(37)

where $\omega ={\displaystyle \frac{3}{10}}$ and $\tau =0$ are the best.

Proof. 

The R.H.S (37) is deduced from $T_{\omega }^{\prime }\left(l\right)<0.$ Using (6) and (7), we obtain

$$\underset{l\to \infty }{lim}{l}^4{T}_{\omega }^{\prime }\left(l\right)=l^4\left[{\displaystyle \frac{1}{12{(l+\omega )}^3}}-{\displaystyle \frac{1}{12l^3}}+{\displaystyle \frac{1}{8{(l+\omega )}^4}}-{\displaystyle \frac{1}{20l^4}}+O\left({\displaystyle \frac{1}{l^5}}\right)\right]={\displaystyle \frac{3}{40}}-{\displaystyle \frac{\omega }{4}}\le 0$$

and then $\omega \ge {\displaystyle \frac{3}{10}}.$ Since $-{\psi }^{\prime \prime \prime }\left(l\right)$ is increasing on ${\mathbb{R}}^{+},$ we deduce that $\omega ={\displaystyle \frac{3}{10}}$ is the best in (37). The other side of (37) for $\tau =0$ is concluded from Theorem 1. We assume that (37) is true for $\tau >0$ and $l\in {\mathbb{R}}^{+},$ and then we obtain $\underset{l\to 0}{lim}{l}^3\left[-\psi \left(l\right)+lnl\right]>\underset{l\to 0}{lim}{l}^3\left[{\displaystyle \frac{1}{2l}}-{\displaystyle \frac{{\psi }^{″}\left(l\right)}{12}}-{\displaystyle \frac{{\psi }^{\prime \prime \prime }(l+\tau )}{24}}\right]$ and this leads to

$$-\underset{l\to 0}{lim}{l}^3\psi \left(l\right)>{\displaystyle \frac{-1}{12}}\underset{l\to 0}{lim}{l}^3{\psi }^{″}\left(l\right)-{\displaystyle \frac{1}{24}}\underset{l\to 0}{lim}{l}^3{\psi }^{\prime \prime \prime }(l+\tau ).$$

(38)

We use (30), (31), and (32), we obtain $\underset{l\to 0}{lim}{l}^3\psi \left(l\right)=0$ and ${\displaystyle \frac{-\underset{l\to 0}{lim}{l}^3{\psi }^{″}\left(l\right)}{12}}-{\displaystyle \frac{\underset{l\to 0}{lim}{l}^3{\psi }^{\prime \prime \prime }(\tau +l)}{24}}=1/6,$ which have a contradiction with (38). Then $\tau =0$ is the best. □

Remark 2.

• Since ${\psi }^{\prime \prime \prime }\left(l\right)$ is completely monotonic on ${\mathbb{R}}^{+},$ we obtain that the upper bound of (37) improves its counterpart of (13) for all $l\in {\mathbb{R}}^{+}.$
• By using (19) and (24), we have that the upper and lower bounds of (37) at $\omega ={\displaystyle \frac{3}{10}}$ and $\tau =0$, respectively, improving their counterparts of (9) for all $l\ge 0.5$ and $l\ge 5.1$, respectively.
• Using (27), we obtain that the upper bound of (37) at $\omega ={\displaystyle \frac{3}{10}}$ improves its counterpart of (16) for all $l\ge 1.3.$

Corollary 4.

Let $0\le \omega ,\tau <\infty ,l\in {\mathbb{R}}^{+}$ and $s=2,3,\cdots .$ Then,

$${\displaystyle \frac{(s-1)!}{2l^s}}+{\displaystyle \frac{{(-1)}^s{\psi }^{(s+1)}\left(l\right)}{12}}+{\displaystyle \frac{{(-1)}^s{\psi }^{(s+2)}(l+\tau )}{24}}<{(-1)}^s{\psi }^{(s-1)}\left(l\right)-{\displaystyle \frac{(s-2)!}{l^{s-1}}}$$

$$<{\displaystyle \frac{(s-1)!}{2l^s}}+{\displaystyle \frac{{(-1)}^s{\psi }^{(s+1)}\left(l\right)}{12}}+{\displaystyle \frac{{(-1)}^s{\psi }^{(s+2)}(l+\omega )}{24}},$$

(39)

where $\omega ={\displaystyle \frac{3}{10}}$ and $\tau =0$ are the best.

Proof. 

The R.H.S. of (39) is deduced from ${(-1)}^s{T}_{\omega }^{\left(s\right)}\left(l\right)>0$ for $s=2,3,\cdots ,$ and by using (7), we obtain

$$\begin{array}{ccc}\hfill \underset{l\to \infty }{lim}{l}^{s+3}{(-1)}^s{T}_{\omega }^{\left(s\right)}\left(l\right)& =& \underset{l\to \infty }{lim}{l}^{s+3}[{\displaystyle \frac{(s+1)!}{24}}\left({\displaystyle \frac{1}{l^{s+2}}}-{\displaystyle \frac{1}{{(l+\omega )}^{s+2}}}\right)+{\displaystyle \frac{(s+2)!}{120l^{s+3}}}\hfill \\ & & -{\displaystyle \frac{(s+2)!}{48{(l+\omega )}^{s+3}}}+O\left({\displaystyle \frac{1}{l^{s+4}}}\right)]={\displaystyle \frac{(s+2)!\omega }{24}}-{\displaystyle \frac{(s+2)!}{80}}\ge 0\hfill \end{array}$$

and then $\omega \ge {\displaystyle \frac{3}{10}}.$ Since ${\psi }^{\prime \prime \prime }\left(l\right)$ is CM on ${\mathbb{R}}^{+},$ we obtain ${(-1)}^s{\psi }^{(s+3)}\left(l\right)>0$ on ${\mathbb{R}}^{+},$ and then ${(-1)}^s{\psi }^{(s+2)}\left(l\right)$ is increasing on ${\mathbb{R}}^{+},$ and then $\omega ={\displaystyle \frac{3}{10}}$ being the best in (39). The other side of (39) for $\tau =0$ is deduced from Theorem 1. We assume that (39) is true for $\tau \in {\mathbb{R}}^{+}$ and $l\in (0,\infty ),$ then we obtain

$$\underset{l\to 0}{lim}{l}^{s+2}\left[{(-1)}^s{\psi }^{(s-1)}\left(l\right)-{\displaystyle \frac{(s-2)!}{l^{s-1}}}\right]>\underset{l\to 0}{lim}{l}^{s+2}\left[{\displaystyle \frac{(s-1)!}{2l^s}}+{\displaystyle \frac{{(-1)}^s{\psi }^{(s+1)}\left(l\right)}{12}}+{\displaystyle \frac{{(-1)}^s{\psi }^{(s+2)}(l+\tau )}{24}}\right]$$

and then

$${(-1)}^s\underset{l\to 0}{lim}{l}^{s+2}{\psi }^{(s-1)}\left(l\right)>{\displaystyle \frac{{(-1)}^s}{12}}\underset{l\to 0}{lim}{l}^{s+2}{\psi }^{(s+1)}\left(l\right)+{\displaystyle \frac{{(-1)}^s}{24}}\underset{l\to 0}{lim}{l}^{s+2}{\psi }^{(s+2)}(l+\tau ).$$

(40)

We use (30), (31), and (32), we obtain $\underset{l\to 0}{lim}{l}^{s+2}{\psi }^{(s-1)}\left(l\right)=0$ and

$${\displaystyle \frac{{(-1)}^s}{12}}\underset{l\to 0}{lim}{l}^{s+2}{\psi }^{(s+1)}\left(l\right)+{\displaystyle \frac{{(-1)}^s}{24}}\underset{l\to 0}{lim}{l}^{s+2}{\psi }^{(s+2)}(l+\tau )={\displaystyle \frac{(s+1)!}{12}}$$

and this in contradiction with (40). Then, $\tau =0$ is the best. □

Remark 3.

• Since $-{\psi }^{″}\left(l\right)$ is CM on ${\mathbb{R}}^{+},$ we have ${(-1)}^s{\psi }^{(s+2)}\left(l\right)<0$ and hence the upper bound of (39) improves its counterpart of (14) for all $s=2,3,\cdots ,$ and $l>0\in {\mathbb{R}}^{+}.$
• We use (20) and (25), we obtain that the upper and lower bounds of (39) at $\omega ={\displaystyle \frac{3}{10}},\tau =0,$ and $s=2$ improve their counterparts of (10) at $s=2$ for all $l\ge 0.5$ and $l\ge 5.1$, respectively.
• Using (28), we obtain that the upper bound of (39) at $\omega ={\displaystyle \frac{3}{10}}$ and $s=2$, refines its counterpart of (17) at $s=2$, for all $l\ge 1.3.$

Corollary 5.

For $0\le \xi <\infty$ and $l\in {\mathbb{R}}^{+},$ we have

$${\displaystyle \frac{1}{\sqrt{l}}}exp\left(-l+{\displaystyle \frac{{\psi }^{\prime }\left(l\right)}{12}}+{\displaystyle \frac{{\psi }^{″}\left(l\right)}{24}}+{\displaystyle \frac{1}{40l^3}}\right)<{\displaystyle \frac{\Gamma \left(l\right)}{\sqrt{2\pi }{l}^l}}$$

$$<{\displaystyle \frac{1}{\sqrt{l}}}exp\left(-l+{\displaystyle \frac{{\psi }^{\prime }\left(l\right)}{12}}+{\displaystyle \frac{{\psi }^{″}\left(l\right)}{24}}+{\displaystyle \frac{1}{40{(l-\xi )}^3}}\right),$$

(41)

where the upper bound is valid for $l>\xi$ and the other bound is valid for $l\in {\mathbb{R}}^{+}$ with $\xi ={\displaystyle \frac{5}{18}}$ is the best.

Proof. 

The R.H.S. of (41) is deduced from $M_{\xi }(l-\xi )>0$ for $l>\xi$ and we use (2) and (7), we obtain

$$\begin{array}{ccc}\hfill \underset{l\to \infty }{lim}{l}^4{M}_{\xi }(l-\xi )& =& \underset{l\to \infty }{lim}{l}^4\left({\displaystyle \frac{1}{40{(l-\xi )}^3}}-{\displaystyle \frac{1}{40l^3}}-{\displaystyle \frac{1}{48l^4}}+O\left({\displaystyle \frac{1}{l^5}}\right)\right)={\displaystyle \frac{3\xi }{40}}-{\displaystyle \frac{1}{48}}\ge 0\hfill \end{array}$$

and then $\xi \ge {\displaystyle \frac{5}{18}}$. As the function ${\displaystyle \frac{1}{y^3}}$ is decreasing on ${\mathbb{R}}^{+},$ we deduce that $\xi ={\displaystyle \frac{5}{18}}$ is the best in (41). The other side (41) is deduced from $M_0\left(l\right)<0$ in Theorem 2. □

Remark 4.

• The lower bound of (41) improves its counterpart of (36) at $\tau =0,$ for all $l\in {\mathbb{R}}^{+}.$
• By using (29), we deduce that the lower bound of (41) improves its counterpart of (18) for all $l\ge 1.3.$

Corollary 6.

For $0\le \xi <\infty$ and $l\in {\mathbb{R}}^{+},$ we have

$${\displaystyle \frac{1}{2l}}-{\displaystyle \frac{{\psi }^{″}\left(l\right)}{12}}-{\displaystyle \frac{{\psi }^{\prime \prime \prime }\left(l\right)}{24}}+{\displaystyle \frac{3}{40l^4}}<lnl-\psi \left(l\right)<{\displaystyle \frac{1}{2l}}-{\displaystyle \frac{{\psi }^{″}\left(l\right)}{12}}-{\displaystyle \frac{{\psi }^{\prime \prime \prime }\left(l\right)}{24}}+{\displaystyle \frac{3}{40{(l-\xi )}^4}},$$

(42)

where the upper bound is valid for $l>\xi$ and the other bound is valid for $l\in {\mathbb{R}}^{+}$ with $\xi ={\displaystyle \frac{5}{18}}$ is the best.

Proof. 

The R.H.S of (42) is deduced from $M_{\xi }^{\prime }(l-\xi )<0$ for $l>\xi$ and then

$$l^5{M}_{\xi }^{\prime }(l-\xi )=l^5\left[lnl-\psi \left(l\right)-{\displaystyle \frac{1}{2l}}+{\displaystyle \frac{1}{12}}{\psi }^{″}\left(l\right)+{\displaystyle \frac{1}{24}}{\psi }^{\prime \prime \prime }\left(l\right)-{\displaystyle \frac{3}{40{(l-\xi )}^4}}\right]<0.$$

(43)

Using (6) and (7), we have

$$\underset{l\to \infty }{lim}{l}^5{M}_{\xi }^{\prime }(l-\xi )=\underset{l\to \infty }{lim}{l}^5\left[{\displaystyle \frac{3}{40l^4}}-{\displaystyle \frac{3}{40{(l-\xi )}^4}}+{\displaystyle \frac{1}{12l^5}}+O\left({\displaystyle \frac{1}{l^6}}\right)\right]={\displaystyle \frac{1}{12}}-{\displaystyle \frac{3\xi }{10}}.$$

From (43), we conclude that $\xi \ge {\displaystyle \frac{5}{18}}.$ Similarly, as before, we obtain $\xi ={\displaystyle \frac{5}{18}}$, which is the best in (42). The other side of (42) is deduced from $M_0^{\prime }\left(l\right)>0$ in Theorem 2. □

Remark 5.

The lower bound of (42) improves its counterpart of (37) at $\tau =0,$ for all $l\in {\mathbb{R}}^{+}.$

Corollary 7.

For $0\le \xi <\infty ,l\in {\mathbb{R}}^{+}$ and $s=2,3,\cdots ,$ we have

$${\displaystyle \frac{(s-1)!}{2l^s}}+{\displaystyle \frac{{(-1)}^s{\psi }^{(s+1)}\left(l\right)}{12}}+{\displaystyle \frac{{(-1)}^s{\psi }^{(s+2)}\left(l\right)}{24}}+{\displaystyle \frac{(s+2)!}{80l^{s+3}}}<{(-1)}^s{\psi }^{(s-1)}\left(l\right)-{\displaystyle \frac{(s-2)!}{l^{s-1}}}$$

$$<{\displaystyle \frac{(s-1)!}{2l^s}}+{\displaystyle \frac{{(-1)}^s{\psi }^{(s+1)}\left(l\right)}{12}}+{\displaystyle \frac{{(-1)}^s{\psi }^{(s+2)}\left(l\right)}{24}}+{\displaystyle \frac{(s+2)!}{80{(l-\xi )}^{s+3}}}$$

(44)

where the upper bound is valid for $l>\xi$ and the other bound is valid for $l\in {\mathbb{R}}^{+}$ with $\xi ={\displaystyle \frac{5}{18}}$ is the best.

Proof. 

The R.H.S of (44) is deduced from

$${(-1)}^s{M}_{\xi }^{\left(s\right)}(l-\xi )={\displaystyle \frac{(s-2)!}{l^{s-1}}}+{(-1)}^{s+1}{\psi }^{(s-1)}\left(l\right)+{\displaystyle \frac{(s-1)!}{2l^s}}+{\displaystyle \frac{{(-1)}^s}{12}}{\psi }^{(s+1)}\left(l\right)$$

$$+{\displaystyle \frac{{(-1)}^s}{24}}{\psi }^{(s+2)}\left(l\right)+{\displaystyle \frac{(s+2)!}{80{(l-\xi )}^{s+3}}}>0,s=2,3,\cdots ,l>\xi .$$

Using (7), we have

$$\begin{array}{ccc}\hfill \underset{l\to \infty }{lim}{l}^{s+4}{(-1)}^s{M}_{\xi }^{\left(s\right)}(l-\xi )& =& \underset{l\to \infty }{lim}{l}^{s+4}\left[{\displaystyle \frac{(s+2)!}{80{(l-\xi )}^{s+3}}}-{\displaystyle \frac{(s+2)!}{80l^{s+3}}}-{\displaystyle \frac{(s+3)!}{288l^{s+4}}}+O\left({\displaystyle \frac{1}{l^{s+5}}}\right)\right]\hfill \\ & =& {\displaystyle \frac{(s+3)!\left(\xi -\frac{5}{18}\right)}{80}}>0\hfill \end{array}$$

and then $\xi \ge {\displaystyle \frac{5}{18}}.$ By the same way, we have that $\xi ={\displaystyle \frac{5}{18}}$ is the best in (44). The other side of (44) is deduced from ${(-1)}^s{M}_0^{\left(s\right)}\left(l\right)<0$ in Theorem 2. □

Remark 6.

The lower bound of (44) improves its counterpart of (39) at $\tau =0,$ for every $s=2,3,\cdots$ and $l>0.$

5. Conclusions

The main conclusions of this paper are stated in Theorems 1 and 2. Concretely speaking, we studied the monotonicity of two functions involving the gamma and polygamma functions to deduce some symmetric bounds for the gamma, psi, and polygamma functions in terms of some exponents of the polygamma functions, which are mentioned in Corollaries 2–7. These results give more accurate bounds for the gamma, psi, and polygamma functions, which are better than the results presented by Alzer and Batir, Batir, Şevli and Batir, and Moustafa, which are mentioned in (8)–(18).