An Asymptotic Expansion for the Generalized Gamma Function

Abstract

The symmetric patterns that inequalities contain are reflected in researchers’ studies in many mathematical sciences. In this paper, we prove an asymptotic expansion for the generalized gamma function ${\mathsf{\Gamma }}_{\mu }\left(v\right)$ and study the completely monotonic (CM) property of a function involving ${\mathsf{\Gamma }}_{\mu }\left(v\right)$ and the generalized digamma function ${\psi }_{\mu }\left(v\right)$. As a consequence, we establish some bounds for ${\mathsf{\Gamma }}_{\mu }\left(v\right)$, ${\psi }_{\mu }\left(v\right)$ and polygamma functions ${\psi }_{\mu }^{\left(r\right)}\left(v\right)$, $r\ge 1$.

1. Introduction

Stirling’s formula is given by

$$\mathsf{\Gamma }(r+1)\sim \sqrt{2\pi r}{\left(\frac{r}{e}\right)}^r,r⟶\infty$$

(1)

where $\mathsf{\Gamma }$ is the classical Gamma function [1]. An elementary and complete proof of this formula is available at [2]. Moreover, many mathematicians have used the logarithm of gamma function to deduce several useful properties of the gamma function, and their powerful tool for such investigations was the digamma function

$$\psi \left(v\right)=\frac{d}{dv}ln\mathsf{\Gamma }\left(v\right)=-\sum _{r=0}^{\infty }\left[\frac{1}{v+r}-\frac{1}{r+1}\right]-\gamma ,v>0$$

where $\gamma =\underset{r\to \infty }{lim}\left({\displaystyle \sum _{k=1}^r}\frac{1}{k}-logr\right)\approx 0.5772156649$ is Euler–Mascheroni’s constant. For more details on bounds of the functions $\mathsf{\Gamma }\left(v\right)$ and $\frac{d^r}{d{v}^r}\psi \left(v\right)$, please refer to [3,4,5,6,7] and the references therein. Many of such bounds deduced from the monotonicity properties of some functions involving $\mathsf{\Gamma }$ or $\psi$. An infinitely differentiable real valued function M defined on $v>0$ is said to be CM if ${(-1)}^r{M}^{\left(r\right)}\left(v\right)\ge 0$ for all $r\ge 0$ on $v>0$. For more details about CM functions and their applications, we refer to [8,9,10,11]. According to Bernstein theorem [12], function M is CM if and only if $M\left(v\right)={\int }_0^{\infty }{e}^{-vu}d\nu \left(u\right)$, where $\nu \left(u\right)$ is a non-negative measure on $u\ge 0$ such that the integral converges for $v>0$.

In 2007, Alzer and Batir [13] studied the completely monotonicity of the function

$$S_{\rho }\left(v\right)=ln\mathsf{\Gamma }\left(v\right)+\frac{1}{2}\psi (v+\rho )+v-\frac{1}{2}ln\left(2\pi \right)-vlnv,\rho \ge 0;v>0$$

(2)

and deduced the following double inequality:

$$\sqrt{2\pi }exp\left[-\frac{1}{2}\psi (a+v)-v\right]<v^{-v}\mathsf{\Gamma }\left(v\right)<\sqrt{2\pi }exp\left[-\frac{1}{2}\psi (b+v)-v\right],v>0$$

(3)

with the constants $a=\frac{1}{3}$ and $b=0$ being the best possible constants. In 2008, Batir [14] modified $S_{\rho }\left(v\right)$ and deduced some bounds for $\mathsf{\Gamma }\left(v\right)$ in terms of digamma and polygamma functions.

Euler [15] originally defined gamma function as $\mathsf{\Gamma }\left(v\right)={lim}_{\mu \to \infty }{\mathsf{\Gamma }}_{\mu }\left(v\right),$ where

$${\mathsf{\Gamma }}_{\mu }\left(v\right)=\frac{{\mu }^v\mu !}{(\mu +v)\cdots (v+2)(v+1)v},v>0,\mu =1,2,\cdots$$

which satisfies the following recurrence relation.

$${\mathsf{\Gamma }}_{\mu }(r+v)=\frac{v{\mu }^r{\mathsf{\Gamma }}_{\mu }\left(v\right)}{v+n+\mu }\prod _{s=1}^{r-1}\left[\frac{s+v}{s+\mu +v}\right],v>0,r\in \mathbb{N}.$$

(4)

In 2010, Krasniqi and Shabani [16] presented the strictly CM property of function ${\psi }_{\mu }^{\prime }$ on $(0,\infty )$, where the following is the case.

$${\psi }_{\mu }\left(v\right)=ln\mu -\sum _{i=0}^{\mu }\frac{1}{v+i}.$$

(5)

Krasniqi and Merovci [17] introduced the following integral representations for ${\psi }_{\mu }$ and its derivatives:

$${\psi }_{\mu }\left(v\right)=ln\mu +{\int }_0^{\infty }\left(\frac{e^{-(\mu +1)u}-1}{1-e^{-u}}\right)e^{-vu}du,v>0,\mu \in \mathbb{N}$$

(6)

and

$${\psi }_{\mu }^{\left(r\right)}\left(v\right)={(-1)}^r{\int }_0^{\infty }\left(\frac{e^{-(\mu +1)u}-1}{1-e^{-u}}\right)u^r{e}^{-vu}du,v>0,\mu ,r\in \mathbb{N}.$$

(7)

Nantomah, Prempeh and Twum [18] presented the following generalization of the functions $\mathsf{\Gamma }$ and $\psi$:

$$\begin{array}{ccc}\hfill {\mathsf{\Gamma }}_{\mu ,k}\left(v\right)& =& \frac{k^{\mu +1}(1+\mu )!{\left(\mu k\right)}^{\frac{v}{k}-1}}{v(v+k)(v+2k)\cdots (v+\mu k)}={\int }_0^{\mu }\frac{u^v}{u}{\left(1-\frac{u^k}{\mu k}\right)}^{\mu }du,v,k>0,\mu \in \mathbb{N}\hfill \end{array}$$

and

$$\begin{array}{ccc}\hfill {\psi }_{\mu ,k}\left(v\right)& =& \frac{1}{k}ln\left(k\mu \right)-\sum _{i=0}^{\mu }\frac{1}{v+ik}=\frac{1}{k}ln\left(k\mu \right)+{\int }_0^{\infty }\left(\frac{e^{-k(\mu +1)u}-1}{1-e^{-ku}}\right)e^{-vu}du.\hfill \end{array}$$

Nantomah, Merovci and Nasiru [19] presented bounds

$$\frac{1}{(k+v+\mu k)}-v^{-1}\le {\psi }_{\mu ,k}\left(v\right)-k^{-1}ln\left(\frac{vk\mu }{v+k+\mu k}\right)\le 0,\mu \in \mathbb{N},v,k>0$$

(8)

and

$$\frac{1}{{(k+\mu k+v)}^2}-\frac{1}{v^2}\le -{\psi }_{\mu ,k}^{\prime }\left(v\right)+k^{-1}\left[v^{-1}-\frac{1}{(v+k+\mu k)}\right]\le 0,\mu \in \mathbb{N},v,k>0.$$

(9)

Recently, several inequalities involving the generalized gamma function have been presented [17,18,19].

Research involving Gamma function is conducted by many authors currently. Results concerning extensions of Gamma function involving Mittag–Leffler function are presented in [20]. Other extensions of Gamma function are investigated in [21] using the two-parameter Mittag–Leffler matrix function and some important properties of these extended matrix functions are proved. A new series representation of the extended $k-$gamma function is provided in [22] and particular cases involving the original gamma function are discussed as corollaries.

In the following, we will present the asymptotic expansion ${\mathsf{\Gamma }}_{\mu }(r+1)\sim \frac{{\mu }^{r+1}{r}^{r+\frac{1}{2}}{e}^{\mu +1}\mu !}{{(r+\mu +1)}^{r+\mu +\frac{3}{2}}}$ for large values of r and we will discuss the property of completely monotonicity of the function

$$S_{\alpha ,\mu }\left(v\right)=ln{\mathsf{\Gamma }}_{\mu }\left(v\right)+1/2{\psi }_{\mu }(\alpha +v)-vln\left(\frac{\mu v}{v+\mu +1}\right)+(\mu +1)ln(v+\mu +1)-ln\left(\mu !\sqrt{\mu }{e}^{\mu +1}\right),$$

for $\mu \in \mathbb{N}$, $v>0$ and different values of $\alpha \ge 0$. As a consequence, we establish some bounds for ${\mathsf{\Gamma }}_{\mu }\left(v\right)$, ${\psi }_{\mu }\left(v\right)$ and ${\psi }_{\mu }^{\left(r\right)}\left(v\right)$, $r\ge 1$.

2. An Asymptotic Expansion for ${\mathsf{\Gamma }}_{\mathbf{\mu }}\left(\mathit{v}\right)$

Theorem 1.

For all $\mu \in \mathbb{N},$

$${\mathsf{\Gamma }}_{\mu }(r+1)\sim \frac{{\mu }^{r+1}{r}^{r+\frac{1}{2}}{e}^{1+\mu }\mu !}{{(r+\mu +1)}^{\frac{3}{2}+\mu +r}},r⟶\infty .$$

(10)

Proof. 

For $\mu =1$, we have

$$\underset{r\to \infty }{lim}\frac{{\mathsf{\Gamma }}_1(1+r)}{\frac{r^{r+\frac{1}{2}}{e}^2}{{(r+2)}^{r+\frac{5}{2}}}}=\frac{1}{e^2}\underset{r\to \infty }{lim}\left[\frac{r!}{(r+2)!}{(r+2)}^2{\left(1+2r^{-1}\right)}^{\frac{1}{2}+r}\right]=1.$$

Similarly, for $\mu =2,$ we have $\underset{r\to \infty }{lim}\frac{{\mathsf{\Gamma }}_2(r+1)}{\frac{2^{r+2}{r}^{r+\frac{1}{2}}{e}^3}{{(r+3)}^{r+\frac{7}{2}}}}=1$. Now, for $\mu \ge 3$, taking logarithm for both sides of (4) at $v=1$ yields the following.

$$ln{\mathsf{\Gamma }}_{\mu }(r+1)=ln\left(\frac{{\mu }^2}{(1+\mu )(2+\mu )}\right)+\sum _{s=2}^{r}ln\left(\frac{\mu s}{\mu +1+s}\right).$$

(11)

The function $A_{\mu }\left(v\right)=\frac{\mu v}{\mu +v+1}$ is strictly increasing on $v>0$ and, hence, $A_{\mu }\left(v\right)>A_{\mu }\left(2\right)\ge 1$ for all $\mu \ge 3.$ Using the relation between the integral and the Riemann sums, we have

$${\int }_2^{r}ln\left(\frac{v\mu }{\mu +v+1}\right)dv<\sum _{s=2}^{r}ln\left(\frac{s\mu }{\mu +s+1}\right)<{\int }_2^{r+1}ln\left(\frac{\mu v}{\mu +v+1}\right)dv$$

and hence, we have the following:

$$ln{u}_{\mu }\left(r\right)<ln{\mathsf{\Gamma }}_{\mu }(1+r)<ln{u}_{\mu }(1+r),$$

(12)

where

$$u_{\mu }\left(r\right)={\left[\frac{r\mu }{1+r+\mu }\right]}^r\left[\frac{{(3+\mu )}^{\mu +3}}{4{(1+r+\mu )}^{1+\mu }(\mu +1)(\mu +2)}\right].$$

Sequence $u_{\mu }\left(r\right)$ satisfies

$$u_{\mu }(r+1)={\left[\frac{1}{r+\mu +1}+1\right]}^{-(r+2+\mu )}\left[\frac{r\mu }{r+\mu +1}\right]{\left[r^{-1}+1\right]}^r\left[r^{-1}+1\right]u_{\mu }\left(r\right)$$

and using inequality ${(r^{-1}+1)}^r<e<{(r^{-1}+1)}^{r+1}$ for $r\in \mathbb{N}$, we have the following.

$$u_{\mu }(r+1)<2\left[\frac{\mu r}{\mu +r+1}\right]u_{\mu }\left(r\right).$$

(13)

Substituting (13) into (12) yields

$${\mathsf{\Gamma }}_{\mu }(1+r)=u_{\mu }\left(r\right){\theta }_{\mu }\left(r\right),$$

where

$$1<{\theta }_{\mu }\left(r\right)<2\left[\frac{\mu r}{r+1+\mu }\right].$$

Let

$$a_{\mu }\left(r\right)=\frac{{\mathsf{\Gamma }}_{\mu }(r+1)}{{\left(r\mu \right)}^{\frac{2r+1}{2}}{\left(\mu +r+1\right)}^{-\mu -r-\frac{3}{2}}},r,\mu \in \mathbb{N}.$$

Using (4) at $v=1$, we have

$$ln{a}_{\mu }\left(r\right)=-(r+1/2)lnr+ln\left(\frac{\sqrt{\mu }}{1+\mu }\right)+\sum _{s=1}^{r}ln\left(\frac{s}{s+1+\mu }\right)+\left(3/2+\mu +r\right)ln(\mu +1+r)$$

and then

$$ln{a}_{\mu }\left(r\right)-ln{a}_{\mu }(1+r)=H\left(r\right)-H(1+\mu +r),$$

(14)

where $H\left(v\right)=(v+\frac{1}{2})ln\left(1+\frac{1}{v}\right)$, $v>0$ is decreasing function on $v>0$ and consequently ${(ln{a}_{\mu }\left(r\right))}_{r\in \mathbb{N}}$ is decreasing. Using (14), we obtain the following.

$$ln{a}_{\mu }\left(1\right)-ln{a}_{\mu }\left(r\right)=\sum _{m=1}^{r-1}[H\left(m\right)-H(m+\mu +1)].$$

(15)

Using well-known series $ln\left(\frac{1+y}{1-y}\right)=2\left(y+\frac{y^3}{3}+\frac{y^5}{5}+\cdots \right)$, $\left|y\right|<1$ and letting $y=\frac{1}{2v+1}$, we obtain the following.

$$H\left(v\right)-1=\sum _{\tau =1}^{\infty }\frac{1}{1+2\tau }{\left(\frac{1}{2v+1}\right)}^{2\tau }<\sum _{\tau =1}^{\infty }\frac{1}{3}{\left(\frac{1}{2v+1}\right)}^{2\tau }=\frac{1}{12v(1+v)}.$$

(16)

As $H\left(v\right)$ is decreasing, we have the following.

$$H\left(v\right)>\underset{v\to \infty }{lim}H\left(v\right)=1.$$

(17)

Substituting (16) and (17) into (15) produces

$$ln{a}_{\mu }\left(1\right)-ln{a}_{\mu }\left(r\right)<\frac{1}{12}\sum _{\tau =1}^{r-1}\left[\frac{1}{\tau }-\frac{1}{(\tau +1)}\right]<\frac{1}{12}$$

and then ${(ln{a}_{\mu }\left(r\right))}_{r\in \mathbb{N}}$ is bounded from below by $(ln{a}_{\mu }\left(1\right)-\frac{1}{12}).$ Hence, ${(ln{a}_{\mu }\left(r\right))}_{r\in \mathbb{N}}$ is convergent to some constant and depends on $\mu .$ Then, the following is obtained.

$${\mathsf{\Gamma }}_{\mu }(1+r)\sim \frac{C_{\mu }{\left(r\mu \right)}^{r+\frac{1}{2}}}{{(\mu +r+1)}^{\mu +r+3/2}}.$$

(18)

Using Stirling’s Formula (1), we obtain $\underset{r\to \infty }{lim}\frac{2^{4r}{(r!)}^4}{\pi r{[\left(2r\right)!]}^2}=1$ and $\underset{r\to \infty }{lim}\frac{\pi (r+\mu ){[(2r+2\mu )!]}^2}{2^{4(r+\mu )}{[(r+\mu )!]}^4}=1$.

Multiplying these two limits produces the following.

$$\underset{r\to \infty }{lim}{\left(\frac{r!}{(\mu +r)!}\right)}^4{\left(\frac{(2\mu +2r)!}{\left(2r\right)!}\right)}^2=2^{4\mu }.$$

(19)

Letting $v=1$ in (4) and using relation $(2r+2\mu )!=(2r+\mu +1)!{\displaystyle \prod _{s=2+\mu }^{2\mu }}(2r+s),$ we have

$${\mathsf{\Gamma }}_{\mu }(2r+1)=\frac{\left(2r\right)!{\mu }^{1+2r}\mu !{\displaystyle \prod _{s=\mu +2}^{2\mu }}(s+2r)}{(2r+2\mu )!}.$$

(20)

Now, using (4) at $v=1$ and inserting (20) into (19) yields

$$\underset{r\to \infty }{lim}\frac{{\left({\mathsf{\Gamma }}_{\mu }(r+1)\right)}^4}{{\left({\mathsf{\Gamma }}_{\mu }(2r+1)\right)}^2}{(r+\mu +1)}^4{\left(\prod _{s=\mu +2}^{2\mu }(2r+s)\right)}^2=2^{4\mu }{\mu }^2{(\mu !)}^2$$

(21)

and using (18), we obtain the following:

$$C_{\mu }^2\underset{r\to \infty }{lim}{\left(\frac{2r+\mu +1}{2r+2\mu +2}\right)}^{4r}{\left(\prod _{s=\mu +2}^{2\mu }\frac{s+2r}{r+1+\mu }\right)}^2={(\mu !)}^2{2}^{2\mu -2}\mu$$

which leads to

$$C_{\mu }=\sqrt{\mu }\mu !e^{\mu +1},\mu \ge 3,$$

which completes the proof. □

In the next part, we provide a double inequality involving ${\mathsf{\Gamma }}_{\mu }(v+1).$

Some Bounds for the Function ${\mathsf{\Gamma }}_{\mu }\left(v\right)$

Theorem 2.

Assume that $\mu \in \mathbb{N}$ and $v\in (0,\infty ).$ Then, we have

$$exp\left[-\lambda (\mu +1+v)+\lambda \left(v\right)\right]<\frac{{\mathsf{\Gamma }}_{\mu }(1+v)}{\left(\frac{{\mu }^{1+v}{v}^{\frac{1}{2}+v}{e}^{\mu +1}\mu !}{{(\mu +v+1)}^{\mu +v+3/2}}\right)}<exp\left[-\beta (\mu +v+1)+\beta \left(v\right)\right],$$

(22)

where $\beta \left(v\right)=\frac{1}{12v}$ and $\lambda \left(v\right)=\beta \left(v\right)-\frac{1}{360v^3}-\frac{1}{120v^4}.$

Proof. 

Let

$$W_{\mu }\left(v\right)=-H(v+\mu +1)+H\left(v\right)+\mathsf{\Delta }\left[\beta (v+\mu +1)-\beta \left(v\right)\right],$$

where $\mathsf{\Delta }g\left(v\right)=g\left(v\right)-g(v+1)$. Then $W_{\mu }^{″}\left(v\right)=-\frac{(1+\mu )(2+\mu +2v)D_{\mu }\left(v\right)}{6v^3{(1+v)}^3{(1+\mu +v)}^3{(2+\mu +v)}^3}<0$, where

$$D_{\mu }\left(v\right)=4+12\mu +13{\mu }^2+6{\mu }^3+{\mu }^4+(14+29\mu +19{\mu }^2+4{\mu }^3)v+(19+23\mu +7{\mu }^2)v^2+(12+6\mu )v^3+3v^4.$$

Hence, $W_{\mu }^{\prime }\left(v\right)$ is decreasing on $v>0$ with $\underset{v\to \infty }{lim}{W}_{\mu }^{\prime }\left(v\right)=0;$ thus, $W_{\mu }^{\prime }\left(v\right)>0$ for all $v>0$.

Hence, $W_{\mu }\left(v\right)$ is increasing on $v>0$ with $\underset{v\to \infty }{lim}{W}_{\mu }\left(v\right)=0.$ Then, we have the following.

$$H\left(v\right)-H(v+\mu +1)<\mathsf{\Delta }\left[\beta \left(v\right)-\beta (v+\mu +1)\right].$$

(23)

Similarly, we obtain the following.

$$\mathsf{\Delta }\left[\lambda \left(v\right)-\lambda (v+\mu +1)\right]<H\left(v\right)-H(v+\mu +1).$$

(24)

Combining (23) with (24) provides the following:

$$\mathsf{\Delta }\left[\lambda \left(v\right)-\lambda (v+\mu +1)\right]<H\left(v\right)-H(v+\mu +1)<\mathsf{\Delta }\left[\beta \left(v\right)-\beta (v+\mu +1)\right]$$

(25)

and using (14), we obtain the following.

$$exp\left(\mathsf{\Delta }\left[\lambda \left(v\right)-\lambda (v+\mu +1)\right]\right)<\frac{a_{\mu }\left(v\right)}{a_{\mu }(v+1)}<exp\left(\mathsf{\Delta }\left[\beta \left(v\right)-\beta (v+\mu +1)\right]\right).$$

It follows that

$$a_{\mu }(v+1)exp\left[\lambda (v+2+\mu )-\lambda (1+v)\right]<a_{\mu }\left(v\right)exp\left[\lambda (v+\mu +1)-\lambda \left(v\right)\right]$$

and consequently, function

$$T_{\mu }\left(v\right)=a_{\mu }\left(v\right)exp\left[\lambda (v+\mu +1)-\lambda \left(v\right)\right]$$

is strictly decreasing on $v>0$ with $\underset{v\to \infty }{lim}{T}_{\mu }\left(v\right)=C_{\mu }.$ Thus, we have the following.

$$a_{\mu }\left(v\right)>C_{\mu }exp\left[\lambda \left(v\right)-\lambda (1+v+\mu )\right].$$

(26)

In a similar way, function

$$L_{\mu }\left(v\right)=a_{\mu }\left(v\right)exp\left[\beta (v+\mu +1)-\beta \left(v\right)\right]$$

is strictly increasing on $v>0$ with $\underset{v\to \infty }{lim}{L}_{\mu }\left(v\right)=C_{\mu }.$ Then,

$$a_{\mu }\left(v\right)<C_{\mu }exp\left[\beta \left(v\right)-\beta (\mu +v+1)\right].$$

(27)

Combining (26) with (27), we obtain

$$exp\left[\lambda \left(v\right)-\lambda (v+\mu +1)\right]<\frac{a_{\mu }\left(v\right)}{C_{\mu }}<exp\left[\beta \left(v\right)-\beta (v+\mu +1)\right],$$

which completes the proof. □

Corollary 1.

$${\mathsf{\Gamma }}_{\mu }\left(v\right)\sim \left(\frac{{\mu }^v{v}^{v-\frac{1}{2}}{e}^{\mu +1}\mu !}{{(v+1+\mu )}^{1/2+v+\mu }}\right)exp\left[\lambda \left(v\right)-\lambda (v+\mu +1)\right],v⟶\infty$$

(28)

where $\lambda \left(v\right)=\frac{1}{12v}-\frac{1}{360v^3}-\frac{1}{120v^4}.$

Proof. 

Using inequality (22) and relation (4) at $r=1$, we obtain the following.

$$1<\frac{{\mathsf{\Gamma }}_{\mu }\left(v\right)}{\left(\frac{{\mu }^v{v}^{v-\frac{1}{2}}{e}^{1+\mu }\mu !}{{(1+v+\mu )}^{v+\mu +\frac{1}{2}}}\right)exp\left[\lambda \left(v\right)-\lambda (1+v+\mu )\right]}$$

$$<exp\left[\frac{1}{360v^3}+\frac{1}{120v^4}-\frac{1}{360{(1+v+\mu )}^3}-\frac{1}{120{(1+v+\mu )}^4}\right].$$

Hence, we have

$$\underset{v\to \infty }{lim}\left[\frac{{\mathsf{\Gamma }}_{\mu }\left(v\right)}{\left(\frac{{\mu }^v{v}^{v-\frac{1}{2}}{e}^{1+\mu }\mu !}{{(v+1+\mu )}^{1/2+\mu +v}}\right)exp\left[\lambda \left(v\right)-\lambda (v+1+\mu )\right]}\right]=1.$$

This completes the proof. □

Corollary 2.

Let μ and s be positive integers. Then,

$$ln{\mathsf{\Gamma }}_{\mu }\left(v\right)\sim -(\mu +1)ln\left(v+\mu +1\right)+\left(\frac{2v-1}{2}\right)ln\left(\frac{\mu v}{v+1+\mu }\right)+ln\left(\sqrt{\mu }\mu !e^{1+\mu }\right)+\left[\lambda \left(v\right)-\lambda (v+1+\mu )\right],$$

(29)

$${\psi }_{\mu }\left(v\right)\sim -\frac{1}{2}\left(\frac{1}{v}-\frac{1}{v+1+\mu }\right)+ln\left(\frac{\mu v}{v+1+\mu }\right)+\left[{\lambda }^{\prime }\left(v\right)-{\lambda }^{\prime }(v+1+\mu )\right],v⟶\infty$$

(30)

and

$${\psi }_{\mu }^{\left(s\right)}\left(v\right)\sim {(-1)}^{s-1}(s-1)!\left(\frac{1}{v^s}-\frac{1}{{(\mu +1+v)}^s}\right)-\frac{{(-1)}^{s}s!}{2}\left(\frac{1}{v^{s+1}}-\frac{1}{{(\mu +1+v)}^{s+1}}\right)$$

$$+{\lambda }^{(s+1)}\left(v\right)-{\lambda }^{(s+1)}(\mu +1+v),v⟶\infty$$

(31)

where

$${\lambda }^{\left(s\right)}\left(v\right)={(-1)}^s\left[\frac{s!}{12v^{1+s}}-\frac{(2+s)!}{720v^{3+s}}-\frac{(3+s)!}{720v^{4+s}}\right]s\in \mathbb{N}.$$

In the next section, we will generalize some results presented by Alzer and Batir [13].

3. Study of a CM Function Involving ${\mathsf{\Gamma }}_{\mathbf{\mu }}$ and ${\mathbf{\psi }}_{\mathbf{\mu }}$ Functions

Theorem 3.

Suppose that $\mu \in \mathbb{N}$ and $v>0.$ Then, the function

$$S_{\alpha ,\mu }\left(v\right)=ln{\mathsf{\Gamma }}_{\mu }\left(v\right)+1/2{\psi }_{\mu }(\alpha +v)-vln\left(\frac{\mu v}{v+\mu +1}\right)+(\mu +1)ln(v+\mu +1)-ln\left(\sqrt{\mu }\mu !e^{1+\mu }\right),\alpha \ge 0$$

is CM on $v>0$ if and only if $\alpha \ge \frac{1}{3}.$ Moreover, $-S_{\alpha ,\mu }\left(v\right)$ is CM on $v>0$ if and only if $\alpha =0.$

Proof. 

From (6), (7) and identity $ln\left(\frac{h}{d}\right)={\int }_0^{\infty }\frac{e^{-dt}-e^{-ht}}{t}dt$ for $h,d>0$ (see [1]), we have

$$S_{\alpha ,\mu }^{\prime }\left(v\right)={\psi }_{\mu }\left(v\right)-ln\left(\frac{\mu v}{v+1+\mu }\right)+\frac{1}{2}{\psi }_{\mu }^{\prime }(\alpha +v)={\int }_0^{\infty }\frac{e^{-(\mu +v+1)u}}{u(e^u-1)}\phi \left(u\right)du,$$

where

$$\phi \left(u\right)=e^{(2+\mu )u}+1-e^{(1+\mu )u}-e^u-u\left[e^{(2+\mu )u}-e^u\right]-\frac{1}{2}{u}^2\left[e^u-e^{(2+\mu )u}\right]e^{-\alpha u}.$$

Let $\alpha \ge \frac{1}{3}$, then we obtain

$$\phi \left(u\right)\le e^{(2+\mu )u}-e^{(1+\mu )u}-e^u+1-u\left[e^{(2+\mu )u}-e^u\right]+\frac{1}{2}{u}^2\left[e^{(\mu +\frac{5}{3})u}-e^{\frac{2}{3}u}\right]$$

and hence,

$$\phi \left(u\right)\le \sum _{r=3}^{\infty }\frac{f_{\mu }\left(r\right)}{(r+2)!}{u}^{r+2}<0,$$

where

$$\begin{array}{ccc}\hfill f_{\mu }\left(r\right)& =& -{(\mu +1)}^{2+r}-1+{(\mu +2)}^{2+r}+\left[1-{(\mu +2)}^{1+r}\right](2+r)\hfill \\ & +& \left[{\left(\mu +5/3\right)}^r-{\left(2/3\right)}^r\right]\frac{(1+r)(2+r)}{2}\hfill \\ & =& \sum _{s=1}^r\left[-(r+2)\left({}_s^{1+r}\right)+\left({}_s^{2+r}\right)+\left({}_s^r\right){\left(2/3\right)}^{r-s}\frac{(2+r)(1+r)}{2}\right]{(1+\mu )}^s\hfill \\ & =& \sum _{s=1}^r\frac{3^{r-s}(2+r)(1+r)}{(2+r-s)}\left({}_s^r\right)\left[-3^{r-s}+2^{r-s}+2^{r-s-1}(r-s)\right]{(\mu +1)}^s\hfill \\ & =& -\sum _{s=1}^r\frac{3^{r-s}(1+r)(2+r)}{(2+r-s)}\left({}_s^r\right)\left[\sum _{l=2}^{r-s}\left({}_l^{r-s}\right)2^{r-s-l}\right]{(1+\mu )}^s.\hfill \end{array}$$

Consequently, $-S_{\alpha ,\mu }^{\prime }\left(v\right)$ is CM on $(0,\infty )$ for $\alpha \ge \frac{1}{3}$. Thus, $S_{\alpha ,\mu }\left(v\right)$ is decreasing and using asymptotic (29) and (30), we have $\underset{v\to \infty }{lim}{S}_{\alpha ,\mu }\left(v\right)=0$ and then $S_{\alpha ,\mu }\left(v\right)>0$. Then, $S_{\alpha ,\mu }\left(v\right)$ is a CM function on $v>0$ for $\alpha \ge \frac{1}{3}$. Conversely, if function $S_{\alpha ,\mu }\left(v\right)$ is CM, then we obtain for $v>0,\mu \in \mathbb{N}$ that

$$\frac{v^2}{(1+\mu )}{S}_{\alpha ,\mu }\left(v\right)=\frac{v^2}{(1+\mu )}\left[ln\left(\frac{{\mathsf{\Gamma }}_{\mu }\left(v\right)}{\frac{v^{v-\frac{1}{2}}{\mu }^v{e}^{1+\mu }\mu !}{{(v+1+\mu )}^{1/2+\mu +v}}}\right)-\frac{1}{2}ln\left(\frac{\mu v}{v+1+\mu }\right)+\frac{1}{2}{\psi }_{\mu }(\alpha +v)\right]>0.$$

(32)

From (29), we have

$$\underset{v\to \infty }{lim}\frac{v^2}{(1+\mu )}ln\left(\frac{{\mathsf{\Gamma }}_{\mu }\left(v\right)}{\frac{{\mu }^v{v}^{v-\frac{1}{2}}{e}^{1+\mu }\mu !}{{(\mu +v+1)}^{\mu +v+\frac{1}{2}}}}\right)=\frac{1}{12}.$$

(33)

Using asymptotic (30), we have $\underset{v\to \infty }{lim}\frac{v^2}{(1+\mu )}\left[ln\left(\frac{v\mu }{\mu +1+v}\right)-{\psi }_{\mu }\left(v\right)\right]=\frac{1}{2}$ and using (5), we obtain $\underset{v\to \infty }{lim}\frac{v^2}{(1+\mu )}\left[{\psi }_{\mu }\left(v\right)-{\psi }_{\mu }(\alpha +v)\right]=-\alpha .$

Hence, we conclude that

$$\underset{v\to \infty }{lim}\frac{v^2}{(\mu +1)}\left[ln\left(\frac{\mu v}{v+\mu +1}\right)-{\psi }_{\mu }(v+\alpha )\right]=\frac{1}{2}-\alpha .$$

From (32), we conclude that $\frac{1}{12}-\frac{1}{2}(\frac{1}{2}-\alpha )\ge 0$ and then $\alpha \ge \frac{1}{3}$. Now, for $\alpha =0,$ we obtain the following.

$$S_{0,\mu }^{\prime }\left(v\right)={\int }_0^{\infty }\frac{e^{-(v+1+\mu )u}}{u(e^u-1)}\left(\sum _{r=2}^{\infty }\left[\sum _{s=1}^r\frac{(r+2)(r+1)(r-s)}{2(2+r-s)}\left({}_s^r\right){(\mu +1)}^s\right]\frac{u^{r+2}}{(r+2)!}\right)du$$

Therefore, $S_{0,\mu }^{\prime }\left(v\right)$ is CM function on $u>0$. Thus, $S_{0,\mu }\left(v\right)$ is an increasing function on $v>0$ with $\underset{v\to \infty }{lim}{S}_{0,\mu }\left(v\right)=0$ and hence, $S_{0,\mu }\left(v\right)<0$. Then, $-S_{0,\mu }\left(v\right)$ is CM on $v>0$. Conversely, if we assume that $-S_{\alpha ,\mu }\left(v\right)$ is CM on $v>0$ with $\alpha >0$, then $S_{\alpha ,\mu }\left(v\right)<0$ on $v>0$. However, this contradicts $\underset{v\to 0}{lim}{S}_{\alpha ,\mu }\left(v\right)=\infty$; hence, $\alpha =0$. □

Corollary 3.

Let $\mu \in \mathbb{N}.$ Then

$$\underset{v\to 0}{lim}{v}^{1+r}{\psi }_{\mu }^{\left(r\right)}\left(v\right)={(-1)}^{1+r}r!,r=0,1,2,\cdots$$

(34)

and

$$\underset{v\to 0}{lim}{v}^r{\psi }_{\mu }^{\left(r\right)}(b+v)=0,r\in \mathbb{N},b>0.$$

(35)

Some Sharp Bounds for ${\mathsf{\Gamma }}_{\mu }$ and ${\psi }_{\mu }^{\left(r\right)}$ Functions

Now, we will present some sharp bounds of ${\mathsf{\Gamma }}_{\mu }$ and ${\psi }_{\mu }^{\left(r\right)}$ depending on Theorem (3).

Corollary 4.

Let two real numbers $a,b\ge 0$. For $\mu \in \mathbb{N}$ and $v>0,$ we have

$$\left(\frac{v^v{\mu }^{\frac{1}{2}+v}\mu !e^{1+\mu }}{{(v+1+\mu )}^{v+1+\mu }}\right)exp\left[-\frac{1}{2}{\psi }_{\mu }(a+v)\right]<{\mathsf{\Gamma }}_{\mu }\left(v\right)<\left(\frac{v^v{\mu }^{\frac{1}{2}+v}\mu !e^{1+\mu }}{{(v+1+\mu )}^{v+1+\mu }}\right)exp\left[-\frac{1}{2}{\psi }_{\mu }(b+v)\right]$$

(36)

with the constants $a=\frac{1}{3}$ and $b=0$ are best possible.

Proof. 

In inequality (36), the left-hand side is equivalent $\frac{v^2}{(\mu +1)}{S}_{a,\mu }\left(v\right)>0$, which leads to $a\ge \frac{1}{3}$ as stated in the proof of Theorem (3). Using the increasing property of the function ${\psi }_{\mu }\left(v\right)$ on $v>0$, we have $e^{-\frac{1}{2}{\psi }_{\mu }(v+a)}\le e^{-\frac{1}{2}{\psi }_{\mu }(v+\frac{1}{3})}$ for $a\ge \frac{1}{3}$. Then, $a=\frac{1}{3}$ is the best possible constant in (36). Moreover, Theorem (3) proves the right-hand side of the inequality (36) at $b=0$. If there exist $b>0$ such that the upper bound of ${\mathsf{\Gamma }}_{\mu }\left(v\right)$ in (36) is valid for $v\in (0,\infty ),$ then we would have

$$\underset{v\to 0}{lim}{\mathsf{\Gamma }}_{\mu }\left(v\right)\le \left(\frac{\sqrt{\mu }{e}^{\mu +1}\mu !}{{(\mu +1)}^{\mu +1}}\right)exp\left[-\frac{1}{2}{\psi }_{\mu }\left(b\right)\right]\underset{v\to 0}{lim}{\left(\frac{v}{v+\mu +1}\right)}^v$$

and hence,

$$\underset{v\to 0}{lim}{\mathsf{\Gamma }}_{\mu }\left(v\right)\le \left(\frac{\sqrt{\mu }\mu !e^{\mu +1}}{{(\mu +1)}^{\mu +1}}\right)exp\left[-\frac{1}{2}{\psi }_{\mu }\left(b\right)\right]$$

which contradicts with $\underset{v\to 0}{lim}{\mathsf{\Gamma }}_{\mu }\left(v\right)=\infty .$ Then, $b=0$ in (36) is the best possible constant. □

Remark 1.

If we let $\mu \to \infty$ in (36), then we obtain (3).

Corollary 5.

Assume that $a,b\in [0,\infty )$ and $\mu \in \mathbb{N}.$ Then, for all $v\in (0,\infty ),$ we have

$$\frac{1}{2}{\psi }_{\mu }^{\prime }(a+v)<-{\psi }_{\mu }\left(v\right)+ln\left(\frac{v\mu }{v+1+\mu }\right)<\frac{1}{2}{\psi }_{\mu }^{\prime }(b+v),$$

(37)

with constants $a=\frac{1}{3}$ and $b=0$ being the best possible.

Proof. 

In inequality (37), the left-hand side is equivalent

$$\frac{v^3}{(1+\mu )}{S}_{a,\mu }^{\prime }\left(v\right)=\frac{v^3}{(1+\mu )}{\sigma }_{\mu }\left(v\right)+\frac{v^3}{2(1+\mu )}\left[{\psi }_{\mu }^{\prime }(a+v)-\frac{(1+\mu )}{v(v+1+\mu )}\right]<0,$$

(38)

where

$${\sigma }_{\mu }\left(v\right)={\psi }_{\mu }\left(v\right)-ln\left(\frac{\mu v}{v+1+\mu }\right)+\frac{(1+\mu )}{2v(v+1+\mu )}.$$

Using asymptotic expansions (30) and (31), we have

$$\underset{v\to \infty }{lim}\frac{v^3}{(\mu +1)}{\sigma }_{\mu }\left(v\right)=\frac{-1}{6}\mathrm{and}\underset{v\to \infty }{lim}\frac{v^3}{(1+\mu )}\left[{\psi }_{\mu }^{\prime }\left(v\right)-\frac{(1+\mu )}{(v+1+\mu )v}\right]=1.$$

Using relation (5), we have

$$\underset{v\to \infty }{lim}\frac{v^3}{(1+\mu )}\left[{\psi }_{\mu }^{\prime }(v+a)-{\psi }_{\mu }^{\prime }\left(v\right)\right]=-2a.$$

Hence, we conclude from inequality (38) that $\frac{-1}{6}+\frac{1}{2}\left(-2a+1\right)\le 0$ or $a\ge \frac{1}{3}.$

Then, the left-hand side of the inequality (37) is satisfied only if $a\ge \frac{1}{3}$. Using the decreasing property of ${\psi }_{\mu }^{\prime }\left(v\right)$ on $(0,\infty )$, we obtain $a=\frac{1}{3}$ in (37), and it is the best possible constant. Moreover, Theorem (3) proves the right-hand side of the inequality (37) for $b=0$. If there exist $b>0$ such that the upper bound of (37) valid for $v>0$, then we obtain

$$\underset{v\to 0}{lim}v\left[ln\left(\frac{\mu v}{v+\mu +1}\right)-{\psi }_{\mu }\left(v\right)\right]<\frac{1}{2}\underset{v\to 0}{lim}v{\psi }_{\mu }^{\prime }(v+b)$$

which leads to

$$-\underset{v\to 0}{lim}v{\psi }_{\mu }\left(v\right)<\frac{1}{2}\underset{v\to 0}{lim}v{\psi }_{\mu }^{\prime }(v+b).$$

(39)

From (34) and (35), we have $\underset{v\to 0}{lim}v{\psi }_{\mu }\left(v\right)=-1$ and $\underset{v\to 0}{lim}v{\psi }_{\mu }^{\prime }(v+b)=0$, which contradict inequality (39). Hence, $b=0$ is the best possible constant in (37). □

Remark 2.

Substituting $k=1$ in (8), we obtain

$$0\le -{\psi }_{\mu }\left(v\right)+ln\left(\frac{v\mu }{v+1+\mu }\right).$$

Using the completely monotonicity property of ${\psi }_{\mu }^{\prime }\left(v\right)$ and inequality (37), we have

$$0<\frac{1}{2}{\psi }_{\mu }^{\prime }(1/3+v)<-{\psi }_{\mu }\left(v\right)+ln\left(\frac{v\mu }{v+1+\mu }\right).$$

Then, the lower bound of (37) improves its counterpart in (8) at $k=1$ for all $v\in (0,\infty ).$

Corollary 6.

Suppose that $a,b\in [0,\infty )$ and $\mu \in \mathbb{N}.$ Then, for all $v\in (0,\infty )$ and $r=2,3,\cdots ,$ we have

$$\frac{{(-1)}^{r+1}}{2}{\psi }_{\mu }^{\left(r\right)}(a+v)<\left[\frac{(r-2)!}{{(1+v+\mu )}^{r-1}}-\frac{(r-2)!}{v^{r-1}}\right]+{(-1)}^r{\psi }_{\mu }^{(r-1)}\left(v\right)<\frac{{(-1)}^{1+r}}{2}{\psi }_{\mu }^{\left(r\right)}(b+v),$$

(40)

with the constants $a=\frac{1}{3}$ and $b=0$ being the best possible.

Proof. 

In inequality (40), the left-hand side is equivalent

$$\begin{gathered} \frac{v^{r+2}}{(1+\mu )}{(-1)}^r{S}_{a,\mu }^{\left(r\right)}\left(v\right)=\frac{v^{2+r}}{(1+\mu )}{F}_{\mu }\left(v\right)+\frac{v^{2+r}}{2(1+\mu )}\left[{(-1)}^r{\psi }_{\mu }^{\left(r\right)}(a+v)\right. \\ \left.+\left(\frac{1}{v^r}-\frac{1}{{(1+v+\mu )}^r}\right)(r-1)!\right]>0, \end{gathered}$$

(41)

where

$$F_{\mu }\left(v\right)={(-1)}^r{\psi }_{\mu }^{(r-1)}\left(v\right)-\frac{(r-1)!}{2}\left[\frac{1}{v^r}-\frac{1}{{(v+1+\mu )}^r}\right]+\left[\frac{(r-2)!}{{(v+1+\mu )}^{r-1}}-\frac{(r-2)!}{v^{r-1}}\right].$$

Using the asymptotic expansion (31), we have

$$\underset{v\to \infty }{lim}\frac{v^{r+2}}{(1+\mu )}{F}_{\mu }\left(v\right)=\frac{(1+r)!}{12}$$

and

$$\underset{v\to \infty }{lim}\frac{v^{r+2}}{(1+\mu )}\left[{(-1)}^r{\psi }_{\mu }^{\left(r\right)}\left(v\right)+\left(\frac{1}{v^r}-\frac{1}{{(v+1+\mu )}^r}\right)(r-1)!\right]=-\frac{(1+r)!}{2}.$$

Using relation (5), we obtain

$$\underset{v\to \infty }{lim}\frac{{(-1)}^r{v}^{2+r}}{(1+\mu )}\left[{\psi }_{\mu }^{\left(r\right)}(a+v)-{\psi }_{\mu }^{\left(r\right)}\left(v\right)\right]=(1+r)!a.$$

Then, we conclude from inequality (3) that $\left[\frac{1}{12}+\frac{1}{2}\left(a-\frac{1}{2}\right)\right](r+1)!\ge 0$ or $a\ge \frac{1}{3}$. Since ${\psi }_{\mu }^{\prime }\left(v\right)$ is strictly CM function on $v>0$, then for $r=0,1,2,\dots$, function ${(-1)}^r{\psi }_{\mu }^{\left(r\right)}\left(v\right)$ is increasing on $v>0$; hence, the best possible constant in (40) is $a=\frac{1}{3}$. Moreover, Theorem (3) proves the right-hand side of inequality (40) for $b=0$. If there exist $b>0$ such that the upper bound of (40) is valid for $v>0$, then we obtain the following.

$$\underset{v\to 0}{lim}{v}^r\left[{(-1)}^r{\psi }_{\mu }^{(r-1)}\left(v\right)-\left(\frac{(r-2)!}{v^{r-1}}-\frac{(r-2)!}{{(v+1+\mu )}^{r-1}}\right)\right]<\underset{v\to 0}{lim}\left[-\frac{{(-1)}^r{v}^r}{2}{\psi }_{\mu }^{\left(r\right)}(b+v)\right]=0.$$

(42)

Using (34), we have

$$\underset{v\to 0}{lim}{v}^r\left[{(-1)}^r{\psi }_{\mu }^{(r-1)}\left(v\right)+\left(\frac{(r-2)!}{{(v+1+\mu )}^{r-1}}-\frac{(r-2)!}{v^{r-1}}\right)\right]=(r-1)!,r=2,3,\cdots$$

which contradicts with inequality (42); hence, $b=0$ is the best possible constant. □

Remark 3.

Inserting $k=1$ in (9), we obtain the following.

$$0\le {\psi }_{\mu }^{\prime }\left(v\right)-\left(\frac{1}{v}-\frac{1}{(1+v+\mu )}\right).$$

Using the completely monotonicity property of ${\psi }_{\mu }^{\prime }\left(v\right)$ and inequality (40), we have the following.

$$0<\frac{-1}{2}{\psi }_{\mu }^{″}(1/3+v)<{\psi }_{\mu }^{\prime }\left(v\right)-\left(\frac{1}{v}-\frac{1}{(1+v+\mu )}\right).$$

Then, for $r=2$, the lower bound of (40) improves its counterpart in (9) at $k=1$ for all $v\in (0,\infty )$.

4. Conclusions

The main conclusions of this paper are stated in Theorems 1–3. We deduced the following asymptotic expansions for the generalized gamma

$${\mathsf{\Gamma }}_{\mu }(r+1)\sim \frac{{\mu }^{r+1}{r}^{r+\frac{1}{2}}{e}^{1+\mu }\mu !}{{(r+\mu +1)}^{\frac{3}{2}+\mu +r}},\mu \in \mathbb{N};r⟶\infty .$$

and the bounds

$$exp\left[\frac{1}{360}\left(\frac{30v^3-v-3}{v^4}+\frac{\mu -30{(\mu +v+1)}^3+v+4}{{(\mu +v+1)}^4}\right)\right]$$

$$<\frac{{\mathsf{\Gamma }}_{\mu }(1+v)}{\left(\frac{{\mu }^{1+v}{v}^{\frac{1}{2}+v}{e}^{\mu +1}\mu !}{{(\mu +v+1)}^{\mu +v+3/2}}\right)}<exp\left[\frac{\mu +1}{12v(\mu +v+1)}\right],\mu \in \mathbb{N};v>0.$$

Moreover, we proved that the function

$$S_{\alpha ,\mu }\left(v\right)=ln{\mathsf{\Gamma }}_{\mu }\left(v\right)+\frac{{\psi }_{\mu }(\alpha +v)}{2}-vln\left(\frac{\mu v}{v+\mu +1}\right)+(\mu +1)ln(v+\mu +1)-ln\left(\sqrt{\mu }\mu !e^{1+\mu }\right),\mu \in \mathbb{N}$$

is CM on $v>0$ if and only if $\alpha \ge \frac{1}{3}$, and $-S_{\alpha ,\mu }\left(v\right)$ is the CM function on $v>0$ if and only if $\alpha =0.$