A class of completely monotonic functions involving the polygamma functions

Abstract

Let $\Gamma (x)$ denote the classical Euler gamma function. We set $\psi _{n}(x)=(-1)^{n-1}\psi ^{(n)}(x)$ ($n\in \mathbb{N}$), where $\psi ^{(n)}(x)$ denotes the nth derivative of the psi function $\psi (x)=\Gamma '(x)/\Gamma (x)$. For λ, α, $\beta \in \mathbb{R}$ and $m,n\in \mathbb{N}$, we establish necessary and sufficient conditions for the functions

$$ L(x;\lambda ,\alpha ,\beta )=\psi _{m+n}(x)-\lambda \psi _{m}(x+ \alpha ) \psi _{n}(x+\beta ) $$

and $-L(x;\lambda ,\alpha ,\beta )$ to be completely monotonic on $(-\min (\alpha ,\beta ,0),\infty )$.

As a result, we generalize and refine some inequalities involving the polygamma functions and also give some inequalities in terms of the ratio of gamma functions.

1 Introduction

We know that Euler’s gamma function is defined by $\Gamma (x)=\int _{0}^{\infty }t^{x-1}e^{-t}\,dt$ for $x>0$. The psi or digamma function is its logarithmic derivative

$$ \psi (x)=\frac{d}{dx}\ln \Gamma (x)= \frac{\Gamma ^{\prime }(x)}{\Gamma (x)}, $$

(1.1)

whose derivatives $\psi '(x)$ and $\psi ''(x)$ are called the trigamma and tetragamma functions, respectively. The polygamma functions are higher-order derivative

$$ \psi ^{(n)}(x)= (-1 )^{n-1} \int _{0}^{\infty }e^{-xt} \frac{t^{n}}{1-e^{-t}}\,dt= (-1 )^{n-1}n!\sum_{k=0}^{ \infty } \frac{1}{(x+k)^{n+1}}, $$

(1.2)

where $n\in \mathbb{N}$. The Gamma function’s history and its development are given in [1].

After Euler discovered the gamma function, some scholars studied the fundamental properties of gamma, digamma, and polygamma functions, see [2–5]. These functions are important in the fields of engineering, physics, inequality theory, or statistics, and many inequalities involving these functions have been obtained through monotonicity or convexity properties, see [6–16].

A function f is said to be completely monotonic on an interval I if f has derivatives of all orders on I and $(-1)^{n}f^{(n)}(x)\geq 0$, $x\in I$, $n\geq 0$ (see [17]). A function f is said to be strictly completely monotonic if $(-1)^{n}f^{(n)}(x)> 0$. The Bernstein–Widder Theorem [17, Theorem 12b, p. 161] states that f is completely monotonic on $(0,\infty )$ if and only if

$$ f(x)= \int _{0}^{\infty }e^{-xt}\,d\alpha (t), $$

where $\alpha (t)$ is nondecreasing such that the integral converges for $x>0$. Completely monotonic functions have attracted the attention of many researchers in various fields (see [8, 18–24]).

The following asymptotic formulas are often encountered in many papers (see [5]).

$$\begin{aligned}& \psi ^{(n)}(x+1)=\psi ^{(n)}(x)+(-1)^{n} \frac{n!}{x^{n+1}},\quad (x>0; n=0,1, \ldots ), \end{aligned}$$

(1.3)

$$\begin{aligned}& \ln \Gamma (x)= \biggl(x-\frac{1}{2} \biggr)\ln x-x+ \frac{1}{2}\ln (2 \pi )+\frac{1}{12x}+\cdots ,\quad (x\rightarrow \infty ), \end{aligned}$$

(1.4)

$$\begin{aligned}& \psi (x)\sim \ln x-\frac{1}{2x}-\frac{1}{12x^{2}}+ \frac{1}{120x^{4}}- \cdots , \quad (x\rightarrow \infty ), \end{aligned}$$

(1.5)

$$\begin{aligned}& (-1)^{n-1}\psi ^{n}(x)\sim \biggl( \frac{(n-1)!}{x^{n}}+ \frac{n!}{2x^{n+1}}+\frac{(n+1)!}{12x^{n+2}}-\cdots \biggr),\quad (x \rightarrow \infty , n=1,2,\ldots ). \end{aligned}$$

(1.6)

For the sake of convenience, we set $\psi _{n}(x)=(-1)^{n-1}\psi ^{(n)}(x)$ for $n\in \mathbb{N}$.

From (1.2) and the Bernstein–Widder Theorem, we know that $\psi _{n}(x)$ is strictly completely monotonic on $(0,\infty )$. From (1.3) and (1.6), we have

$$\begin{aligned}& \lim_{x \rightarrow 0^{+}}x^{n+1}\psi _{n}(x)=n!, \end{aligned}$$

(1.7)

$$\begin{aligned}& \lim_{x \rightarrow \infty }x^{n}\psi _{n}(x)=(n-1)!, \end{aligned}$$

(1.8)

which easily yields that $\lim_{x \rightarrow 0^{+}}\psi _{n}(x)=\infty$ and $\lim_{x \rightarrow \infty }\psi _{n}(x)=0$.

In order to prove [25, Theorem 4.8], Alzer provided

$$ \psi _{1}^{2}(x)-\psi _{2}(x)>0,\quad x>0, $$

(1.9)

which was verified in a distinct way in [26, Lemma 1.1]. Furthermore, it is worthwhile to notice that [27, Lemma 1.2] is a generalization of the inequality (1.9) and is used to establish many interesting results (see [26, 27]). From [27, Theorem 2.2], it follows that

$$ \psi _{1}^{2} \biggl(x+\frac{1}{2} \biggr)-\psi _{2}(x)< 0,\quad x>0. $$

(1.10)

In light of (1.9) and (1.10), a novel question was raised in [27], which asks whether it is possible that there exist constants α and β such that

$$ \psi _{1}^{2} (x+\alpha )-\psi _{2}(x)>0, $$

(1.11)

and

$$ \psi _{1}^{2} (x+\beta )-\psi _{2}(x)< 0,\quad x>0. $$

(1.12)

Recently, Qi and Guo showed in [28, Theorem 1] that for $\alpha \in \mathbb{R}$, the function

$$ f(x;\alpha )=\psi _{1}^{2} (x+\alpha )-\psi _{2}(x) $$

(1.13)

is completely monotonic on $(-\min (0,\alpha ), \infty )$ if and only if $\alpha \leq 0$, and so is the function $-f(x;\alpha )$ if

$$ \alpha \geq \sup_{x\in (0,\infty )} \frac{x}{\rho ^{-1}(e^{2x}(2(x+1)^{2}-1))}, $$

where $\rho (x)=x\coth x$ for $x>0$ and $\rho ^{-1}(x)$ is the inverse function of $\rho (x)$.

In addition, it was shown in [28, Theorem 3] that the function

$$ f_{\lambda }(x)=\psi _{1}^{2}(x)- \lambda \psi _{2}(x) $$

(1.14)

is completely monotonic on $(0,\infty )$ if and only if $\lambda \leq 1$.

Besides the preceding conclusions invoked, we can refer to more references on results extending (1.9) or (1.14) (see [19, 20, 24, 29–35]).

In view of (1.13), we define the function $L(x;\lambda ,\alpha ,\beta )$ for $\lambda , \alpha , \beta \in \mathbb{R}$, $\eta =\min (\alpha ,\beta ,0)$ and $m,n \in \mathbb{N}$ as follows:

$$ L(x;\lambda ,\alpha ,\beta )=\psi _{m+n}(x)-\lambda \psi _{m}(x+ \alpha ) \psi _{n}(x+\beta ) $$

(1.15)

with respect to $x\in (-\eta , \infty )$.

Then it is a question to put forward: Do sufficient and necessary conditions exist such that $L(x;\lambda ,\alpha ,\beta )$ is completely monotonic?

The aim of this paper is to solve this question and then apply it to obtain more inequalities involving ratios, differences of digamma and polygamma functions.

A detailed plan of this paper is as follows: In Sect. 2, we give detailed proof of our main results. In Sect. 3, some more inequalities for ratios of gamma functions are obtained with the aid of Theorem 3.1.

2 A lemma

In order to prove our main results, we need the following:

Lemma 1

For α, $\beta \in \mathbb{R}$ and $t >0$, let the function $\phi (x)$ be defined on $(0,1)$ by

$$ \phi (x)=\frac{t (1-e^{-t})}{1-e^{-t x}}\frac{x(1-x)}{1-e^{-t (1-x)}}e^{- \alpha t x}e^{-\beta t (1-x)}. $$

(2.1)

Then the following statements are true:

(1) For $\beta -\alpha \geq \frac{1}{2}$, the function $\phi (x)$ is increasing from $(0,1)$ onto $(e^{-\beta t},e^{-\alpha t})$;

(2) For $\beta -\alpha \leq -\frac{1}{2}$, the function $\phi (x)$ is decreasing from $(0,1)$ onto $(e^{-\alpha t},e^{-\beta t})$;

(3) For $-1/2<\alpha -\beta <0$, there exists $t_{0}\geq 0$ such that when $0< t< t_{0}$, the function $\phi (x)$ is increasing from $(0,1)$ onto $(e^{-\beta t},e^{-\alpha t})$;

(4) For $0<\alpha -\beta <1/2$, there exists $t_{0}\geq 0$ such that when $0< t< t_{0}$, the function $\phi (x)$ is decreasing from $(0,1)$ onto $(e^{-\alpha t},e^{-\beta t})$;

(5) For $|\alpha -\beta |<\frac{1}{2}$, there exists $t_{0}\geq 0$ such that when $t>t_{0}$, the function $\phi (x)$ has a unique maximum point $x_{0}(t)$ on $(0,1)$, that is, $\phi (x)$ is increasing on $(0,x_{0}(t))$ and decreasing on $(x_{0}(t),1)$. In particular, if $\alpha =\beta$, then $x_{0}(t)=1/2$.

Proof

Differentiating $\upsilon (x)=\ln \phi (x)$ yields

$$ \upsilon ' (x)=t \bigl(\omega (t x)-\omega \bigl(t(1-x)\bigr)+\beta -\alpha \bigr), $$

(2.2)

where

$$ \omega (x)=\frac{1}{x}-\frac{1}{e^{x}-1},\quad x>0. $$

It is not difficult to show that $\omega (x)$ is decreasing from $(0,\infty )$ onto $(0, \frac{1}{2})$ by noting that

$$ \omega '(x)= \frac{ [(\frac{x}{2})^{2}-(\sinh (\frac{x}{2}))^{2} ]}{ (x \sinh (\frac{x}{2}) )^{2}}< 0. $$

(2.3)

Apparently, we have

$$ \upsilon '' (x)=t^{2} \bigl(\omega '(t x)+\omega ' \bigl(t(1-x)\bigr)\bigr)< 0, $$

so that

$$ \upsilon ' (1)=t \bigl(-\bigl(\omega (0)-\omega (t) \bigr)+\beta -\alpha \bigr)< \upsilon ' (x) < t \bigl(\omega (0)-\omega (t)+\beta -\alpha \bigr)=\upsilon ' (0). $$

(2.4)

For $\beta -\alpha \geq \frac{1}{2}$, from $\omega '(x)<0$ and $0<\omega (x)<1/2$, we see that $\upsilon '(1)>0$, that is, $\upsilon (x)$ is increasing on $(0,1)$, which immediately yields

$$ e^{-\beta t}< \phi (x)< e^{-\alpha t}. $$

(2.5)

For $\beta -\alpha \leq -\frac{1}{2}$, a similar argument yields $\upsilon '(0)<0$, and therefore $\upsilon (x)$ is decreasing on $(0,1)$, which leads to the inversed inequality of (2.5).

If $0<|\alpha -\beta |<\frac{1}{2}$, then there exists $t_{0}>0$ satisfying $\omega (0)-\omega (t_{0})=|\beta -\alpha |$.

Case 1. $0<\alpha -\beta <1/2$. Since $\omega (x)$ is decreasing on $(0,\infty )$, we obtain

$$ \upsilon ' (0)< 0, \quad \text{for } 0< t< t_{0}, $$

(2.6)

and

$$ \upsilon ' (0)>0, \quad \text{for } t>t_{0}. $$

(2.7)

Hence (2.4) and (2.6) imply that $\upsilon ' (x)<0$ for $0< t< t_{0}$ on $(0,1)$, namely, $\phi (x)$ is decreasing from $(0,1)$ onto $(e^{-\alpha t},e^{-\beta t})$.

Case 2. $-1/2<\alpha -\beta <0$. By the same argument, assertion (4) can be proved.

Simultaneously, we observe that $\upsilon ' (1)<0$ for $t>t_{0}$. This in combination with (2.7) and $\upsilon ''(x)<0$ suggests that $\upsilon ' (x)$ is strictly decreasing and therefore has a unique zero point $x_{0}(t)$, that is, $\upsilon (x)$ is increasing on $(0,x_{0}(t))$ and decreasing on $(x_{0}(t),1)$. Moreover, for $\alpha = \beta$, it follows from (2.2) that $\upsilon ' (x)$ has a unique zero point at $x=\frac{1}{2}$. This completes the proof. □

3 Main results

For $x,y\in \mathbb{R}$, let

$$\begin{aligned}& D_{1}=\bigl\{ (x,y)|x\leq 0, y\leq 0\bigr\} \quad \text{and}\quad D_{2}=\bigl\{ (x,y)|x> 0, y> 0 \bigr\} , \\& D_{3}=\bigl\{ (x,y)|x\geq M, y\geq M\bigr\} \cup \biggl\{ (x,y)\Big| \vert x-y \vert \geq \frac{1}{2},x \geq 0, y\geq 0\biggr\} , \end{aligned}$$

where

$$ M=\max_{x>0}\biggl\{ G(x)=\frac{\ln x(1-e^{-x})-2\ln (1-e^{-x/2})-\ln 4}{x} \biggr\} < \frac{1}{2}. $$

(3.1)

We point here that M in (3.1) is well defined since $\lim_{x\rightarrow 0}G(x)=0$ and $\lim_{x\rightarrow \infty }G(x)=0$. In fact, $G(x)$ reaches the maximum at $x_{0}=10.042944\ldots$ , that is, $M=\max_{x>0}G(x)=0.09297\ldots$ .

Theorem 3.1

For $\lambda , \alpha , \beta \in \mathbb{R}$, $\eta =\min (\alpha ,\beta ,0)$ and $m,n \in \mathbb{N}$, let the function $L(x;\lambda ,\alpha ,\beta )$ be defined by (1.15). Then we have

(1) For $(\alpha ,\beta )\in D_{1}$, $-L(x;\lambda ,\alpha ,\beta )$ is completely monotonic on $(-\eta ,\infty )$ if and only if $\lambda \geq \frac{(m+n-1)!}{(m-1)!(n-1)!}$, and so is the function $L(x;\lambda ,\alpha ,\beta )$ if and only if $\lambda \leq 0$;

(2) For $(\alpha ,\beta )\in D_{2}$, $L(x;\lambda ,\alpha ,\beta )$ is completely monotonic on $(0,\infty )$ if and only if $\lambda \leq \inf_{t>0} 1/W(t)$. In particular, if $(\alpha ,\beta )\in D_{3}$, then $\inf_{t>0} 1/W(t)=\frac{(m+n-1)!}{(m-1)!(n-1)!}$, where

$$ W(t)= \int _{0}^{1} \frac{t(1-e^{-t})}{1-e^{-tx}} \frac{x^{m}(1-x)^{n}}{1-e^{-t(1-x)}}e^{-\alpha tx}e^{-\beta t(1-x)}\,dx, \quad t>0. $$

(3.2)

Proof

Using the well-known formula (1.2) and applying the convolution theorem for the Laplace transform, we have

$$ L(x;\lambda ,\alpha ,\beta )= \int _{0}^{\infty } e^{-xt}P(t)\,dt, $$

where

$$ P(t)= \int _{0}^{t} \biggl(\frac{t^{m+n-1}}{1-e^{-t}}-\lambda \frac{s^{m}}{1-e^{-s}}\frac{(t-s)^{n}}{1-e^{-(t-s)}}e^{-\alpha s}e^{- \beta (t-s)} \biggr)\,ds. $$

Changing of variable $s=tx$ yields

$$ P(t)=\frac{t^{m+n} \lambda }{1-e^{-t}} \biggl(\frac{1}{\lambda }- W(t) \biggr). $$

(3.3)

Using the integral representation

$$ \int _{0}^{1}x^{m}(1-x)^{n}\,dx=\frac{n!m!}{(n+m+1)!}, $$

(3.4)

for $\lambda =\frac{(m+n-1)!}{(m-1)!(n-1)!}$, the expression (3.3) can be written as

$$ P(t)=\frac{\lambda t^{m+n}}{1-e^{-t}} \int _{0}^{1}x^{m-1}(1-x)^{n-1} U(x)\,dx, $$

where

$$ U(x)=1- \frac{t(1-e^{-t})}{1-e^{-tx}}\frac{x(1-x)}{1-e^{-t(1-x)}}e^{- \alpha tx}e^{-\beta t(1-x)}. $$

(3.5)

Case 1. $(\alpha ,\beta )\in D_{1}$. First of all, we shall show that

$$ 1< \frac{t(1-e^{-t})}{1-e^{-tx}}\frac{x(1-x)}{1-e^{-t(1-x)}}, \quad \text{for } t>0 \text{ and } 0< x< 1, $$

which is equivalent to

$$ V(t)=\bigl(1-e^{-tx}\bigr) \bigl(1-e^{-t(1-x)}\bigr)-x(1-x)t \bigl(1-e^{-t}\bigr)< 0. $$

A simple computation gives $e^{t}V'(t)=V_{1}(t)$ and

$$ V'_{1}(t)=x(1-x)\sum_{k=1}^{\infty } \frac{t^{k}}{k!}\bigl(x^{k}+(1-x)^{k}-1\bigr)< 0, $$

which together with $V_{1}(0)=0$ yields that $V_{1}(t)<0$ for $t>0$ and $0< x<1$. Furthermore, combining this with $V(0)=0$ and $e^{t}V'(t)=V_{1}(t)$, we have $V(t)<0$. Hence for $(\alpha ,\beta )\in D_{1}$, we see that

$$ 1< \frac{t(1-e^{-t})}{1-e^{-tx}}\frac{x(1-x)}{1-e^{-t(1-x)}}e^{- \alpha tx}e^{-\beta t(1-x)}, \quad \text{for } t>0 \text{ and } 0< x< 1, $$

(3.6)

that is $U(x)>0$ for $t>0$ and $0< x<1$.

From (3.2), (3.4) and (3.6), we conclude that $W(t)> \frac{(m-1)!(n-1)!}{(m+n-1)!}$ for $t>0$. For $(\alpha ,\beta )\in D_{1}$, we also observe that $\lim_{t\rightarrow 0}W(t)=\frac{(m-1)!(n-1)!}{(m+n-1)!}$, and $\lim_{t\rightarrow \infty }W(t)=+\infty$. Hence we have the sharp inequality

$$ 0< \frac{1}{W (t)}< \frac{(m+n-1)!}{(m-1)!(n-1)!},\quad \text{for all } t>0. $$

(3.7)

Finally, according to (3.3), (3.7) and the Bernstein–Widder Theorem, we complete the proof of assertion (1).

Case 2. $(\alpha ,\beta )\in D_{2}$. Since $\lim_{t\rightarrow 0}W(t)=\frac{(m-1)!(n-1)!}{(m+n-1)!}$ and $\lim_{t\rightarrow \infty }W(t)=0$, then $\lambda \leq \inf_{t>0} 1/W(t)$ is well defined. Once more using the Bernstein–Widder Theorem and (3.3), we know that $L(x;\lambda ,\alpha ,\beta )$ is completely monotonic on $(0,\infty )$ if and only if $\lambda \leq \inf_{t>0} 1/W(t)$.

In particular, we consider the case $(\alpha ,\beta )\in D_{3}$. If we prove

$$ 1\geq \frac{t(1-e^{-t})}{1-e^{-tx}}\frac{x(1-x)}{1-e^{-t(1-x)}}e^{- \alpha tx}e^{-\beta t(1-x)},\quad t>0,0< x< 1, $$

(3.8)

we get $\inf_{t>0} 1/W(t)=\frac{(m+n-1)!}{(m-1)!(n-1)!}$ according to (3.2), (3.4) and $\lim_{t\rightarrow 0} W(t)=\frac{(m-1)!(n-1)!}{(m+n-1)!}$.

For $\alpha =\beta \geq M$, $U(x)$ is reduced to

$$ U_{1}(x)=1-\frac{t(1-e^{-t})}{1-e^{-tx}}\frac{x(1-x)}{1-e^{-t(1-x)}}e^{- \alpha t}. $$

(3.9)

In virtue of Lemma 1, we know that $U_{1}(x)$ is decreasing on $(0,1/2)$ and increasing on $(1/2,1)$, that is, $U_{1}(x)\geq U_{1}(\frac{1}{2})$. Since $\alpha \geq M$ is equivalent to $U_{1}(\frac{1}{2})\geq 0$, we have $U_{1}(x) \geq 0$ for $t>0$ and $0< x<1$.

If $\beta \geq M$ and $\alpha \geq \beta$, then we write

$$ U(x)=1-\frac{t(1-e^{-t})}{1-e^{-tx}}\frac{x(1-x)}{1-e^{-t(1-x)}}e^{- \beta t}e^{(\beta -\alpha )tx}. $$

Together with $U_{1}(x) \geq 0$, it leads to (3.8). Similarly, we can prove that (3.8) is still valid for the case $\alpha \geq M$ and $\beta \geq \alpha$.

If $\beta -\alpha \geq \frac{1}{2}$ and $\alpha \geq 0$ or $\beta -\alpha \leq -\frac{1}{2} \text{and} \beta \geq 0$, in view of Lemma 1, we can prove (3.8).

The proof is completed. □

Remark 1

Obviously, Theorem 3.1 is a generalization of [28, Theorem 3] for higher derivatives of $\psi (x)$.

Corollary 1

For $\alpha \in \mathbb{R}$, $m,n\in \mathbb{N}$ and $\lambda =\frac{(m+n-1)!}{(m-1)!(n-1)!}$, the functions

$$\begin{aligned}& f_{1}(x)=\lambda \psi _{n}(x+\alpha )\psi _{m}(x+\alpha )-\psi _{m+n}(x), \\& f_{2}(x)=\lambda \psi _{n}(x)\psi _{m}(x+\alpha )-\psi _{m+n}(x) \end{aligned}$$

are completely monotonic on $(-\alpha ,\infty )$ if and only if $\alpha \leq 0$.

Proof

The sufficient conditions of the assertion is proved in the proof of Theorem 3.1.

Next we shall prove the necessary conditions.

Suppose that $\alpha >0$. Since $f_{1}(x)$ and $f_{2}(x)$ are completely monotonic on $(0,\infty )$, we have $f_{1}(x),f_{2}(x) \geq 0$. On the other hand, it is easy to check that

$$\begin{aligned}& \lim_{x\rightarrow 0+}f_{1}(x) = \lambda \psi _{m} ( \alpha ) \psi _{n} (\alpha ) -\lim_{x\rightarrow 0+}\psi _{m+n}(x)=-\infty , \\& \lim_{x\rightarrow 0+}\frac{f_{2}(x)}{\psi _{n}(x)}= \lambda \psi _{n}( \alpha )-\lim_{x\rightarrow 0+}\frac{\psi _{m+n}(x)}{\psi _{m}(x)}=- \infty , \end{aligned}$$

which yields contradictions.

The proof is completed. □

Remark 2

The function $f_{1}(x)$ can be written equivalently as $h_{1}(t)=f_{1}(t-\alpha )$ $(t>0)$. By hypothesis, we get

$$ \alpha \leq t-\psi _{m+n}^{-1} \bigl[\lambda \psi _{n}(t)\psi _{m}(t) \bigr] $$

(3.10)

which yields $\alpha \leq 0$ as $t\rightarrow 0$. Furthermore, Corollary 1 clearly strengthens [28, Theorem 1].

Remark 3

In [27, Theorem 2.2], Batir proved the inequality

$$ \biggl(\frac{\psi _{n}(x+1/2)}{(n-1)!} \biggr)^{\frac{m}{n}}< \frac{ \psi _{m}(x)}{(m-1)!}< \biggl(\frac{\psi _{n}(x)}{(n-1)!} \biggr)^{\frac{m}{n}} $$

(3.11)

for $m\in \mathbb{N}$, $n=1,2,\ldots ,m-1$ and $x>0$. By Theorem 3.1, inequality (3.11) can be refined partially. Taking the logarithm in (3.11) yields

$$ \frac{1}{n}\ln \biggl(\frac{\psi _{n}(x+1/2)}{(n-1)!} \biggr)< \frac{1}{m}\ln \biggl(\frac{\psi _{m}(x)}{(m-1)!} \biggr)< \frac{1}{n} \ln \biggl(\frac{\psi _{n}(x)}{(n-1)!} \biggr). $$

(3.12)

Corollary 2

For m and $n\in \mathbb{N}$, we have the following inequalities

$$\begin{aligned}& \begin{aligned} \biggl(\frac{\psi _{n}(x+1/2)}{(n-1)!} \biggr)^{ \frac{m+n}{n}}& < \frac{\psi _{m}(x)}{(m-1)!} \frac{\psi _{n}(x+1/2)}{(n-1)!}< \frac{ \psi _{m+n}(x)}{(m+n-1)!} \\ &< \frac{\psi _{m}(x)}{(m-1)!}\frac{\psi _{n}(x)}{(n-1)!}< \biggl( \frac{\psi _{n}(x)}{(n-1)!} \biggr)^{\frac{m+n}{n}},\quad (m> n), \end{aligned} \end{aligned}$$

(3.13)

$$\begin{aligned}& \begin{aligned} \biggl(\frac{\psi _{n}(x+M)}{(n-1)!} \biggr)^{ \frac{m+n}{n}}& < \frac{\psi _{m}(x+M)}{(m-1)!} \frac{\psi _{n}(x+M)}{(n-1)!}< \frac{ \psi _{m+n}(x)}{(m+n-1)!} \\ &< \biggl(\frac{\psi _{n}(x)}{(n-1)!} \biggr)^{\frac{m+n}{n}}< \frac{\psi _{m}(x)}{(m-1)!} \frac{\psi _{n}(x)}{(n-1)!}, \quad (m< n) \end{aligned} \end{aligned}$$

(3.14)

for $x>0$, where M is defined by (3.1).

Proof

On the one hand, if $m> n$, a simple calculation shows that the right-hand side of (3.12) is equivalent to

$$ \frac{\psi _{m}(x)}{(m-1)!}\frac{\psi _{n}(x)}{(n-1)!}< \biggl( \frac{\psi _{n}(x)}{(n-1)!} \biggr)^{\frac{m+n}{n}}. $$

(3.15)

Similarly, the left-hand side of (3.12) is equivalent to

$$ \biggl(\frac{\psi _{n}(x+1/2)}{(n-1)!} \biggr)^{\frac{m+n}{n}}< \frac{\psi _{m}(x)}{(m-1)!}\frac{\psi _{n}(x+1/2)}{(n-1)!}. $$

(3.16)

By (3.15), (3.16) and Theorem 3.1, we see that (3.13) is proved.

On the other hand, if $m< n$, inequality (3.15) is reversed by a similar calculation. From the reversed inequality of (3.15), it follows that

$$ \biggl(\frac{\psi _{n}(x+M)}{(n-1)!} \biggr)^{\frac{m+n}{n}}< \frac{\psi _{m}(x+M)}{(m-1)!}\frac{\psi _{n}(x+M)}{(n-1)!}. $$

(3.17)

Taking into account the right-hand side of (3.12), the reversed inequality of (3.15), (3.17), and Theorem 3.1, we prove (3.14).

Consequently, the proof of the two inequalities is complete. □

4 Application

In [36], Elezović et al. derived that

$$ \psi _{1}(x)< e^{-\psi (x)} $$

(4.1)

by the fact that the function $e^{\psi (x+t)}-x$ is decreasing on $(0,\infty )$ for all $t>0$. In addition, [26, Lemma 1.2] provides a different proof of (4.1). Some extensions of (4.1) for higher-order derivatives of $\psi (x)$ can be found in [25, 27]. For example, it is given in [27, Theorem 2.1] that the inequality (4.1) was generalized to

$$ e^{-n\psi (x+\beta )}< \frac{\psi _{n}(x)}{(n-1)!}< e^{-n\psi (x+ \alpha )} $$

(4.2)

for $x>0$ and $n\in \mathbb{N}$, where $\beta =1/2$ and $\alpha =0$. In particular, inequality (4.2) was proved again by using monotonicity of functions involving the polygamma functions (see [37, Corollary 1]).

We introduce the divided differences of psi and polygamma functions (see [38]). For $n\in \mathbb{N}$, $s,t\in \mathbb{R}$, $r=\min \{s,t\}$ and $x\in (-r,\infty )$, we define

$$ \phi _{n}(x)= \textstyle\begin{cases} \frac{\psi _{n-1}(x+s)-\psi _{n-1}(x+t)}{t-s}, & t\neq s; \\ \psi _{n}(x+t), & t=s. \end{cases} $$

(4.3)

For the sake of consistency, we set $\psi _{0}(x)=-\psi (x)$.

Using Theorem 3.1 and inequality (4.2), we establish the following result.

Corollary 3

For $\beta \in \mathbb{R}$ and $n\in \mathbb{N}$, let the function $f_{3}(x)=(n-1)!e^{-n\psi (x+\beta )}-\psi _{n}(x)$ be defined on $(\max (-\beta ,0),\infty )$. Then the function $f_{3}(x)$ is decreasing on $(-\beta ,\infty )$ if $\beta \leq 0$; and is increasing on $(0,\infty )$ if $\beta \geq \frac{1}{2}$.

Proof

A simple computation gives

$$ f'_{3}(x)=\psi _{n+1}(x)-n(n-1)!\psi _{1}(x+\beta )e^{-n\psi (x+ \beta )}. $$

For $\beta \leq 0$, from the right-hand side of (4.2), we get

$$ f'_{3}(x)< \psi _{n+1}(x)-n\psi _{1}(x+\beta )\psi _{n}(x+\beta ), $$

and therefore, in the view of Theorem 3.1, we have $f'_{3}(x)<0$. By the same spirit, the left-hand side of (4.2) and Theorem 3.1 imply the case $\beta \geq \frac{1}{2}$.

This completes the proof. □

Remark 4

For $\lambda \neq 0$, $s,t\in \mathbb{R}$ and $r=\min \{s,t\}$, define the function Ψ for $x\in (-r,\infty )$

$$ \Psi (x;\lambda ,s,t)= \textstyle\begin{cases} [\frac{\Gamma (x+t)}{\Gamma (x+s)} ]^{ \frac{1}{\lambda (t-s)}}, & t\neq s; \\ e^{\frac{1}{\lambda }\psi (x+s)}, & t=s. \end{cases} $$

(4.4)

It was shown in [36] that the function $\Psi (x;1,s,t)$ is convex on $(-r,\infty )$ for $|t-s|<1$ and concave on the same interval for $|t-s|>1$. Since

$$ \Psi ''(x;\lambda ,s,t)= \frac{1}{\lambda ^{2}}\Psi (x; \lambda ,s,t) \bigl( \phi _{1}^{2}(x)-\lambda \phi _{2}(x)\bigr), $$

we deduce from [39, Theorem 3.1]:

(1) For $0<|t-s|<1$, the function $\Psi (x;\lambda ,s,t)$ is convex on $(-r,\infty )$ if and only if $\lambda \neq 0 \leq 1$ and concave on the same interval if and only if $\lambda \geq \frac{1}{|t-s|}$;

(2) For $|t-s|>1$, the function $\Psi (x;\lambda ,s,t)$ is convex on $(-r,\infty )$ if and only if $\lambda \neq 0 \leq \frac{1}{|t-s|}$ and concave on the same interval if and only if $\lambda \geq 1$;

(3) For $|t-s|=1$, the function $\Psi (x;\lambda ,s,t)$ is convex on $(-r,\infty )$ if and only if $\lambda \neq 0 \leq 1$ and concave on the same interval if and only if $\lambda \geq 1$;

(4) For $s=t$, the function $\Psi (x;\lambda ,s,t)$ is convex on $(-r,\infty )$ if and only if $\lambda \neq 0 \leq 1$.

In addition, it was proved in [36] that

$$ \Psi (x;1,s,t) \phi _{1}(x)< 1 $$

(4.5)

holds for $x>-r$ if $|t-s|<1$ and its reversed inequality is valid on $(-r,\infty )$ if $|t-s|>1$. Obviously, (4.5) is a generalization of (4.1).

In the following, we will prove the monotonicity of the function $z(x;\lambda ,s,t)=\Psi (x;\lambda ,s,t) \phi _{n}(x)$ and therefore extend (4.5) or the right-hand side of (4.2).

Theorem 4.1

For $\lambda \neq 0$, $s, t \in \mathbb{R}$, $r=\min (s,t)$ and $n \in \mathbb{N}$, the function $z(x;\lambda ,s,t)$ has the following monotonic properties:

(1) For $0<|t-s|<1$, the function $z(x;\lambda ,s,t)$ is increasing on $(-r,\infty )$ if and only if $1/\lambda \geq n$ and decreasing on the same interval if and only if $1/\lambda \leq n|t-s|$;

(2) For $|t-s|>1$, the function $z(x;\lambda ,s,t)$ is increasing on $(-r,\infty )$ if and only if $1/\lambda \geq n|t-s|$ and decreasing on the same interval if and only if $1/\lambda \leq n$;

(3) For $|t-s|=1$, the function $z(x;\lambda ,s,t)$ is increasing on $(-r,\infty )$ if and only if $1/\lambda \geq n$ and decreasing on the same interval if and only if $1/\lambda \leq n$;

(4) For $s=t$, the function $z(x;\lambda ,s,t)$ is increasing on $(-r,\infty )$ if and only if $1/\lambda \geq n$ and decreasing on the same interval if and only if $1/\lambda \leq 0$.

Proof

Differentiating $z(x;\lambda ,s,t)$ yields

$$ z'(x;\lambda ,s,t)=\Psi (x;\lambda ,s,t) \biggl(\frac{1}{\lambda } \phi _{1}(x) \phi _{n}(x)-\phi _{n+1}(x) \biggr). $$

This in combination with Theorem [39, Theorem 3.1] easily establishes the Theorem. □

Using Theorem 4.1, we have the following:

Corollary 4

For $s,t\in \mathbb{R}$, $r=\min \{s,t\}$ and $n \in \mathbb{N}$, we have the inequality

$$ \Psi \biggl(x;\frac{1}{n},s,t \biggr) \phi _{n}(x)< (n-1)! $$

(4.6)

for $x>-r$ if $|t-s|<1$ and its reversed inequality is valid on $(-r,\infty )$ if $|t-s|>1$.

Proof

Obviously, we only assume $s\neq t$. In view of Theorem 4.1, we only need to check

$$ \lim_{x \rightarrow \infty } \Psi \biggl(x;\frac{1}{n},s,t \biggr) \phi _{n}(x)=(n-1)!. $$

(4.7)

Applying the asymptotic formula (1.5), we obtain

$$ \lim_{x \rightarrow \infty }\frac{e^{n\psi (x)}}{x^{n}}=1. $$

Therefore, this together with [24, Lemma 4] establishes

$$ \lim_{x \rightarrow \infty }e^{n\psi (x+c)} \phi _{n}(x)=(n-1)!, \quad \text{for all } c\in \mathbb{R}. $$

(4.8)

According to [13, Corollary 1.4], the inequality

$$ e^{\psi (x+r)}< \biggl[\frac{\Gamma (x+t)}{\Gamma (x+s)} \biggr]^{1/(t-s)}< e^{ \psi (x+\frac{s+t}{2} )} $$

(4.9)

holds for $x>-r$, so that this combined with (4.8) yields (4.7).

Hence we complete the proof of this Theorem. □

Theorem 4.2

For $s,t\in \mathbb{R}$, $r=\min \{s,t\}$ and $c\in (-r,\infty )$, we have the double inequality

$$ \frac{e^{G_{s,t}(X_{s,t})}}{e^{H_{s,t}(X_{s,t})}}< \frac{e^{G_{s,t}(x)}}{e^{H_{s,t}(x)}}< \frac{\sqrt{2\pi e} e^{A_{c,s,t}-(s+t)/2}}{\Gamma (c+\frac{s+t}{2} )} $$

for $x>X_{s,t}$ if $|t-s|<1$ and its reversed inequality is valid on $(X_{s,t}, \infty )$ if $|t-s|>1$, where $X_{s,t}$ is the only zero of $1+\ln \Psi (x;1,s,t)$ on $(-r,\infty )$,

$$\begin{aligned}& G_{s,t}(x)= \textstyle\begin{cases} \frac{1}{t-s}\int _{c}^{x} \ln [ \frac{\Gamma (u+t)}{\Gamma (u+s)} ]\,du, & t\neq s, \\ \int _{c}^{x} \psi (u+s)\,du, & t=s; \end{cases}\displaystyle \\& A_{c,s,t}= \textstyle\begin{cases} \int _{c}^{\infty } \frac{1}{t-s}\ln [ \frac{\Gamma (u+t)}{\Gamma (u+s)} ]-\psi (u+\frac{s+t}{2} )\,du, & t\neq s, \\ 0, & t=s; \end{cases}\displaystyle \end{aligned}$$

(4.10)

and

$$ H_{s,t}(x)=\Psi (x;1,s,t)\ln \Psi (x;1,s,t)-x. $$

Proof

Let $g_{s,t}(x)=e^{G_{s,t}(x)}$, $h_{s,t}(x)=\ln \Psi (x;1,s,t)$ and $f_{s,t}(x)=g_{s,t}(x)e^{x-h_{s,t}(x) e^{h_{s,t}(x)}}$. Since $g_{s,t}'(x)=g_{s,t}(x) h_{s,t}(x)$ and $h_{s,t}'(x)=\phi _{1}(x)$, we obtain

$$ f_{s,t}'(x)=g_{s,t}(x)e^{x-h_{s,t}(x) e^{h_{s,t}(x)}} \bigl(1-\phi _{1}(x)e^{h_{s,t}(x)}\bigr) \bigl(1+h_{s,t}(x) \bigr). $$

(4.11)

Using the asymptotic formula (see [4])

$$ \frac{\Gamma (x+t)}{\Gamma (x+s)}=x^{t-s} \biggl(1- \frac{(s-t)(s+t-1)}{2x}+O \biggl(\frac{1}{x^{2}} \biggr) \biggr), \quad x \rightarrow \infty , $$

(4.12)

we get $\lim_{x\rightarrow \infty }h_{s,t}(x)=\infty$, and therefore by $h_{s,t}'(x)>0$ and $\lim_{x\rightarrow -r}h_{s,t}(x)=-\infty$, we conclude that $1+h_{s,t}(x)$ has a unique zero on $(-r,\infty )$.

Hence thanks to $h_{s,t}'(x)>0$, Corollary 4 and (4.11), we have the following statements:

(i) For $|t-s|<1$, $f_{s,t}(x)$ is increasing on $(X_{s,t},\infty )$ and decreasing on $(-r,X_{s,t})$,

(ii) For $|t-s|>1$, $f_{s,t}(x)$ is decreasing on $(X_{s,t},\infty )$ and increasing on $(-r,X_{s,t})$.

On the one hand, we check that

$$\begin{aligned}& \lim_{x\rightarrow \infty } \int _{c}^{x} \psi \biggl(u+\frac{s+t}{2} \biggr)\,du +x- h_{s,t}(x)e^{h_{s,t}(x)} \\& \quad =\frac{1}{2}+ \frac{1}{2} \ln (2\pi )- \frac{s+t}{2}- \ln \Gamma \biggl(c+ \frac{s+t}{2} \biggr). \end{aligned}$$

(4.13)

Case 1. $s\neq t$. Now we derive the asymptotic formula of $h_{s,t}(x)e^{h_{s,t}(x)}$. Taking the logarithm in (4.12), we get

$$ \begin{aligned} h_{s,t}(x)&= \frac{ [\ln \Gamma (x+t)-\ln \Gamma (x+s) ]}{t-s} \\ & =\ln x+\frac{1}{t-s}\ln \biggl(1-\frac{(s-t)(s+t-1)}{2x}+O \biggl( \frac{1}{x^{2}} \biggr) \biggr),\quad x\rightarrow \infty . \end{aligned} $$

(4.14)

Together with

$$ \ln (1+x)=x-\frac{x^{2}}{2}+O\bigl(x^{3}\bigr),\quad x\rightarrow 0, $$

we can rewrite (4.14) as

$$\begin{aligned} h_{s,t}(x)&= \frac{ [\ln \Gamma (x+t)-\ln \Gamma (x+s) ]}{t-s} \\ &=\ln x+ \frac{t+s-1}{2x}+O \biggl(\frac{1}{x^{2}} \biggr),\quad x\rightarrow \infty , \end{aligned}$$

(4.15)

which implies that

$$ h_{s,t}(x)e^{h_{s,t}(x)}=x \biggl(\ln x+ \frac{t+s-1}{2x}+O \biggl( \frac{1}{x^{2}} \biggr) \biggr)e^{\frac{t+s-1}{2x}+O ( \frac{1}{x^{2}} )}, \quad x \rightarrow \infty . $$

(4.16)

Therefore, by the aid of

$$ e^{x}=1+x+O\bigl(x^{2}\bigr),\quad x\rightarrow 0, $$

we obtain

$$ \begin{aligned} h_{s,t}(x)e^{h_{s,t}(x)}={}&x \ln x+\frac{t+s-1}{2}\ln x+ \frac{t+s-1}{2}+\frac{(t+s-1)^{2}}{4x} \\ &{} +O \biggl(\frac{\ln x}{x} \biggr)+O \biggl(\frac{1}{x^{2}} \biggr),\quad x \rightarrow \infty . \end{aligned} $$

(4.17)

Combining (1.4) with (4.17), we deduce that

$$\begin{aligned} &\ln \Gamma \biggl(x+\frac{s+t}{2} \biggr)+x- h_{s,t}(x)e^{h_{s,t}(x)} \\ \begin{aligned}&\quad =x \ln \biggl(1+\frac{s+t}{2x} \biggr) + \frac{t+s-1}{2} \ln \biggl(1+ \frac{s+t}{2x} \biggr) \\ &\qquad {}+\frac{1}{2}-(t+s) +\frac{1}{2} \ln (2\pi )+\frac{1}{12} \frac{1}{x+\frac{s+t}{2}} \end{aligned} \\ &\qquad {}-\frac{(t+s-1)^{2}}{4x}+O \biggl(\frac{\ln x}{x} \biggr)+O \biggl( \frac{1}{x^{2}} \biggr),\quad x\rightarrow \infty , \end{aligned}$$

(4.18)

which implies (4.13).

Case 2. $s=t$. Using (1.4) and the asymptotic formula (see [4])

$$ \psi \biggl(x+\frac{s+t}{2} \biggr)=\ln x+ \frac{s+t-1}{2x}+O \biggl( \frac{1}{x^{2}} \biggr),\quad x\rightarrow \infty , $$

(4.19)

we can easily prove (4.13).

On the other hand, we show that

$$ \lim_{x\rightarrow \infty } \biggl( G_{s,t}(x)- \int _{c}^{x} \psi \biggl(u+ \frac{s+t}{2} \biggr)\,du \biggr)= \textstyle\begin{cases} A_{c,s,t}, & t\neq s, \\ 0, & t=s; \end{cases} $$

(4.20)

where

$$ \int _{c}^{\infty } \frac{1}{t-s}\ln \biggl[ \frac{\Gamma (u+t)}{\Gamma (u+s)} \biggr]-\psi \biggl(u+\frac{s+t}{2}\biggr)\,du=A_{c,s,t}. $$

Note that the case $t=s$ is obvious. Then using (4.15) and (4.19), we get

$$ \frac{1}{t-s}\ln \biggl[\frac{\Gamma (x+t)}{\Gamma (x+s)} \biggr]- \psi \biggl(x+ \frac{s+t}{2} \biggr)=O \biggl(\frac{1}{x^{2}} \biggr),\quad x \rightarrow \infty , $$

which implies the exitance of constants C and $X>0$ such that

$$ \biggl\vert \frac{1}{t-s}\ln \biggl[\frac{\Gamma (x+t)}{\Gamma (x+s)} \biggr]-\psi \biggl(x+\frac{s+t}{2} \biggr) \biggr\vert \leq C \biggl\vert \frac{1}{x^{2}} \biggr\vert $$

for all $x>X$. It follows that

$$ \lim_{x\rightarrow \infty } \int _{x}^{\infty }\frac{1}{t-s}\ln \biggl[ \frac{\Gamma (u+t)}{\Gamma (u+s)} \biggr]-\psi \biggl(u+\frac{s+t}{2} \biggr)\,du=0, $$

so that $A_{c,s,t}$ is well defined. Hence, (4.20) is proved.

Finally, taking into consideration (4.13) and (4.20), we have

$$\begin{aligned}& \lim_{x\rightarrow \infty } \bigl( G_{s,t}(x)+x- h_{s,t}(x)e^{h_{s,t}(x)} \bigr) \\& \quad = \textstyle\begin{cases} \frac{1}{2}+\frac{1}{2} \ln (2\pi )-\frac{s+t}{2}- \ln \Gamma (c+ \frac{s+t}{2} )+A_{c,s,t}, \quad t\neq s; \\ \frac{1}{2}+\frac{1}{2} \ln (2\pi )-s- \ln \Gamma (c+s ),\quad t=s. \end{cases}\displaystyle \end{aligned}$$

(4.21)

Applying the monotonicity of $f_{s,t}(x)$ and (4.21), we complete the proof of this Theorem. □

Remark 5

Let $0.785003\leq s < t$. Using the inequality (see [13, Corollary 1.4])

$$ \psi (x+s)< \frac{1}{t-s} \bigl[\ln \Gamma (x+t)-\ln \Gamma (x+s) \bigr]< \psi \biggl(x+\frac{s+t}{2} \biggr),\quad x>-s, $$

we have $1+h_{s,t}(0)>1+\psi (s)\geq 0$, so that by $h_{s,t}'(x)>0$, Corollary 4 and (4.11) again, we conclude that $f_{s,t}(x)$ is increasing on $(0,\infty )$ if $|t-s|<1$ and decreasing on the same interval if $|t-s|>1$. Similarly, we have the inequality

$$ \begin{aligned} \biggl[\frac{\Gamma (x+t)}{\Gamma (x+s)} \biggr]^{ \frac{1}{t-s}}\ln \biggl[\frac{\Gamma (x+t)}{\Gamma (x+s)} \biggr]^{ \frac{1}{t-s}}< {}& \biggl[\frac{\Gamma (t)}{\Gamma (s)} \biggr]^{ \frac{1}{t-s}}\ln \biggl[\frac{\Gamma (t)}{\Gamma (s)} \biggr]^{ \frac{1}{t-s}}+x \\ &{} + \frac{1}{t-s} \int _{0}^{x} \ln \biggl[ \frac{\Gamma (u+t)}{\Gamma (u+s)} \biggr]\,du \end{aligned} $$

for $x>0$ if $|t-s|<1$ and its reversed inequality is valid on $(0,\infty )$ if $|t-s|>1$.

5 Discussion

Observing that Corollary 4 generalizes the right-hand side of (4.2), we conjecture that the left-hand side of (4.2) might be generalized to

$$ (n-1)!< \Psi \biggl(x+\frac{1}{2};\frac{1}{n},s,t \biggr) \phi _{n}(x) $$

for $x>-r$ if $|t-s|<1$ and that its reversed inequality might be valid on $(-r,\infty )$ if $|t-s|>1$, where $s,t\in \mathbb{R}$ and $r=\min \{s,t\}$.

We turn to pay attention to the class of strongly completely monotonic functions, which are introduced in [40]. A function $f: (0,\infty )\rightarrow \mathbb{R}$ is called strongly completely monotonic if it satisfies the more restrictive condition that $(-1)^{n} x^{n+1} f^{(n)}(x)$ is nonnegative and decreasing on $(0,\infty )$ for all $n\in \mathbb{N}$. Note that [40, Theorem 1] gives a characterization of strongly completely monotonic functions.

It was shown in [20] that the function $\psi _{1}^{2}(x)-\psi _{2}(x)$ is strongly completely monotonic on $(0,\infty )$. Inspired by this, we will determine necessary and sufficient conditions for λ such that the function $\Phi (x;\lambda ,s,t)$ is strongly completely monotonic on $(-r,\infty )$ for all fixed $s,t\in \mathbb{R}$ and $r=\min \{s,t\}$ in the future work.