1 Introduction

The classical Euler’s gamma and psi (or called digamma) functions are defined for \(x>0 \) by

$$ \Gamma ( x ) = \int_{0}^{\infty}e^{-t}t^{x-1}\,dt,\qquad \psi ( x ) =\frac{\Gamma^{\prime} ( x ) }{\Gamma ( x ) }, $$

respectively. Furthermore, the derivatives \(\psi^{\prime}, \psi ^{\prime\prime},\ldots, \psi^{ ( i ) }\) for \(i=1,2,\ldots \) , are called polygamma functions.

For convenience, we denote \(\psi_{n}(x)= ( -1 ) ^{n-1}\psi ^{ ( n ) }(x)\). It is well known that \(\psi_{n}(x)\) is strictly complete monotonic on \(( 0,\infty ) \); namely, \(( -1 ) ^{n-1}\psi^{ ( n ) } ( x ) >0\) for \(x>0\) and \(n\in\mathbb{N}\). Note that for the following integral and series representations (see [1], Sections 6.3, 6.4):

$$\begin{aligned}& \psi_{0} ( x ) =-\psi ( x ) =\gamma+ \int _{0}^{\infty} \frac{e^{-xt}-e^{-t}}{1-e^{-t}}\,dt=\gamma+ \frac{1}{x}-\sum_{k=1}^{\infty} \frac{x}{k ( x+k ) }, \end{aligned}$$
(1.1)
$$\begin{aligned}& \psi_{n} ( x ) = ( -1 ) ^{n-1}\psi^{ ( n ) } ( x ) = \int_{0}^{\infty}\frac{t^{n}}{1-e^{-t}} e^{-xt}\,dt=n! \sum_{k=0}^{\infty}\frac{1}{ ( x+k ) ^{n+1}}, \end{aligned}$$
(1.2)

it is easy to see that \(\psi_{n} ( 0^{+} ) =\infty\) for \(n\geq0\), \(\psi_{n}( \infty) =0\) for \(n\geq1\), and \(\psi_{0} ( \infty ) =-\infty\). Moreover, \(\psi_{n}^{\prime}=-\psi_{ ( n+1 ) }(x)<0\).

Let \(f:I\rightarrow\mathbb{R}\) be strictly monotone and \(a,b\in I\). Then the so-called integral f-mean of a and b is defined in [2] by

$$ I_{f} ( a,b ) =f^{-1} \biggl( \frac{\int_{a}^{b}f ( x )\,dx}{ b-a} \biggr) \quad\mbox{if }a\neq b,\quad\mbox{and}\quad I_{f} ( a,a ) =a. $$

For \(f=\psi\), Elezović and Pečarić [2], Theorem 6, proved an interesting result as follows.

Theorem EP

For \(x,a,b>0\), the digamma function ψ has the following properties:

  1. (i)

    \(I_{\psi^{\prime}} ( a,b ) \leq I_{\psi} ( a,b ) \); namely,

    $$ \bigl( \psi^{\prime} \bigr) ^{-1} \biggl( \frac{\int_{a}^{b}\psi^{\prime } ( x )\,dx}{b-a} \biggr) \leq\psi^{-1} \biggl( \frac{ \int_{a}^{b}\psi ( x )\,dx}{b-a} \biggr). $$
  2. (ii)

    \(x\mapsto I_{\psi} ( x+a,x+b ) -x\) is increasing concave, and

    $$ \lim_{x\rightarrow\infty} \bigl[ I_{\psi} ( x+a,x+b ) -x \bigr] = \frac{a+b}{2}. $$

Remark 1.1

It should be noted that, for \(a,b\in I\), if \(A ( a,b ) \) is a mean of a and b, then for \(x+a,x+b\in I\) the function \(x\mapsto A ( x+a,x+b ) -x\) is still a mean of a and b, which is due to the following relations:

$$\begin{aligned} \min ( a,b ) =&\min ( x+a,x+b ) -x\leq A ( x+a,x+b ) -x\\ \leq&\max ( x+a,x+b ) -x=\max ( a,b ) . \end{aligned}$$

Further, Batir [3], Theorem 2.7, gave a nice double inequality for \(I_{\psi_{n}} ( a,b ) \) as follows.

Theorem B

Let a and b be distinct positive real numbers and n be a positive integer. Then we have

$$ ( -1 ) ^{n}\psi^{ ( n+1 ) } \biggl( \frac{a+b}{2} \biggr) < ( -1 ) ^{n}\frac{\psi^{ ( n ) }(a)-\psi^{ ( n ) }(b)}{a-b}< ( -1 ) ^{n}\psi^{ ( n+1 ) } \bigl( S_{- ( n+1 ) } ( a,b ) \bigr) , $$

or, equivalently,

$$ S_{- ( n+1 ) } ( a,b ) < I_{\psi_{n+1}} ( a,b ) =\psi_{n+1}^{-1} \biggl( \frac{\int_{a}^{b}\psi_{n+1} ( t )\,dt}{b-a} \biggr) < \frac{a+b}{2}, $$

where

$$ S_{p} ( a,b ) =\left \{ \textstyle\begin{array}{@{}l@{\quad}l} ( \frac{a^{p}-b^{p}}{p ( a-b ) } ) ^{1/ ( p-1 ) },& \textit{if }p\neq0,1, \\ \frac{a-b}{\ln a-\ln b}, & \textit{if }p=0, \\ e^{-1} ( \frac{a^{a}}{b^{b}} ) ^{1/ ( a-b ) }, & \textit{if }p=1, \end{array}\displaystyle \right . $$
(1.3)

is the generalized logarithmic mean of a and b.

An improvement of Theorem B was given in [4], Theorem 1, and [5], Theorem 1, by Qi as follows.

Theorem Q1

For real numbers \(a,b>0\) with \(a\neq b\) and an integer \(n\geq0\), the inequality

$$ ( -1 ) ^{n}\psi^{ ( n ) } \bigl( S_{p} ( a,b ) \bigr) < ( -1 ) ^{n}\frac{\int_{a}^{b}\psi^{ ( n ) } ( t )\,dt}{b-a}\leq ( -1 ) ^{n} \psi^{ ( n ) } \bigl( S_{q} ( a,b ) \bigr) $$

or

$$ S_{p} ( a,b ) < I_{\psi_{n}} ( a,b ) =\psi _{n}^{-1} \biggl( \frac{\int_{a}^{b}\psi_{n} ( t )\,dt}{b-a} \biggr) \leq S_{q} ( a,b ) $$

holds if \(p\leq-n\) and \(q\geq-n+1\), where \(S_{p} ( a,b ) \) is given in (1.3).

Motivated by the results just mentioned, the main aim of this paper is to continue the study of some further properties of the mean \(I_{\psi_{n}} ( a,b ) \) and \(I_{\psi _{n}} ( x+a,x+b ) -x\). More precisely, we have the following.

Theorem 1.2

For \(a,b>0\) with \(a\neq b\), the sequence \(\{I_{\psi _{n}} ( a,b ) \}_{n\geq0}\) is strictly decreasing, and

$$ \lim_{n\rightarrow\infty}I_{\psi_{n}} ( a,b ) =\min ( a,b ) . $$

Theorem 1.3

Let a and b be distinct real numbers, and \(n\geq0\) be an integer. If \(\psi^{-1}_{n}\) is strictly decreasing with respected to x, then the function \(x\mapsto A_{\psi _{n}} ( x ) \) with

$$ A_{\psi_{n}} ( x ) =I_{\psi_{n}} ( x+a,x+b ) -x=\psi _{n}^{-1} \biggl( \frac{\int_{a}^{b}\psi_{n}(x+t)\,dt}{b-a} \biggr) -x $$
(1.4)

is strictly increasing from \(( -\min ( a,b ) ,\infty ) \) onto \(( \min ( a,b ) , ( a+b ) /2 ) \).

As a direct consequence, noting that \(\psi_{n}^{-1}\) is strictly decreasing, by Theorem 1.3 we have the following.

Corollary 1.4

Let a and b be distinct real numbers and \(n\geq0\) be an integer. Then for \(x>-\min ( a,b ) \) we have

$$ \psi_{n} \biggl( x+\frac{a+b}{2} \biggr) < \frac{\int_{a}^{b}\psi_{n}(x+t)\,dt}{ b-a}< \psi_{n} \bigl( x+\min ( a,b ) \bigr) , $$

where \(\min ( a,b ) \) and \(( a+b ) /2\) are the best constants. In particular, note that \(\psi_{0}=-\psi\), the double inequality

$$ \psi \bigl( x+\min ( a,b ) \bigr) < \frac{\int_{a}^{b}\psi(x+t)\,dt }{b-a}< \psi \biggl( x+ \frac{a+b}{2} \biggr) $$

or

$$ \exp\psi \bigl( x+\min ( a,b ) \bigr) < \biggl[ \frac{\Gamma ( x+b ) }{\Gamma ( x+a ) } \biggr] ^{1/ ( b-a ) }< \exp\psi \biggl( x+\frac{a+b}{2} \biggr) $$
(1.5)

holds for \(x>-\min ( a,b ) \) with the best constants \(\min ( a,b ) \) and \(( a+b ) /2\).

Suppose that \(a,b>0\) with \(a\neq b\) in Theorem 1.3. Utilizing the strictly increasing property of \(x\mapsto A_{\psi_{n}} ( x ) \) on \(( 0,\infty ) \), we have \(A_{\psi_{n}} ( 0 ) < A_{\psi _{n}} ( x ) < A_{\psi_{n}} ( \infty ) \); namely,

$$ I_{\psi_{n}} ( a,b ) =\psi_{n}^{-1} \biggl( \frac{\int _{a}^{b}\psi _{n}(t)\,dt}{b-a} \biggr) < \psi_{n}^{-1} \biggl( \frac{\int_{a}^{b}\psi _{n}(x+t)\,dt}{b-a} \biggr) -x< \frac{a+b}{2}. $$

Therefore, we conclude the following.

Corollary 1.5

Let \(a,b>0\) with \(a\neq b\) and \(n\geq0\) be an integer. Then for \(x>0\) we have

$$ \psi_{n} \biggl( x+\frac{a+b}{2} \biggr) < \frac{\int_{a}^{b}\psi_{n}(x+t)\,dt}{ b-a}< \psi_{n} \bigl( x+I_{\psi_{n}} ( a,b ) \bigr) , $$

where \(I_{\psi_{n}} ( a,b ) \) and \(( a+b ) /2\) are the best constants. Particularly, noting that \(\psi_{0}=-\psi\), the double inequality

$$ \psi \bigl( x+I_{\psi} ( a,b ) \bigr) < \frac{\int _{a}^{b}\psi (x+t)\,dt}{b-a}< \psi \biggl( x+ \frac{a+b}{2} \biggr) $$

or

$$\exp\psi \bigl( x+I_{\psi} ( a,b ) \bigr) < \biggl[ \frac {\Gamma ( x+b ) }{\Gamma ( x+a ) } \biggr] ^{1/ ( b-a ) }< \exp\psi \biggl( x+\frac{a+b}{2} \biggr) $$
(1.6)

holds for \(x>0\) with the best constants \(I_{\psi} ( a,b ) \) and \(( a+b ) /2\).

We would think it worth noticing that the double inequality (1.6) was first proved in [6] by Elezović et al.

Remark 1.6

The second Kershaw double inequality [7] states that

$$ \exp \bigl[ ( 1-s ) \psi ( x+\sqrt{s} ) \bigr] < \frac{\Gamma ( x+1 ) }{\Gamma ( x+s ) }< \exp \biggl[ ( 1-s ) \psi \biggl( x+\frac{s+1}{2} \biggr) \biggr] $$
(1.7)

for \(s\in ( 0,1 ) \) and \(x\geq0\). Some of the refinements, extensions, and generalizations of the double inequality (1.7) can be found in Qi’s review paper [8] and the references therein. It seems that our double inequality (1.5) may be the best second Kershaw type inequality, since the ranges of a and b in (1.5) are arbitrary real numbers, and the lower and upper bounds are sharp.

As an application of Theorem 1.3, we use it to prove a necessary and sufficient condition for the functions \(x\mapsto F_{a,b,c} ( x ) \) defined by (3.1) and \(x\mapsto 1/F_{a,b,c} ( x ) \) to be logarithmically monotonic on \(( -\rho,\infty ) \) with \(\rho=\min ( a,b,c ) \), which improves a well-known result.

2 Proofs of main results

This section we devote to the proof of our main results. First of all, let us give the following assertion, which is an improvement of Theorem 4 in [2].

Lemma 2.1

Let \(f\in C^{ ( 2 ) } ( I ) \). If f is strictly monotone, then the mean function

$$ A_{f} ( x ) =I_{f} ( a+x,b+x ) -x=f^{-1} \biggl( \frac{ \int_{a}^{b}f(x+t)\,dt}{b-a} \biggr) -x $$
(2.1)

is strictly increasing (decreasing) according to \(f^{\prime\prime }/f^{\prime}\) being strictly increasing (decreasing).

Proof

By the Jensen inequality we have

$$ f^{\prime} \biggl( f^{-1} \biggl( \frac{\int_{a}^{b}f(x+t)\,dt}{b-a} \biggr) \biggr) < ( > ) \frac{\int_{a}^{b}f^{\prime}(x+t)\,dt}{b-a} $$
(2.2)

if \(f^{\prime}\circ f^{-1}\) is strictly convex (concave).

Differentiation yields

$$ \frac{df^{\prime}(f^{-1}(x))}{dx}=f^{\prime\prime}\bigl(f^{-1}(x)\bigr) \frac{ d(f^{-1}(x))}{dx}=\frac{f^{\prime\prime}(f^{-1}(x))}{f^{\prime}(f^{-1}(x)) }=\frac{f^{\prime\prime}(u)}{f^{\prime}(u)}, $$

where \(u=f^{-1}(x)\). This shows that \(f^{\prime}\circ f^{-1}\) is strictly convex if and only if both f and \(f^{\prime\prime}/f^{\prime}\) are either increasing or decreasing, and concave if and only if one of f and \(f^{\prime\prime}/f^{\prime}\) is increasing, while the other is decreasing.

Case 1: Both f and \(f^{\prime\prime}/f^{\prime}\) are increasing. Then \(f^{\prime}>0\) and \(f^{\prime}\circ f^{-1}\) is convex, and it follows from (2.2) that

$$ \frac{dA_{f} ( x ) }{dx}\mathbf{=} \frac{\int_{a}^{b}f^{ \prime}(x+t)\,dt}{b-a} \Big/ f^{\prime} \biggl( f^{-1} \biggl( \frac{ \int_{a}^{b}f(x+t)\,dt}{b-a} \biggr) \biggr) -1>0. $$

Case 2: f is decreasing and \(f^{\prime\prime}/f^{\prime}\) is increasing. Then \(f^{\prime}<0\) and \(f^{\prime}\circ f^{-1}\) is concave and by (2.2) we also have \(dA_{f} ( t ) /dt>0\).

Case 3: Both f and \(f^{\prime\prime}/f^{\prime}\) are decreasing. Then \(f^{\prime}<0\) and \(f^{\prime}\circ f^{-1}\) is convex. Similarly, we have \(dA_{f} ( t ) /dt<0\).

Case 4: f is increasing and \(f^{\prime\prime}/f^{\prime}\) is decreasing. Then \(f^{\prime}>0\) and \(f^{\prime}\circ f^{-1}\) is concave. Obviously, we see that \(dA_{f} ( t ) /dt<0\).

To sum up, if \(f^{\prime\prime}/f^{\prime}\) is increasing (decreasing), then so is \(A_{f}\), which completes the proof. □

The following lemma is useful for our main proof, which is a generalization of Lemma 1.4 in [3] and Lemma 4 in [9].

Lemma 2.2

Let \(A: ( 0,\infty ) \times ( 0,\infty ) \rightarrow ( 0,\infty ) \) be a differentiable one-order homogeneous mean. Then, for all \(x+t,y+t\in ( 0,\infty ) \), we have

$$ \lim_{t\rightarrow\infty} \bigl( A(x+t,y+t)-t \bigr) =px+ ( 1-p ) y, $$
(2.3)

where \(p=A_{x} ( 1,1 ) \in [ 0,1 ] \). In particular, if \(A ( x,y ) \) is symmetric with respect to x and y, then

$$ \lim_{t\rightarrow\infty} \bigl( A(x+t,y+t)-t \bigr) = \frac{x+y}{ 2}. $$
(2.4)

Proof

Using homogeneity of \(A(x,y)\) and the L’Hospital rule yield

$$\begin{aligned} \lim_{t\rightarrow\infty} \bigl( A(x+t,y+t)-t \bigr) &= \lim_{t\rightarrow\infty}\frac{A(t^{-1}x+1,t^{-1}y+1)-1}{t^{-1}} \\ &\stackrel{t^{-1}=u}{=} \lim_{u\rightarrow0} \frac{A(ux+1,uy+1)-1}{u} \\ &=\lim_{u\rightarrow0}\frac{\partial A\mathbf{(}ux+1,uy+1)}{ \partial u}=xA_{x} ( 1,1 ) +yA_{y} ( 1,1 ) . \end{aligned}$$

In addition, it follows from [10] that

$$ A_{x} ( x,x ) ,A_{y} ( x,x ) \in [ 0,1 ] \quad\mbox{and}\quad A_{x} ( x,x ) +A_{y} ( x,x ) =1. $$
(2.5)

Putting the above together, we get (2.3).

In particular, if A is symmetric, that is, \(A ( x,y ) =A ( y,x ) \), then we clearly see that \(A_{x} ( x,y ) =A_{y} ( y,x ) \), and so \(A_{x} ( x,x ) =A_{y} ( x,x ) \). It follows from (2.5) that \(A_{x} ( x,x ) =A_{y} ( x,x ) =1/2\), and then (2.4) holds. The proof is complete. □

Lemma 2.3

Let \(\psi_{n}= ( -1 ) ^{n-1}\psi^{ ( n ) }\) for \(n\in\mathbb{N}\). Then all the following statements are true, and mutually equivalent.

  1. (i)

    the sequence \(\{\psi_{n+1}/\psi_{n}\}_{n\in\mathbb{N}}\) is strictly increasing;

  2. (ii)

    the function \(x\mapsto\psi_{n+1} ( x ) /\psi _{n} ( x ) \) is strictly decreasing on \(( 0,\infty ) \);

  3. (iii)

    the function \(x\mapsto\psi_{n} ( x ) \) is log-convex on \(( 0,\infty ) \).

Proof

(i) It suffices to prove \(\psi_{n+2}/\psi_{n+1}>\psi_{n+1}/\psi _{n}\) for \(n\in\mathbb{N}\), which is equivalent to \(\psi_{n+2}\psi_{n}-\psi _{n+1}^{2}>0\). By virtue of the integral representation given in (1.2), we get

$$\begin{aligned} \psi_{n+2}\psi_{n}-\psi_{n+1}^{2} =& \int_{0}^{\infty}\frac{t^{n+2}}{ 1-e^{-t}}e^{-xt}\,dt \int_{0}^{\infty}\frac{t^{n}}{1-e^{-t}}e^{-xt}\,dt- \biggl( \int_{0}^{\infty}\frac{t^{n+1}}{1-e^{-t}}e^{-xt}\,dt \biggr) ^{2} \\ =&\frac{1}{2} \int_{0}^{\infty} \int_{0}^{\infty}\frac{t^{n}s^{n} ( t-s ) ^{2} }{ ( 1-e^{-t} ) ( 1-e^{-s} ) }e^{-x ( t+s ) }\,dt\,ds>0, \end{aligned}$$

which proves assertion (i).

(ii) Note that \(\psi_{n}^{\prime}=-\psi_{n+1}\), we have

$$ \biggl( \frac{\psi_{n+1}}{\psi_{n}} \biggr) ^{\prime}=\frac{\psi _{n+1}^{\prime}\psi_{n}-\psi_{n+1}\psi_{n}^{\prime}}{\psi_{n}^{2}}= \frac{-\psi_{n+2}\psi_{n}+\psi_{n+1}^{2}}{\psi_{n}^{2}}< 0, $$

which implies that the second assertion is true.

(iii) Differentiation gives

$$ ( \ln\psi_{n} ) ^{\prime}=\frac{\psi_{n}^{\prime}}{\psi_{n}} =- \frac{\psi_{n+1}}{\psi_{n}}, \qquad ( \ln\psi_{n} ) ^{\prime\prime}=- \biggl( \frac{\psi_{n+1}}{\psi_{n}} \biggr) ^{\prime}>0, $$

which completes the proof. □

Now we are in a position to prove our main results.

Proof of Theorem 1.2

We first prove that the sequence \(\{I_{\psi_{n}} ( a,b ) \} _{n\geq 0}\) is strictly decreasing, which means that for \(n\geq0\) the inequality

$$ \psi_{n}^{-1} \biggl( \frac{\int_{a}^{b}\psi_{n} ( x )\,dx}{b-a} \biggr) > \psi_{n+1}^{-1} \biggl( \frac{\int_{a}^{b}\psi_{n+1} ( x )\,dx}{b-a} \biggr) $$
(2.6)

holds for \(a,b>0\) with \(a\neq b\). By the Jensen inequality, it suffices to check that \(\psi_{n+1}\circ\psi_{n}^{-1}\) is convex on \(( 0,\infty ) \). In fact, by Lemma 2.3 we have

$$\begin{aligned}& \frac{d}{dx}\psi_{n+1} \bigl( \psi_{n}^{-1} ( x ) \bigr) =\frac{ \psi_{n+1}^{\prime} ( \psi_{n}^{-1} ( x ) ) }{\psi _{n}^{\prime} ( \psi_{n}^{-1} ( x ) ) }=\frac{\psi _{n+2} ( \psi_{n}^{-1} ( x ) ) }{\psi_{n+1} ( \psi _{n}^{-1} ( x ) ) }, \\& \frac{d^{2}}{dx^{2}}\psi_{n+1} \bigl( \psi_{n}^{-1} ( x ) \bigr) = \biggl( \frac{\psi_{n+2} ( u ) }{\psi_{n+1} ( u ) } \biggr) ^{\prime} \frac{1}{\psi_{n}^{\prime} ( u ) }=- \biggl( \frac{\psi_{n+2} ( u ) }{\psi_{n+1} ( u ) } \biggr) ^{\prime} \frac{1}{\psi_{n+1} ( u ) }>0, \end{aligned}$$

where \(u=\psi_{n}^{-1} ( x ) \). This means that \(\psi _{n+1}\circ \psi_{n}^{-1}\) is convex, which proves inequality (2.6).

Taking \(p=-n\) and \(q=-n+1\) in Theorem Q1 gives

$$ S_{-n} ( a,b ) < \psi_{n}^{-1} \biggl( \frac{\int_{a}^{b}\psi _{n} ( t )\,dt}{b-a} \biggr) < S_{-n+1} ( a,b ) . $$
(2.7)

Considering that \(\lim_{p\rightarrow-\infty}S_{p} ( a,b ) =\min ( a,b ) \) in [11], then we get

$$ \lim_{n\rightarrow\infty}\psi_{n}^{-1} \biggl( \frac{\int_{a}^{b}\psi _{n} ( t )\,dt}{b-a} \biggr) =\min ( a,b ) , $$

which completes the proof. □

Proof of Theorem 1.3

To prove \(x\mapsto A_{\psi_{n}} ( x ) \) is strictly increasing on \(( -\min ( a,b ) ,\infty ) \), by Lemma 2.1 it suffices to check that \(\psi_{n}^{\prime\prime}/\psi_{n}^{\prime}\) is strictly increasing on \(( 0,\infty ) \). In fact, since \(\psi _{n}^{\prime}=-\psi_{n+1}\) we see that \(\psi_{n}^{\prime\prime }/\psi _{n}^{\prime}=-\psi_{n+2}/\psi_{n+1}\) is strictly increasing on \(( 0,\infty ) \) by the second assertion of Lemma 2.3. Thus, the increasing property of \(A_{\psi_{n}}\) follows.

As mentioned in the introduction, we see that \(\psi_{n} ( 0^{+} ) =\infty\) for \(n\geq0\), and so \(\psi_{n}^{-1} ( \infty ) =0\). Note that the symmetry of a and b, without loss of generality we may assume that \(b>a\). Then we have

$$ \lim_{x\rightarrow-a^{+}}\frac{\int_{a}^{b}\psi_{n} ( x+t )\,dt}{ b-a}=\lim_{x\rightarrow-a^{+}} \frac{ ( -1 ) ^{n-1} ( \psi ^{ ( n-1 ) } ( x+b ) -\psi^{ ( n-1 ) } ( x+a ) ) }{b-a}=\infty, $$

which implies

$$\begin{aligned} \lim_{x\rightarrow-a^{+}}A_{\psi_{n}} ( x ) =&\lim_{x\rightarrow -a^{+}} \psi_{n}^{-1} \biggl( \frac{\int_{a}^{b}\psi_{n} ( x+t )\,dt}{ b-a} \biggr) -\lim _{x\rightarrow-a^{+}}x \\ =&\psi_{n}^{-1} \biggl( \lim_{x\rightarrow-a^{+}} \frac{\int_{a}^{b}\psi _{n} ( x+t )\,dt}{b-a} \biggr) +a=\psi_{n}^{-1} ( \infty ) +a=a. \end{aligned}$$

To obtain \(\lim_{x\rightarrow\infty}A_{\psi_{n}} ( x ) = ( a+b ) /2\), we use (2.7) to derive that

$$ S_{-n} ( x+b,x+a ) -x< \psi_{n}^{-1} \biggl( \frac{\int _{a}^{b}\psi _{n} ( x+t )\,dt}{b-a} \biggr) -x< S_{-n+1} ( x+b,x+a ) -x. $$

Note that the generalized logarithmic mean \(S_{p} ( x,y ) \) is homogeneous and symmetric, it follows from Lemma 2.2 that

$$ \lim_{x\rightarrow\infty} \bigl( S_{p} ( x+b,x+a ) -x \bigr) = \frac{a+b}{2}. $$

Therefore, we conclude that \(\lim_{x\rightarrow\infty}A_{\psi _{n}} ( x ) = ( a+b ) /2\), which completes the proof. □

3 An application

A function f is said to be completely monotonic on an interval I if f has derivatives of all orders on I and \(( -1 ) ^{n} ( f ( x ) ) ^{ ( n ) }\geq0\) for \(x\in I\) and n ≥0 (see [12]). A positive function f is called logarithmically completely monotonic on an interval I if f has derivatives of all orders on I and its logarithm lnf satisfies \(( -1 ) ^{n} ( \ln f ( x ) ) ^{ ( n ) }\geq0\) for all \(n\in\mathbb{N}\) on I (see [13]). For convenience, we denote the sets of the completely monotonic functions and the logarithmically completely monotonic functions on I by \(\mathcal{C} [ I ] \) and \(\mathcal{L} [ I ] \), respectively. Qi in [14], Theorem 1, [15], Theorem 1, investigated the logarithmically complete monotonicity of the functions

$$ x\mapsto F_{a,b,c} ( x ) =\left \{ \textstyle\begin{array}{@{}l@{\quad}l} ( \frac{\Gamma ( x+b ) }{\Gamma ( x+a ) } ) ^{1/ ( a-b ) }e^{\psi ( x+c ) }, & \mbox{if }a\neq b, \\ e^{\psi ( x+c ) -\psi ( x+a ) }, & \mbox{if }a=b, \end{array}\displaystyle \right . $$
(3.1)

and \(x\mapsto1/F_{a,b,c} ( x ) \). Furthermore, he concluded the following result.

Theorem Q2

Let \(a,b\), and c be real numbers and \(\rho=\min ( a,b,c ) \). If \(\theta ( t ) \) is an implicit function defined by

$$ e^{t}-t=e^{\theta ( t ) }-\theta ( t ) $$

on \(( -\infty,\infty ) \), then \(\theta ( t ) \) is decreasing and \(t\theta ( t ) <0\) for \(\theta ( t ) \neq t\). Moreover:

  1. (1)

    \(F_{a,b,c} ( x ) \in\mathcal{L} [ ( -\rho,\infty ) ] \) if

    $$\begin{aligned} ( a,b,c ) \in& \{ c\geq a,c\geq b \} \cup \bigl\{ c\geq a,0\geq c-b\geq\theta ( c-a ) \bigr\} \\ &{}\cup \bigl\{ c\leq a,c-b\geq\theta ( c-a ) \bigr\} \backslash \{ a=b=c \} . \end{aligned}$$
  2. (2)

    \(1/F_{a,b,c} ( x ) \in\mathcal{L} [ ( -\rho,\infty ) ] \) if

    $$\begin{aligned} ( a,b,c ) \in& \{ c\leq a,c\leq b \} \cup \bigl\{ c\geq a,c-b\leq\theta ( c-a ) \bigr\} \\ &{}\cup \bigl\{ c\leq a,0\leq c-b\leq\theta ( c-a ) \bigr\} \backslash \{ a=b=c \} . \end{aligned}$$

Later, Qi and Guo in [16], Theorem 1, [17], Theorem 1, proved another result concerning the logarithmically complete monotonicity of the functions \(x\mapsto F_{a,b,c} ( x ) \) and \(x\mapsto1/F_{a,b,c} ( x ) \) for \(x>-\min ( a,b,c ) \), where \(c=c ( a,b ) \) is a constant depending on a and b. More precisely, they showed the following.

Theorem QG

Let a and b be two real numbers with \(a\neq b\) and \(c ( a,b ) \) be a constant depending on a and b.

  1. (1)

    If \(c ( a,b ) \leq\min ( a,b ) \), then \(1/F_{a,b,c} ( x ) \in\mathcal{L} [ ( -c ( a,b ) ,\infty ) ] \).

  2. (2)

    \(F_{a,b,c} ( x ) \in\mathcal{L} [ ( -\min ( a,b ) ,\infty ) ] \) if and only if \(c ( a,b ) \geq ( a+b ) /2\).

We would like to remark that the result in Theorem Q2 is rather interesting but somewhat complicated. Theorem QG shows that c is a constant depending on a and b, and \(c ( a,b ) \leq\min ( a,b ) \) is only sufficient for \(1/F_{a,b,c} ( x ) \in\mathcal{L} [ ( -c ( a,b ) ,\infty ) ] \). Here, we apply Theorem 1.3 to deduce that c is a constant independent of a and b, and \(c\leq\min ( a,b ) \) is also necessary for \(1/F_{a,b,c} ( x ) \in\mathcal{L} [ ( -c ( a,b ) ,\infty ) ] \). This improved result can be restated as follows.

Theorem 3.1

Let \(a,b\), and c be real numbers, and \(\rho=\min ( a,b,c ) \). Then \(1/F_{a,b,c} ( x ) \in\mathcal{L} [ ( -\rho,\infty ) ] \) if and only if \(c\leq\min ( a,b ) \), while \(F_{a,b,c} ( x ) \in\mathcal{L} [ ( -\rho,\infty ) ] \) if and only if \(c\geq ( a+b ) /2\).

Proof

For \(a\neq b\), we have

$$ \ln F_{a,b,c} ( x ) =\psi ( x+c ) -\frac{\ln\Gamma ( x+b ) -\ln\Gamma ( x+a ) }{b-a}=\psi ( x+c ) - \frac{\int_{a}^{b}\psi ( x+t )\,dt}{b-a} $$

and

$$\begin{aligned} ( -1 ) ^{n} \bigl( \ln F_{a,b,c} ( x ) \bigr) ^{ ( n ) } =& ( -1 ) ^{n}\psi^{ ( n ) } ( x+c ) -\frac{ ( -1 ) ^{n}\int_{a}^{b}\psi^{ ( n ) } ( x+t )\,dt}{b-a} \\ =&\frac{\int_{a}^{b}\psi_{n} ( x+t )\,dt}{b-a}-\psi_{n} ( x+c ) \\ =&\frac{ ( b-a ) ^{-1}\int_{a}^{b}\psi_{n} ( x+t )\,dt-\psi_{n} ( x+c ) }{\psi_{n}^{-1} ( ( b-a ) ^{-1}\int_{a}^{b}\psi_{n} ( x+t )\,dt ) - ( x+c ) } \bigl( A_{\psi_{n}} ( x ) -c \bigr) , \end{aligned}$$

where \(\psi_{n}= ( -1 ) ^{n-1}\psi^{ ( n ) }\) and \(A_{\psi_{n}} ( x ) \) is defined by (1.4).

Since \(\psi_{n}^{\prime}=-\psi^{ ( n+1 ) }<0\), \(( \psi _{n}^{-1} ) ^{\prime}<0\), which means that \(\psi_{n}^{-1}\) is strictly decreasing on \(( 0,\infty ) \). This yields

$$ \frac{ ( b-a ) ^{-1}\int_{a}^{b}\psi_{n} ( x+t )\,dt-\psi _{n} ( x+c ) }{\psi_{n}^{-1} ( ( b-a ) ^{-1}\int_{a}^{b}\psi_{n} ( x+t )\,dt ) -\psi_{n}^{-1} ( \psi_{n} ( x+c ) ) }< 0 $$

for \(x\in ( -\rho,\infty ) \). Therefore, we have

$$ \operatorname{sgn} \bigl( ( -1 ) ^{n} \bigl( \ln F_{a,b,c} ( x ) \bigr) ^{ ( n ) } \bigr) =\operatorname{sgn} \bigl( c-A_{\psi_{n}} ( x ) \bigr) . $$

Theorem 1.3 tells that \(x\mapsto A_{\psi_{n}} ( x ) =I_{\psi_{n}} ( x+a,x+b ) -x\) is strictly increasing from \(( -\min ( a,b ) ,\infty ) \) onto \(( \min ( a,b ) , ( a+b ) /2 ) \), which implies that

$$ \operatorname{sgn} \bigl( c-A_{\psi_{n}} ( x ) \bigr) \leq0\quad\iff\quad c\leq \inf A_{\psi_{n}} ( x ) =\min ( a,b ) $$

and

$$ \operatorname{sgn} \bigl( c-A_{\psi_{n}} ( x ) \bigr) \geq0\quad\iff\quad c\geq \sup A_{\psi_{n}} ( x ) =\frac{a+b}{2}. $$

It is obvious that these are also true for \(a=b\). This completes the proof. □