Monotonicity of a mean related to polygamma functions with an application

Abstract

(ProQuest: ... denotes formulae and/or non-USASCII text omitted; see image)

Let ...... ...... (......), where ...... denotes the psi and polygamma functions. We prove that for ...... and two different real numbers a and b, the function ...... is strictly increasing from ...... onto ......, which generalizes a well-known result. As an application, the complete monotonicity for a ratio of gamma functions is improved.

Yang and Zheng Journal of Inequalities and Applications ( 2016) 2016:216 DOI 10.1186/s13660-016-1155-4

Monotonicity of a mean related to polygamma functions with an application

Zhen-Hang Yang1,2 and Shen-Zhou Zheng1*

*Correspondence: [email protected]

1Department of Mathematics, Beijing Jiaotong University, Beijing, 100044, ChinaFull list of author information is available at the end of the article

1 Introduction

The classical Eulers gamma and psi (or called digamma) functions are dened for x > by

(x) =

ettx dt,

(x) =

(x)

(x) ,

respectively. Furthermore, the derivatives , , . . . , (i) for i = , , . . . , are called poly-gamma functions.

For convenience, we denote n(x) = ()n(n)(x). It is well known that n(x) is strictly complete monotonic on (, ); namely, ()n(n)(x) > for x > and n

N. Note that

for the following integral and series representations (see [], Sections ., .):

(x) = (x) = +

ext et

et dt =

+

x

k=xk(x + k), (.)

n(x) = ()n(n)(x) =

tn et ext dt = n!

k=(x + k)n+ , (.)

it is easy to see that n(+) = for n , n() = for n , and () = . Moreover, n = (n+)(x) < .

Let f : I

R be strictly monotone and a, b I. Then the so-called integral f -mean of a and b is dened in [] by

2016 Yang and Zheng. This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.

Yang and Zheng Journal of Inequalities and Applications ( 2016) 2016:216 Page 2 of 10

If (a, b) = f

ba f (x) dx b a

if a = b, and If (a, a) = a.

For f = , Elezovi and Peari [], Theorem , proved an interesting result as follows.

Theorem EP For x, a, b > , the digamma function has the following properties:(i) I (a, b) I(a, b); namely,

ba (x) dx b a

ba (x) dx b a

.

(ii) x I(x + a, x + b) x is increasing concave, and

lim

x

I(x + a, x + b) x

= a + b

.

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

min(a, b) = min(x + a, x + b) x A(x + a, x + b) x

max(x + a, x + b) x = max(a, b).

Further, Batir [], Theorem ., gave a nice double inequality for In(a, b) as follows.

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

()n(n+)

a + b

< ()n

(n)(a) (n)(b)a b < ()n

(n+)

S(n+)(a, b)

,

or, equivalently,

S(n+)(a, b) < In+(a, b) = n+

ba n+(t) dt

b a

< a + b

,

where

Sp(a, b) =

( apbpp(ab))/(p), if p = , ,

ab

ln aln b , if p = , e(aabb )/(ab), if p = ,

(.)

is the generalized logarithmic mean of a and b.

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

Theorem Q For real numbers a, b > with a = b and an integer n , the inequality

()n(n)

Sp(a, b)

< ()n

ba (n)(t) dtb a ()n

(n)

Sq(a, b)

Yang and Zheng Journal of Inequalities and Applications ( 2016) 2016:216 Page 3 of 10

or

Sp(a, b) < In(a, b) = n

ba n(t) dt b a

Sq(a, b)

holds if p n and q n + , where Sp(a, b) is given in (.).

Motivated by the results just mentioned, the main aim of this paper is to continue the study of some further properties of the mean In(a, b) and In(x + a, x + b) x. More precisely, we have the following.

Theorem . For a, b > with a = b, the sequence {In(a, b)}n is strictly decreasing, and

lim

n In(a, b) =

min(a, b).

Theorem . Let a and b be distinct real numbers, and n be an integer. If n is strictly decreasing with respected to x, then the function x An(x) with

An (x) = In(x + a, x + b) x = n

ba n(x + t) dt b a

x (.)

is strictly increasing from ( min(a, b), ) onto (min(a, b), (a + b)/).

As a direct consequence, noting that n is strictly decreasing, by Theorem . we have the following.

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

n x + a + b

<

ba n(x + t) dtb a <

n

x + min(a, b)

,

where min(a, b) and (a + b)/ are the best constants. In particular, note that = , the double inequality

x + min(a, b) <

ba (x + t) dtb a <

x + a + b

or

exp

x + min(a, b) <

(x + b) (x + a)

/(ba)< exp

x + a + b

(.)

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

Suppose that a, b > with a = b in Theorem .. Utilizing the strictly increasing property of x An(x) on (, ), we have An() < An(x) < An(); namely,

In (a, b) = n

ba n(t) dt b a

< n

ba n(x + t) dt b a

x < a + b

.

Therefore, we conclude the following.

Yang and Zheng Journal of Inequalities and Applications ( 2016) 2016:216 Page 4 of 10

Corollary . Let a, b > with a = b and n be an integer. Then for x > we have

n x + a + b

<

ba n(x + t) dtb a <

n

x + In(a, b)

,

where In(a, b) and (a + b)/ are the best constants. Particularly, noting that = , the double inequality

x + I(a, b)

<

ba (x + t) dtb a <

x + a + b

or

exp

x + I(a, b)

<

(x + b) (x + a)

/(ba)< exp

x + a + b

(.)

holds for x > with the best constants I(a, b) and (a + b)/.

We would think it worth noticing that the double inequality (.) was rst proved in [] by Elezovi et al.

Remark . The second Kershaw double inequality [] states that

exp

( s)(x + s)

<

(x + ) (x + s) <

exp

( s)

x + s +

(.)

for s (, ) and x . Some of the renements, extensions, and generalizations of the double inequality (.) can be found in Qis review paper [] and the references therein.

It seems that our double inequality (.) may be the best second Kershaw type inequality, since the ranges of a and b in (.) are arbitrary real numbers, and the lower and upper bounds are sharp.

As an application of Theorem ., we use it to prove a necessary and sucient condition for the functions x Fa,b,c(x) dened by (.) and x /Fa,b,c(x) to be logarithmically monotonic on (, ) with = 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 in [].

Lemma . Let f C()(I). If f is strictly monotone, then the mean function

Af (x) = If (a + x, b + x) x = f

ba f (x + t) dt b a

x (.)

is strictly increasing (decreasing) according to f /f being strictly increasing (decreasing).

Yang and Zheng Journal of Inequalities and Applications ( 2016) 2016:216 Page 5 of 10

Proof By the Jensen inequality we have

f

f ba f (x + t) dt b a

< (>)

ba f (x + t) dtb a (.)

if f f is strictly convex (concave).Dierentiation yields

df (f (x))

dx = f

f (x)

d(f (x))

dx =

f (f (x))

f (f (x)) =

f (u)

f (u) ,

where u = f (x). This shows that f f is strictly convex if and only if both f and f /f are either increasing or decreasing, and concave if and only if one of f and f /f is increasing,

while the other is decreasing.

Case : Both f and f /f are increasing. Then f > and f f is convex, and it follows from (.) that

dAf (x)

dx =

ba f (x + t) dt b a

f

f ba f (x + t) dt b a

> .

Case : f is decreasing and f /f is increasing. Then f < and f f is concave and by (.) we also have dAf (t)/dt > .

Case : Both f and f /f are decreasing. Then f < and f f is convex. Similarly, we have dAf (t)/dt < .

Case : f is increasing and f /f is decreasing. Then f > and f f is concave. Obviously, we see that dAf (t)/dt < .

To sum up, if f /f is increasing (decreasing), then so is Af , which completes the proof.

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

Lemma . Let A : (, ) (, ) (, ) be a dierentiable one-order homogeneous mean. Then, for all x + t, y + t (, ), we have

lim

t

A(x + t, y + t) t = px + ( p)y, (.)

where p = Ax(, ) [, ]. In particular, if A(x, y) is symmetric with respect to x and y, then

lim

t

A(x + t, y + t) t = x + y . (.)

Proof Using homogeneity of A(x, y) and the LHospital rule yield

lim

t

A(x + t, y + t) t = limt

A(tx + , ty + ) t

t=u

= lim

u

A(ux + , uy + ) u

= lim

u

A(ux + , uy + )

u = xAx(, ) + yAy(, ).

Yang and Zheng Journal of Inequalities and Applications ( 2016) 2016:216 Page 6 of 10

In addition, it follows from [] that

Ax(x, x), Ay(x, x) [, ] and Ax(x, x) + Ay(x, x) = . (.)

Putting the above together, we get (.).In particular, if A is symmetric, that is, A(x, y) = A(y, x), then we clearly see that Ax(x, y) =

Ay(y, x), and so Ax(x, x) = Ay(x, x). It follows from (.) that Ax(x, x) = Ay(x, x) = /, and then (.) holds. The proof is complete.

Lemma . Let n = ()n(n) for n

N. Then all the following statements are true, and

mutually equivalent.

(i) the sequence {n+/n}nN is strictly increasing;

(ii) the function x n+(x)/n(x) is strictly decreasing on (, );

(iii) the function x n(x) is log-convex on (, ).

Proof (i) It suces to prove n+/n+ > n+/n for n

N, which is equivalent to n+n n+ > . By virtue of the integral representation given in (.), we get

n+n n+ =

et ext dt

tn+

tn et ext dt

et ext dt

tn+

=

tnsn(t s)

( et)( es)ex(t+s) dt ds > ,

which proves assertion (i).(ii) Note that n = n+, we have

n+ n

= n+n n+ nn =n+n + n+n < ,

which implies that the second assertion is true.(iii) Dierentiation gives

(ln n) =

nn =

n+

n , (

ln n) =

n+ n

> ,

which completes the proof.

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

Proof of Theorem . We rst prove that the sequence {In(a, b)}n is strictly decreasing, which means that for n the inequality

n

ba n(x) dx b a

> n+

ba n+(x) dx

b a

(.)

holds for a, b > with a = b. By the Jensen inequality, it suces to check that n+ n is convex on (, ). In fact, by Lemma . we have

d dx

n+

n(x)

n+(n(x))= n(n(x)) =

n+(n(x)) n+(n(x)) ,

Yang and Zheng Journal of Inequalities and Applications ( 2016) 2016:216 Page 7 of 10

d dx

n+(u) > ,

where u = n(x). This means that n+ n is convex, which proves inequality (.).Taking p = n and q = n + in Theorem Q gives

Sn(a, b) < n

ba n(t) dt b a

n+

n(x)

=

n+(u) n+(u)

n(u) = n+(u) n+(u)

< Sn+(a, b). (.)

Considering that limp Sp(a, b) = min(a, b) in [], then we get

lim

n

n

ba n(t) dt b a

= min(a, b),

which completes the proof.

Proof of Theorem . To prove x An(x) is strictly increasing on ( min(a, b), ), by Lemma . it suces to check that n/ n is strictly increasing on (, ). In fact, since

n = n+ we see that n/ n = n+/n+ is strictly increasing on (, ) by the second assertion of Lemma .. Thus, the increasing property of An follows.

As mentioned in the introduction, we see that n(+) = for n , and so n() = . Note that the symmetry of a and b, without loss of generality we may assume that b > a.

Then we have

lim

xa

+

ba n(x + t) dtb a =

lim

xa

+

()n((n)(x + b) (n)(x + a))b a = ,

which implies

lim

xa

+ An(x) = lim

xa

+ n

ba n(x + t) dt b a

lim

xa

+ x

= n

lim

xa

+

ba n(x + t) dt b a

+ a = n() + a = a.

To obtain limx An(x) = (a + b)/, we use (.) to derive that

Sn(x + b, x + a) x < n

ba n(x + t) dt b a

x < Sn+(x + b, x + a) x.

Note that the generalized logarithmic mean Sp(x, y) is homogeneous and symmetric, it follows from Lemma . that

lim

x

Sp(x + b, x + a) x

= a + b

.

Therefore, we conclude that limx An(x) = (a + b)/, which completes the proof.

Yang and Zheng Journal of Inequalities and Applications ( 2016) 2016:216 Page 8 of 10

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 ()n(f (x))(n) for x I and n (see []). 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 ln f satises ()n(ln f (x))(n) for all n

N on I (see

[]). For convenience, we denote the sets of the completely monotonic functions and the logarithmically completely monotonic functions on I by C[I] and L[I], respectively. Qi in [], Theorem , [], Theorem , investigated the logarithmically complete monotonicity of the functions

x Fa,b,c(x) =

( (x+b) (x+a))/(ab)e(x+c), if a = b,e(x+c)(x+a), if a = b, (.)

and x /Fa,b,c(x). Furthermore, he concluded the following result.

Theorem Q Let a, b, and c be real numbers and = min(a, b, c). If (t) is an implicit function dened by

et t = e(t) (t)

on (, ), then (t) is decreasing and t(t) < for (t) = t. Moreover:

() Fa,b,c(x) L[(, )] if

(a, b, c) {c a, c b}

c a, c b (c a)

c a, c b (c a) \{a = b = c}.

() /Fa,b,c(x) L[(, )] if

(a, b, c) {c a, c b}

c a, c b (c a)

c a, c b (c a) \{a = b = c}.

Later, Qi and Guo in [], Theorem , [], Theorem , proved another result concerning the logarithmically complete monotonicity of the functions x Fa,b,c(x) and x /Fa,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 = b and c(a, b) be a constant depending on a and b.

() If c(a, b) min(a, b), then /Fa,b,c(x) L[(c(a, b), )].

() Fa,b,c(x) L[( min(a, b), )] if and only if c(a, b) (a + b)/.

We would like to remark that the result in Theorem Q is rather interesting but somewhat complicated. Theorem QG shows that c is a constant depending on a and b, and c(a, b) min(a, b) is only sucient for /Fa,b,c(x) L[(c(a, b), )]. Here, we apply Theorem . to deduce that c is a constant independent of a and b, and c min(a, b) is also necessary for /Fa,b,c(x) L[(c(a, b), )]. This improved result can be restated as follows.

Yang and Zheng Journal of Inequalities and Applications ( 2016) 2016:216 Page 9 of 10

Theorem . Let a, b, and c be real numbers, and = min(a, b, c). Then /Fa,b,c(x)

L[(, )] if and only if c min(a, b), while Fa,b,c(x) L[(, )] if and only if c (a + b)/.

Proof For a = b, we have

ln Fa,b,c(x) = (x + c)

ln (x + b) ln (x + a)b a =

(x + c)

ba (x + t) dt b a

and

()n

ln Fa,b,c(x)

(n) = ()n(n)(x + c) ()n

ba (n)(x + t) dt b a

=

ba n(x + t) dtb a

n(x + c)

= (b a)

ba n(x + t) dt n(x + c)

n((b a)

ba n(x + t) dt) (x + c)

An(x) c

,

where n = ()n(n) and An(x) is dened by (.).

Since n = (n+) < , (n) < , which means that n is strictly decreasing on (, ).

This yields

(b a)

ba n(x + t) dt n(x + c)

n((b a)

ba n(x + t) dt) n(n(x + c))

<

for x (, ). Therefore, we have

sgn

()n

ln Fa,b,c(x)

(n)

= sgn

c An(x)

.

Theorem . tells that x An(x) = In(x + a, x + b) x is strictly increasing from ( min(a, b), ) onto (min(a, b), (a + b)/), which implies that

sgn

c An(x)

c inf An(x) = min(a, b)

and

sgn

c An(x)

c sup An(x) =

a + b

.

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

Competing interests

The authors declare that they have no competing interests.

Authors contributions

All authors contributed to each part of this work equally, and they all read and approved the nal manuscript.

Author details

1Department of Mathematics, Beijing Jiaotong University, Beijing, 100044, China. 2Power Supply Service Center, ZPEPC Electric Power Research Institute, Hangzhou, Zhejiang 310009, China.

Yang and Zheng Journal of Inequalities and Applications ( 2016) 2016:216 Page 10 of 10

Funding

This paper is partially supported by the National Natural Science Foundation of China with grant No. 11371050.

Received: 12 July 2016 Accepted: 26 August 2016

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