A Completely Monotonic Function Involving Divided Differences of Psi and Polygamma Functions and an Application

(1)

A COMPLETELY MONOTONIC FUNCTION INVOLVING DIVIDED DIFFERENCES OF PSI AND POLYGAMMA

FUNCTIONS AND AN APPLICATION

FENG QI

Abstract. A function involving the divided differences of the psi function and the polygamma functions is proved to be completely monotonic. As an appli-cation of this result, the monotonicity and convexity of a function originated from establishing the best upper and lower bounds in Kershaw’s inequality is deduced.

1. Introduction

Recall [5] that a functionf is said to be completely monotonic on an intervalI iff has derivatives of all orders onI and (−1)n_f(n)₍_x₎_≥_{0 for} _x_∈ _I _and_n _≥_0.

For information about the history, applications and recent developments on the completely monotonic function, please refer to the expository article [5] and the references therein.

The Kershaw’s inequality [4] states that

µ

x+s 2

¶1−s

< Γ(x+ 1) Γ(x+s) <

Ã

x−1

2 +

r

s+1 4

!1−s

(1)

for 0< s <1 andx≥1, where Γ denotes the classical Euler’s gamma function and the middle term in (1) is a special case of the Wallis’ function Γ(_Γ(x_x+₊p_q)₎ forx+p >0 andx+q >0. It is clear that inequality (1) can be rearranged as

s 2 <

·

Γ(x+ 1) Γ(x+s)

¸1/(1−s) −x <

r

s+1 4 −

1

2. (2)

Letsandtbe nonnegative numbers andα= min{s, t}. Define

zs,t(x) =

  

  ·

Γ(x+t) Γ(x+s)

¸1/(t−s)

−x, s6=t eψ(x+s)₋_x, _s₌_t

(3)

inx∈(−α,∞). Standard differentiating and simplifying yields

zs,t0 (x) = [zs,t(x) +x]

ψ(x+t)−ψ(x+s)

t−s −1, (4)

2000Mathematics Subject Classification. Primary 33B15; Secondary 26A48, 26A51, 26D15.

Key words and phrases. completely monotonic function, divided difference, psi function, polygamma function, Kershaw’s inequality.

The author was supported in part by the Science Foundation of the Project for Fostering Innovation Talents at Universities of Henan Province, China.

This paper was typeset usingAMS-LA_TEX.

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z00

s,t(x) = [zs,t(x) +x]

½·

ψ(x+t)−ψ(x+s) t−s

¸2

+ψ

0₍_x₊_t₎₋_ψ0₍_x₊_s₎ t−s

¾

(5)

= zs,t(x) +x (t−s)2

½

[ψ(x+t)−ψ(x+s)]2_{+ (}_t₋_s_)[_ψ0₍_x₊_t₎₋_ψ0₍_x₊_s_)]

¾

. (6)

In order to obtain the best upper and lower bounds for double inequality (1) or (2), the monotonicity and convexity properties of the function zs,t(x) inx∈ (−α,∞) is showed in [2, 3, 7] by using Laplace transform and other complicated techniques respectively.

Let

Θs,t(x) = [ψ(x+t)−ψ(x+s)]2_{+ (}_t₋_s_)[_ψ0₍_x₊_t₎₋_ψ0₍_x₊_s_)] ₍₇₎ and

∆s,t(x) =

  

  ·

ψ(x+t)−ψ(x+s) t−s

¸2

+ψ

0₍_x₊_t₎₋_ψ0₍_x₊_s₎

t−s , s6=t [ψ0₍_x₊_s_)]2₊_ψ00₍_x₊_s₎_, _s₌_t

(8)

inx∈(−α,∞). It is clear from (5) and (6) that

z00

s,t(x) = [zs,t(x) +x]∆s,t(x) =

zs,t(x) +x

(t−s)2 Θs,t(x) (9)

fort6=s.

The aim of this paper is to prove the completely monotonic property of the functions Θs,t(x) and ∆s,t(x) in (−α,∞).

Theorem 1. The functions Θs,t(x) for |t−s| < 1 and −Θs,t(x) for |t−s| > 1

are completely monotonic in (−α,∞). The functions ∆s,t(x) for |t−s| < 1 and −∆s,t(x)for|t−s|>1 are completely monotonic inx∈(−α,∞).

Remark 1. Note that, among other things, the positivity of the function ∆0,0(x) =

[ψ0₍_x_)]2₊_ψ00₍_x_{) in (8) has been verified in [1].}

As a straightforward application of Theorem 1, the monotonicity and convexity of the functionzs,t(x) is obtained.

Theorem 2 ([2, 3, 7]). The functionzs,t(x) in (−α,∞) is either convex and de-creasing for|t−s|<1 or concave and increasing for |t−s|>1.

2. Proofs of theorems

The basic tool of this paper is the following lemma.

Lemma 1. Let f(x)be defined in an infinite interval I. Iflimx→∞f(x) = 0 and f(x)−f(x+ε)>0for any given ε >0, thenf(x)>0 in I.

Proof. By induction, for anyx∈I, we have

f(x)> f(x+ε)> f(x+ 2ε)>· · ·> f(x+kε)→0

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2.1. Proof of Theorem 1. It is well known that for any positive integer n∈N

the psi function ψ(x) and the polygamma or multigamma functions ψ(n)₍_x_{) have}

the following integral expressions

ψ(x) = lnx+

Z ∞

0 ·

1 u−

1 1−e−u

¸

e−xu_d_u ₍₁₀₎

and

ψ(n)(x) = (−1)n+1

Z _∞

0

un 1−e−ue

−xu_d_u. ₍₁₁₎

Using

ψ(i−1)(x+ 1) =ψ(i−1)(x) +(−1)

i−1₍_i₋_1)!

xi (12)

fori∈Nandx >0 and direct computing gives

Θs,t(x)−Θs,t(x+ 1) =©[ψ(x+t) +ψ(x+t+ 1)]−[ψ(x+s) +ψ(x+s+ 1)]ª

×©[ψ(x+t)−ψ(x+t+ 1)]−[ψ(x+s)−ψ(x+s+ 1)]ª

+ (t−s)©[ψ0₍_x₊_t₎₋_ψ0₍_x₊_t_{+ 1)]}₋_[_ψ0₍_x₊_s₎₋_ψ0₍_x₊_s_{+ 1)]}ª

=

½

[ψ(x+t+ 1) +ψ(x+t)]−[ψ(x+s+ 1) +ψ(x+s)] t−s

− 2x+s+t

(x+s)(x+t)

¾

(t−s)2

(x+s)(x+t) ,Λs,t(x)

(t−s)2

(x+s)(x+t)

(13)

and

Λs,t(x)−Λs,t(x+ 1) = 1 t−s

µ

1 x+s+

1 x+s+ 1 −

1 x+t −

1 x+t+ 1

¶

−2x

2_{+ 2(}_s₊_t_{+ 1)}_x₊_s2₊_t2₊_s₊_t

(x+s)(x+s+ 1)(x+t)(x+t+ 1)

= 1−(s−t)2

(x+s)(x+s+ 1)(x+t)(x+t+ 1).

Since limx→∞Λ(s,ti)(x) = 0 for any nonnegative integeri by (10) and (11), and the function Λs,t(x)−Λs,t(x+1)

1−(s−t)2 is completely monotonic, that is,

(−1)i[Λs,t(x)−Λs,t(x+ 1)]

(i)

1−(s−t)2 =

(−1)i_Λ(i)

s,t(x)−(−1)iΛ

(i)

s,t(x+ 1) 1−(s−t)2 ≥0,

in (−α,∞), then (−1)iΛ(s,ti)(x)

1−(s−t)2 ≥0 follows from Lemma 1. This means the function Λs,t(x)

1−(s−t)2 is completely monotonic in (−α,∞).

Since the function ₍_x₊(t_s−₎₍s_x)2₊_t₎ is completely monotonic and a product of two completely monotonic functions is also completely monotonic, then the function

Θs,t(x)−Θs,t(x+1)

1−(s−t)2 is completely monotonic in (−α,∞) by considering (13), which is

equivalent to

(−1)k

·

Θs,t(x)−Θs,t(x+ 1) 1−(s−t)2

¸(k)

= (−1) k_Θ(k)

s,t(x)−(−1)kΘ

(k)

s,t(x+ 1) 1−(s−t)2 ≥0

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that (−1)kΘ

(k)

s,t(x)

1−(s−t)2 ≥0 for any nonnegative integerk. This implies (−1)kΘ (k)

s,t(x)R0 if and only if |t−s| ≶ 1. Therefore, the functions Θs,t(x) for |t−s| < 1 and

−Θs,t(x) for|t−s|>1 are completely monotonic in (−α,∞).

Since Θs,t(x) = (t−s)2_∆s,t(_x_{), the function ∆s,t(}_x_{) has the same monotonicity}

property as Θs,t(x) in (−α,∞). The proof of Theorem 1 is complete.

2.2. Proof of Theorem 2. By Theorem 1, it is easy to see that Θs,t(x)R0 and ∆s,t(x)R0 in (−α,∞) if and only if |t−s|≶1. Then z00

s,t(x)R0 for|t−s|≶1 follows from formula (9). The convexity and concavity of the function zs,t(x) is proved.

In [6], the inequality

exp [(s−r)ψ(s)]> Γ(s)

Γ(r) >exp [(s−r)ψ(r)] (14) fors > r >0 was obtained, which is equivalent to

Γ(s) Γ(r)

¸1/(s−r)

for any positive numberss >0 andt >0. This implies

z0s,t(x) =

·

Γ(x+t) Γ(x+s)

¸1/(t−s)

ψ(x+t)−ψ(x+s)

t−s −1

< eψ(x+t)ψ(x+t)−ψ(x+s)

t−s −1

=eψ(x+t)_ψ0₍_x₊_ξ₎₋₁_{< ψ}0₍_x₊_t₎_eψ(x+t)₋₁

(15)

and

zs,t0 (x)> eψ(x+s)

ψ(x+t)−ψ(x+s)

t−s −1

=eψ(x+s)_ψ0₍_x₊_ξ₎₋₁ > ψ0₍_x₊_s₎_eψ(x+s)₋₁_,

(16)

if assumingt > s >0 without loss of generality, whereξ∈(s, t). By inequality

lnx−1

x< ψ(x)<lnx− 1

2x (17)

forx >0, we obtain

xψ0(x)e−1/x< ψ0(x)eψ(x)< xψ0(x)e−1/2x (18) forx >0. Using the asymptotic representation

ψ0₍_x₎_∼ 1 x+

1

2x2 +· · · (19)

asx→ ∞yields lim x→∞

£

xψ0₍_x₎_e−1/x¤_{= 1} _and _lim x→∞

£

xψ0₍_x₎_e−1/2x¤_{= 1}_. ₍₂₀₎

Hence,

lim x→∞

£

(5)

Combining (21) with (15) and (16) leads to lim

x→∞z 0

s,t(x)≤_xlim_→∞

£

ψ0(x+t)eψ(x+t)¤−1 = lim x+t→∞

£

ψ0(x+t)eψ(x+t)¤−1 = 0

and

lim x→∞z

0

s,t(x)≥_xlim_→∞

£

ψ0₍_x₊_s₎_eψ(x+s)¤₋_{1 =} _lim

x+s→∞

£

ψ0₍_x₊_s₎_eψ(x+s)¤₋_{1 = 0}_.

Thus, it is concluded that limx→∞z0s,t(x) = 0. Since z00

s,t(x) R 0 in x ∈ (−α,∞) for |t−s| ≶ 1, then the function zs,t0 (x) is increasing/decreasing in x ∈ (−α,∞) for |t−s| ≶ 1. Thus, it follows that z0

s,t(x)Q0 andzs,t(x) is decreasing/increasing inx∈(−α,∞) for|t−s|≶1. The monotonicity of the functionzs,t(x) is proved.

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In-equal. Appl.3(2000), 239–252. Available online at

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Available online at

(F. Qi)College of Mathematics and Information Science, Henan Normal University, Xinxiang City, Henan Province, 453007, China; Research Institute of Mathematical In-equality Theory, Henan Polytechnic University, Jiaozuo City, Henan Province, 454010, China

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