# Logarithmically Complete Monotonicity Properties Relating to the Gamma Function

Authors: Tie-Hong Zhao; Yu-Ming Chu; Hua Wang
Journal: Abstract and Applied Analysis
 (2011)

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## Abstract

We prove that the functionfα,β(x)=Γβ(x+α)/xαΓ(βx) is strictly logarithmically completely monotonic on (0,∞) if (α,β)∈{(α,β):1/α≤β≤1,α≠1}∪{(α,β):0<β≤1,φ1(α,β)≥0,φ2(α,β)≥0} and [fα,β(x)]-1 is strictly logarithmically completely monotonic on (0,∞) if (α,β)∈{(α,β):0<α≤1/2,0<β≤1}∪{(α,β):1≤β≤1/α≤2,α≠1}∪{(α,β):1/2≤α<1,β≥1/(1-α)}, where φ1(α,β)=(α2+α-1)β2+(2α2-3α+1)β-α and φ2(α,β)=(α-1)β2+(2α2-5α+2)β-1.


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## Body

## 1. Introduction

It is well known that the classical Euler’s gamma functionΓ(x) is defined for x>0 as(1.1)Γ(x)=∫0∞tx-1e-tdt.
The logarithmic derivative of Γ(x) defined by(1.2)ψ(x)=Γ′(x)Γ(x)
is called the psi or digamma function and ψi(x) for i∈ℕ are known as the polygamma or multigamma functions.  These functions play central roles in the theory of special functions and have lots of extensive applications in many branches, for example, statistics, physics, engineering, and other mathematical sciences.For extension of these functions to complex variable and for basic properties, see [1]. Over the past half century, many authors have established inequalities and monotonicity for these functions (see [2–22]).Recall that a real-valued functionf:I→ℝ is said to be completely monotonic on I if f has derivatives of all orders on I and(1.3)(-1)nf(n)(x)≥0
for all x∈I and n≥0. Moreover, f is said to be strictly completely monotonic if inequality (1.3) is strict.Recall also that a positive real-valued functionf:I→(0,∞) is said to be logarithmically completely monotonic on I if f has derivatives of all orders on I and its logarithm logf satisfies(1.4)(-1)k[log⁡f(x)](k)≥0
for all x∈I and k∈ℕ. Moreover, f is said to be strictly logarithmically completely monotonic if inequality (1.4) is strict.Recently, the completely monotonic or logarithmically completely monotonic functions have been the subject of intensive research. There has been a lot of literature about the (logarithmically) completely monotonic functions related to the gamma function, psi function, and polygamma function, for example, [17, 18, 23–37] and the references therein. In 1997, Merkle [38] proved that F(x)=Γ(2x)/Γ2(x) is strictly log-concave on (0,∞). Later, Chen [39] showed that [F(x)]-1=Γ2(x)/Γ(2x) is strictly logarithmically completely monotonic on (0,∞). In [40], Li and Chen proved that Fβ(x)=Γβ(x)/Γ(βx) is strictly logarithmically completely monotonic on (0,∞) for β>1, and [Fβ(x)]-1 is strictly logarithmically completely monotonic on (0,∞) for 0<β<1. Qi et al. in their article [41] showed that fα(x)=Γ(x+α)/xαΓ(x) is strictly logarithmically complete monotonic on (0,∞) for α>1, and [fα(x)]-1 is strictly logarithmically complete monotonic on (0,∞) for 0<α<1.The aim of this paper is to discuss the logarithmically complete monotonicity properties of the functions(1.5)fα,β(x)=Γβ(x+α)xαΓ(βx)
and [fα,β(x)]-1 on (0,∞) where α>0 and β>0. The function fα,β(x) is the deformation of the functions in [40, 41] with respect to the parameters α and β. We show that the properties of logarithmically complete monotonic are also true for suitable extensions of (α,β) near by two lines α=0 and β=1, which generalizes the results of [40, 41].For(x,y)∈(0,∞)×(0,∞), we define two binary functions as follows:(1.6)φ1(x,y)=(x2+x-1)y2+(2x2-3x+1)y-x,φ2(x,y)=(x-1)y2+(2x2-5x+2)y-1.For convenience, we need to define five subsets of(0,∞)×(0,∞) and refer to Figure 2, (1.7)Ω1={(α,β):1α≤β≤1,α≠1},Ω2={(α,β):0<β≤1,φ1(α,β)≥0,φ2(α,β)≥0},Ω3={(α,β):0<α≤12,0<β≤1},Ω4={(α,β):1≤β≤1α≤2,α≠1},Ω5={(α,β):12≤α<1,β≥11-α}.We summarize the result as follows.Theorem 1.1.
Letα>0, β>0, and fα,β(x) be defined as (1.5); then the following statements are true:(1)
fα,β(x) is strictly logarithmically completely monotonic on (0,∞) if (α,β)∈Ω1∪Ω2;(2)
[fα,β(x)]-1 is strictly logarithmically completely monotonic on (0,∞) if (α,β)∈Ω3∪Ω4∪Ω5.
Note thatfα,β(x) is the constant 1 for α=β=1 since Γ(x+1)=xΓ(x).


## 2. Lemmas

In order to prove our Theorem1.1, we need two lemmas which we present in this section.We considerφ1(x,y) and φ2(x,y) defined as (1.6) and discuss the properties for these functions, see Figure 1 more clearly.Figure 1
The blue curve is the graph of the equationφ1(x,y)=0 with the vertical asymptotic line x=(5-1)/2 and the green curve is the graph of φ2(x,y)=0 with the vertical asymptotic line x=1.Figure 2
The shading areas are respectively denoted by the subsetsΩi  for i=1,2,…,5. The function fα,β(x) is strictly logarithmically completely monotonic on (0,∞) if (α,β)∈Ω1∪Ω2, and [fα,β(x)]-1 is strictly logarithmically completely monotonic on (0,∞) if (α,β)∈Ω3∪Ω4∪Ω5.


### 2.1. The Properties of Functionφ1(x,y)

The functionφ1(x,y) can be interpreted as a quadric equation with respect to y. Let (2.1)φ1(x,y)=a1(x)y2+b1(x)y+c1(x),
where a1(x)=x2+x-1, b1(x)=2x2-3x+1,c1(x)=-x, and its discriminant function (2.2)Δ1(x)=b12(x)-4a1(x)c1(x)=4x4-8x3+17x2-10x+1.Ifx=(5-1)/2, then it is easy to see that
(2.3)φ1(5-12,y)=11-552y-5-12<0
for y>0.Letx1, x2 be two real roots of Δ1(x) with x1<x2; then we claim that 0<x1<x2<(5-1)/2. Indeed,
(2.4)Δ1(0)=1,lim⁡x→∞Δ1(x)=+∞,(2.5)Δ1′(0)=-10,(2.6)Δ1′(x)=16x3-24x2+34x-10,(2.7)Δ1′′(x)=48x2-48x+34>0.
From (2.5)–(2.7), we know that Δ1′(x) has only one root ξ, which is (2.8)ξ=12+(-27+8715)1/3262/3-112[6(-27+8715)]1/3≈0.365….
Moreover, Δ1′(x)<0 for x∈(0,ξ) and Δ1′(x)>0 for x∈(ξ,∞), which implies that Δ1(x) is strictly decreasing on (0,ξ) and strictly increasing on (ξ,∞).  An easy computation shows that ξ<(5-1)/2, Δ1(ξ)<0, and Δ1((5-1)/2)>0. Combining with (2.4), there exist two real roots x1,x2 such that 0<x1<x2<(5-1)/2. Furthermore, we conclude that Δ1(x)>0 for 0<x<x1 or x>x2 and Δ1(x)<0 for x1<x<x2.Ifx1<x<x2, then φ1(x,y)<0 since Δ1(x)<0 and x2+x-1<0.Ifx2<x<(5-1)/2, then a1(x)<0, b1(x)<0,c1(x)<0, which implies φ1(x,y)<0.If0<x≤x1 or x>(5-1)/2, then Δ1(x)≥0. We can solve two roots of the equation φ1(x,y)=0, which are (2.9)ỹ1(x)=-2x2+3x-1-4x4-8x3+17x2-10x+12(x2+x-1),y1(x)=-2x2+3x-1+4x4-8x3+17x2-10x+12(x2+x-1).
For 0<x≤x1, we know that φ1(x,y)>0 for y1(x)<y<ỹ1(x) and φ1(x,y)<0 for 0<y<y1(x) or y>ỹ1(x). For x>(5-1)/2, we know that φ1(x,y)<0 for 0<y<y1(x) and φ1(x,y)>0 for y>y1(x). Moreover, we see that y1(x)→+∞ as x→(5-1)/2 and y1(x)→0 as x→+∞.


### 2.2. The Properties of Functionφ2(x,y)

The functionφ2(x,y) can also be interpreted as a quadric equation with respect to y. Let (2.10)φ2(x,y)=a2(x)y2+b2(x)y+c2(x),
where a2(x)=x-1, b2(x)=2x2-5x+2,c2(x)=-1, and its discriminant function (2.11)Δ2(x)=b22(x)-4a2(x)c2(x)=4x4-20x3+33x2-16x.Ifx=1, then we have φ2(1,y)=-y-1<0 for y>0.Ifx<1, then a simple calculation leads to Δ2(x)<0 for 0<x<(1/6)[10-1/(53-678)1/3-(53-678)1/3]≈0.8427…. This implies that φ2(x,y)<0. Notice that a2(x)<0, b2(x)<0, and c2(x)=-1; for 1/2<x<1, then we have φ2(x,y)<0.Ifx>1, then we can solve the roots of the equation φ2(x,y)=0 but only one of the roots is positive, that is,(2.12)y2(x)=-2x2+5x-2+4x4-20x3+33x2-16x2(x-1).Therefore, we conclude thatφ2(x,y)<0 for 0<y<y2(x) and φ2(x,y)>0 for y>y2(x). Moreover, it is easy to see that y2(x)→+∞ as x→1 and y2(x)→0 as x→+∞.Finally, we calculate an intersection point ofφ1(x,y)=0 and φ2(x,y)=0, that is, the point (2.13)(233-3,2-3).Lemma 2.1.
The psi or digamma function, the logarithmic derivative of the gamma function, and the polygamma functions can be expressed as(2.14)ψ(x)=Γ′(x)Γ(x)=-γ+∫0∞e-t-e-xt1-e-tdt,(2.15)ψ(n)(x)=(-1)n+1∫0∞tn1-e-te-xtdt
for x>0 and n∈ℕ:={1,2,…}, where γ=0.5772… is Euler’s constant.Lemma 2.2.
Let(α,β)∈(0,∞)×(0,∞) and
(2.16)r(t)=(1-e-t)(βe-αβt-αe-βt)+e-βt-αe-t+α-1.
Then the following statements are true: (1)
if(α,β)∈Ω1∪Ω2, then r(t)>0 for t∈(0,∞);(2)
if(α,β)∈Ω3∪Ω4∪Ω5, then r(t)<0 for t∈(0,∞);(3)
if0<α<1/2, β>1 or 1/2<α<1, 0<β<1, then there exist δ2≫δ1>0 such that r(t)>0 for t∈(0,δ1) and r(t)<0 for t∈(δ2,∞);(4)
ifα>1, β>1, then there exist δ4≫δ3>0 such that r(t)<0 for t∈(0,δ3) and r(t)>0 for t∈(δ4,∞).Proof.
Letr1(t)=etr′(t),r2(t)=(1/β)e(αβ-1)tr1′(t),r3(t)=etr2′(t), and r4(t)=e(β-αβ)tr3′(t). Then simple calculations lead to
(2.17)r(0)=0,r′(t)=(β+αβ2)e-(αβ+1)t-(α+αβ)e-(β+1)t-αβ2e-αβt+(αβ-β)e-βt+αe-t,(2.18)r1(0)=r′(0)=0,(2.19)r1(t)=β(1+αβ)e-αβt-α(1+β)e-βt-αβ2e-(αβ-1)t+β(α-1)e-(β-1)t+α,(2.20)r1′(t)=-αβ2(1+αβ)e-αβt+αβ(1+β)e-βt+αβ2(αβ-1)e-(αβ-1)t-β(α-1)(β-1)e-(β-1)t,(2.21)r2(0)=1βr1′(0)=(β-1)(1-2α),r2(t)=-αβ(1+αβ)e-t+α(1+β)e(αβ-β-1)t-(α-1)(β-1)e(α-1)βt+αβ(αβ-1),(2.22)r2′(t)=αβ(1+αβ)e-t+α(1+β)(αβ-β-1)e(αβ-β-1)t-β(α-1)2(β-1)e(α-1)βt,(2.23)r3(0)=r2′(0)=φ1(α,β),r3(t)=α(β+1)(αβ-β-1)e(α-1)βt-β(α-1)2(β-1)e(αβ-β+1)t+αβ(1+αβ),(2.24)r3′(t)=αβ(α-1)(β+1)(αβ-β-1)e(α-1)βt+β(α-1)2(β-1)(β-αβ-1)e(αβ-β+1)t,(2.25)r4(0)=r3′(0)=β(α-1)φ2(α,β),r4(t)=β(α-1)2(β-1)(β-αβ-1)et+αβ(α-1)(β+1)(αβ-β-1),(2.26)r4′(t)=β(α-1)2(β-1)(β-αβ-1)et.
(1) If(α,β)∈Ω1∪Ω2, then we divide the proof into two cases. Note that Ω1∩Ω2={(α,β):max{1/α,y2(α)}≤β≤1}, see Figure 2.
Case 1.
If(α,β)∈Ω1, then 1/α≤β≤1, α≠1, and it follows from (2.21) that
(2.27)r2(t)=-αβ(1+αβ)e-t+e(α-1)βt[α(1+β)e-t+(α-1)(1-β)]+αβ(αβ-1)>α(1-αβ2)e-t+(α-1)(1-β)+αβ(αβ-1)≥α(1-αβ2)+(α-1)(1-β)+αβ(αβ-1)=(β-1)(1-2α)≥0.
Therefore,r(t)>0 for t∈(0,∞) follows from (2.17), (2.18) together with (2.27).Case 2.
If(α,β)∈Ω2, then 0<β≤1, φ1(α,β)≥0, and φ2(α,β)≥0. It follows from φ2(α,β)≥0 that α>1 and then (2.20) and (2.22) together with (2.24) lead to
(2.28)r2(0)≥0,(2.29)r3(0)=φ1(α,β)≥0,(2.30)r4(0)=β(α-1)φ2(α,β)≥0,(2.31)r4′(t)≥0.
This could not happen together for all qualities of (2.28)–(2.31) since the qualities of (2.29) and (2.30) hold only for α=23/(3-3), β=2-3 while the qualities of (2.29) and (2.30) hold only for β=1.
Therefore,r(t)>0 for t∈(0,∞) follows from (2.17) and (2.18) together with (2.28)–(2.31).
(2) If(α,β)∈Ω3∪Ω4∪Ω5, then we divide the proof into three cases.
Case 1.
If(α,β)∈Ω3, then 0<α≤1/2 and 0<β≤1<1/(1-α). From (2.26), we clearly see that
(2.32)r4′(t)≥0.
In terms of the properties ofφ2(x,y), we know that φ2(α,β)<0 for (α,β) lying on the left-side of the green curve, see Figure 1. From (2.24), we see that
(2.33)r4(0)=β(α-1)φ2(α,β)>0.
Combining (2.32) with (2.33) we get that r3(t) is strictly increasing on (0,∞).
Ifφ1(α,β)≥0, then 0<β<1 and r3(t)>0 follow from (2.22), which implies that r2(t) is strictly increasing in (0,∞). Thus we can obtain
(2.34)r2(t)<lim⁡t→∞r2(t)=αβ(αβ-1)<0.
Ifφ1(α,β)<0, then it follows from limt→∞r3(t)=+∞ or αβ(1+αβ)>0 that there exists σ1>0 such that r3(t)<0 for t∈(0,σ1) and r3(t)>0 for t∈(σ1,∞). Hence, r2(t) is strictly decreasing in (0,σ1) and strictly increasing in (σ1,∞). Then we can obtain
(2.35)r2(t)<max⁡{r2(0),lim⁡t→∞r2(t)}≤0.
Finally, we conclude that r(t)<0 for t∈(0,∞) follows from (2.17), (2.18) together with (2.34), (2.35).Case 2.
If(α,β)∈Ω4, then 1/2≤α<1 and 1≤β≤1/α. It follows from (2.21) that
(2.36)r2(t)=-αβ(1+αβ)e-t+e(α-1)βt[α(1+β)e-t+(1-α)(β-1)]+αβ(αβ-1)<α(1-αβ2)e-t+(1-α)(β-1)+αβ(αβ-1)≤α(1-αβ2)+(1-α)(β-1)+αβ(αβ-1)=(β-1)(1-2α)≤0.
Therefore,r(t)<0 for t∈(0,∞) follows from (2.17), (2.18) together with (2.36).Case 3.
If(α,β)∈Ω5, then 1/2≤α<1 and β-αβ-1≥0. From (2.26), we know that
(2.37)r4′(t)≥0.
In terms of the location of Ω3, we know that φ2(α,β)<0. From (2.24), we see that
(2.38)r4(0)=β(α-1)φ2(α,β)>0.
It follows from (2.37) and (2.38) that r3(t) is strictly increasing on (0,∞).
Ifφ1(α,β)≥0, then 1/2<α<1 and r3(t)>0 follow that from (2.22), which implies that r2(t) is strictly increasing on (0,∞). From (2.20) and (2.21), we see that
(2.39)r2(0)=(β-1)(1-2α)<0,lim⁡t→+∞r2(t)=αβ(αβ-1)>0.
Thus there exists σ2>0 such that r2(t)<0 for t∈(0,σ2) and r2(t)>0 for t∈(σ2,∞), which implies that r1(t) is strictly decreasing on (0,σ2) and strictly increasing on (σ2,∞). It follows from (2.18) and limt→∞r1(t)=α>0 that σ3>σ2 such that r1(t)<0 for t∈(0,σ3) and r1(t)>0 for t∈(σ3,∞), which implies that r(t) is strictly decreasing on (0,σ3) and strictly increasing on (σ3,∞). Therefore, it follows from (2.17) and limt→∞r(t)=α-1<0 that
(2.40)r(t)<max⁡{r(0),lim⁡t→∞r(t)}=0
for t∈(0,∞).
Ifφ1(α,β)<0, then there exists σ4>0 such that r3(t)<0 for t∈(0,σ4) and r3(t)>0 for t∈(σ4,∞) follows from limt→∞r3(t)=αβ(1+αβ)>0 or limt→∞r3(t)=β[(α-1/2)2+β(2α-1)+3/4]>0. This leads to r2(t) being strictly decreasing in (0,σ4) and strictly increasing in (σ4,∞). From (2.20), we clearly see that
(2.41)r2(0)≤0.
For special case ofαβ=1, that is, α=1/2 and β=2, it follows from (2.41) and (2.21) that
(2.42)r2(t)<max⁡{r2(0),lim⁡t→∞r2(t)}=0,
which implies that r(t)<0 for t∈(0,∞) follows from (2.17) and (2.18).
Forαβ>1, it follows from (2.38) and limt→∞r2(t)=αβ(αβ-1)>0 that there exists σ5>σ4>0 such that r2(t)<0 for t∈(0,σ5) and r2(t)>0 for t∈(σ5,∞). Making use of the same arguments as the case of φ1(α,β)≥0, then r(t)<0 for t∈(0,∞) follows from (2.17).
(3) If0<α<1/2, β>1 or 1/2<α<1, 0<β<1, then we have
(2.43)lim⁡t→∞r(t)=α-1<0.
From (2.20), we know that
(2.44)r2(0)=(β-1)(1-2α)>0.
It follows from (2.44) that there exists δ1>0 such that r2(t)>0 for t∈(0,δ1), which implies that r1(t) is strictly increasing on (0,δ1). Therefore, r(t)>0 for t∈(0,δ1) follows from (2.17) and (2.18).
From (2.43), we know that there exists δ2≫δ1>0 such that r(t)<0 for t∈(δ2,∞).
(4) Ifα>1, β>1, then we have
(2.45)lim⁡t→∞r(t)=α-1>0.
From (2.15), we know that
(2.46)r2(0)=(β-1)(1-2α)<0.
Making use of (2.45) and (2.46) together with the same arguments as in Lemma 2.2(3), we know that there exist δ4≫δ3>0 such that r2(t)<0 for t∈(0,δ3) and r(t)>0 for t∈(δ4,∞).


## 2.1. The Properties of Functionφ1(x,y)

The functionφ1(x,y) can be interpreted as a quadric equation with respect to y. Let (2.1)φ1(x,y)=a1(x)y2+b1(x)y+c1(x),
where a1(x)=x2+x-1, b1(x)=2x2-3x+1,c1(x)=-x, and its discriminant function (2.2)Δ1(x)=b12(x)-4a1(x)c1(x)=4x4-8x3+17x2-10x+1.Ifx=(5-1)/2, then it is easy to see that
(2.3)φ1(5-12,y)=11-552y-5-12<0
for y>0.Letx1, x2 be two real roots of Δ1(x) with x1<x2; then we claim that 0<x1<x2<(5-1)/2. Indeed,
(2.4)Δ1(0)=1,lim⁡x→∞Δ1(x)=+∞,(2.5)Δ1′(0)=-10,(2.6)Δ1′(x)=16x3-24x2+34x-10,(2.7)Δ1′′(x)=48x2-48x+34>0.
From (2.5)–(2.7), we know that Δ1′(x) has only one root ξ, which is (2.8)ξ=12+(-27+8715)1/3262/3-112[6(-27+8715)]1/3≈0.365….
Moreover, Δ1′(x)<0 for x∈(0,ξ) and Δ1′(x)>0 for x∈(ξ,∞), which implies that Δ1(x) is strictly decreasing on (0,ξ) and strictly increasing on (ξ,∞).  An easy computation shows that ξ<(5-1)/2, Δ1(ξ)<0, and Δ1((5-1)/2)>0. Combining with (2.4), there exist two real roots x1,x2 such that 0<x1<x2<(5-1)/2. Furthermore, we conclude that Δ1(x)>0 for 0<x<x1 or x>x2 and Δ1(x)<0 for x1<x<x2.Ifx1<x<x2, then φ1(x,y)<0 since Δ1(x)<0 and x2+x-1<0.Ifx2<x<(5-1)/2, then a1(x)<0, b1(x)<0,c1(x)<0, which implies φ1(x,y)<0.If0<x≤x1 or x>(5-1)/2, then Δ1(x)≥0. We can solve two roots of the equation φ1(x,y)=0, which are (2.9)ỹ1(x)=-2x2+3x-1-4x4-8x3+17x2-10x+12(x2+x-1),y1(x)=-2x2+3x-1+4x4-8x3+17x2-10x+12(x2+x-1).
For 0<x≤x1, we know that φ1(x,y)>0 for y1(x)<y<ỹ1(x) and φ1(x,y)<0 for 0<y<y1(x) or y>ỹ1(x). For x>(5-1)/2, we know that φ1(x,y)<0 for 0<y<y1(x) and φ1(x,y)>0 for y>y1(x). Moreover, we see that y1(x)→+∞ as x→(5-1)/2 and y1(x)→0 as x→+∞.


## 2.2. The Properties of Functionφ2(x,y)

The functionφ2(x,y) can also be interpreted as a quadric equation with respect to y. Let (2.10)φ2(x,y)=a2(x)y2+b2(x)y+c2(x),
where a2(x)=x-1, b2(x)=2x2-5x+2,c2(x)=-1, and its discriminant function (2.11)Δ2(x)=b22(x)-4a2(x)c2(x)=4x4-20x3+33x2-16x.Ifx=1, then we have φ2(1,y)=-y-1<0 for y>0.Ifx<1, then a simple calculation leads to Δ2(x)<0 for 0<x<(1/6)[10-1/(53-678)1/3-(53-678)1/3]≈0.8427…. This implies that φ2(x,y)<0. Notice that a2(x)<0, b2(x)<0, and c2(x)=-1; for 1/2<x<1, then we have φ2(x,y)<0.Ifx>1, then we can solve the roots of the equation φ2(x,y)=0 but only one of the roots is positive, that is,(2.12)y2(x)=-2x2+5x-2+4x4-20x3+33x2-16x2(x-1).Therefore, we conclude thatφ2(x,y)<0 for 0<y<y2(x) and φ2(x,y)>0 for y>y2(x). Moreover, it is easy to see that y2(x)→+∞ as x→1 and y2(x)→0 as x→+∞.Finally, we calculate an intersection point ofφ1(x,y)=0 and φ2(x,y)=0, that is, the point (2.13)(233-3,2-3).Lemma 2.1.
The psi or digamma function, the logarithmic derivative of the gamma function, and the polygamma functions can be expressed as(2.14)ψ(x)=Γ′(x)Γ(x)=-γ+∫0∞e-t-e-xt1-e-tdt,(2.15)ψ(n)(x)=(-1)n+1∫0∞tn1-e-te-xtdt
for x>0 and n∈ℕ:={1,2,…}, where γ=0.5772… is Euler’s constant.Lemma 2.2.
Let(α,β)∈(0,∞)×(0,∞) and
(2.16)r(t)=(1-e-t)(βe-αβt-αe-βt)+e-βt-αe-t+α-1.
Then the following statements are true: (1)
if(α,β)∈Ω1∪Ω2, then r(t)>0 for t∈(0,∞);(2)
if(α,β)∈Ω3∪Ω4∪Ω5, then r(t)<0 for t∈(0,∞);(3)
if0<α<1/2, β>1 or 1/2<α<1, 0<β<1, then there exist δ2≫δ1>0 such that r(t)>0 for t∈(0,δ1) and r(t)<0 for t∈(δ2,∞);(4)
ifα>1, β>1, then there exist δ4≫δ3>0 such that r(t)<0 for t∈(0,δ3) and r(t)>0 for t∈(δ4,∞).Proof.
Letr1(t)=etr′(t),r2(t)=(1/β)e(αβ-1)tr1′(t),r3(t)=etr2′(t), and r4(t)=e(β-αβ)tr3′(t). Then simple calculations lead to
(2.17)r(0)=0,r′(t)=(β+αβ2)e-(αβ+1)t-(α+αβ)e-(β+1)t-αβ2e-αβt+(αβ-β)e-βt+αe-t,(2.18)r1(0)=r′(0)=0,(2.19)r1(t)=β(1+αβ)e-αβt-α(1+β)e-βt-αβ2e-(αβ-1)t+β(α-1)e-(β-1)t+α,(2.20)r1′(t)=-αβ2(1+αβ)e-αβt+αβ(1+β)e-βt+αβ2(αβ-1)e-(αβ-1)t-β(α-1)(β-1)e-(β-1)t,(2.21)r2(0)=1βr1′(0)=(β-1)(1-2α),r2(t)=-αβ(1+αβ)e-t+α(1+β)e(αβ-β-1)t-(α-1)(β-1)e(α-1)βt+αβ(αβ-1),(2.22)r2′(t)=αβ(1+αβ)e-t+α(1+β)(αβ-β-1)e(αβ-β-1)t-β(α-1)2(β-1)e(α-1)βt,(2.23)r3(0)=r2′(0)=φ1(α,β),r3(t)=α(β+1)(αβ-β-1)e(α-1)βt-β(α-1)2(β-1)e(αβ-β+1)t+αβ(1+αβ),(2.24)r3′(t)=αβ(α-1)(β+1)(αβ-β-1)e(α-1)βt+β(α-1)2(β-1)(β-αβ-1)e(αβ-β+1)t,(2.25)r4(0)=r3′(0)=β(α-1)φ2(α,β),r4(t)=β(α-1)2(β-1)(β-αβ-1)et+αβ(α-1)(β+1)(αβ-β-1),(2.26)r4′(t)=β(α-1)2(β-1)(β-αβ-1)et.
(1) If(α,β)∈Ω1∪Ω2, then we divide the proof into two cases. Note that Ω1∩Ω2={(α,β):max{1/α,y2(α)}≤β≤1}, see Figure 2.
Case 1.
If(α,β)∈Ω1, then 1/α≤β≤1, α≠1, and it follows from (2.21) that
(2.27)r2(t)=-αβ(1+αβ)e-t+e(α-1)βt[α(1+β)e-t+(α-1)(1-β)]+αβ(αβ-1)>α(1-αβ2)e-t+(α-1)(1-β)+αβ(αβ-1)≥α(1-αβ2)+(α-1)(1-β)+αβ(αβ-1)=(β-1)(1-2α)≥0.
Therefore,r(t)>0 for t∈(0,∞) follows from (2.17), (2.18) together with (2.27).Case 2.
If(α,β)∈Ω2, then 0<β≤1, φ1(α,β)≥0, and φ2(α,β)≥0. It follows from φ2(α,β)≥0 that α>1 and then (2.20) and (2.22) together with (2.24) lead to
(2.28)r2(0)≥0,(2.29)r3(0)=φ1(α,β)≥0,(2.30)r4(0)=β(α-1)φ2(α,β)≥0,(2.31)r4′(t)≥0.
This could not happen together for all qualities of (2.28)–(2.31) since the qualities of (2.29) and (2.30) hold only for α=23/(3-3), β=2-3 while the qualities of (2.29) and (2.30) hold only for β=1.
Therefore,r(t)>0 for t∈(0,∞) follows from (2.17) and (2.18) together with (2.28)–(2.31).
(2) If(α,β)∈Ω3∪Ω4∪Ω5, then we divide the proof into three cases.
Case 1.
If(α,β)∈Ω3, then 0<α≤1/2 and 0<β≤1<1/(1-α). From (2.26), we clearly see that
(2.32)r4′(t)≥0.
In terms of the properties ofφ2(x,y), we know that φ2(α,β)<0 for (α,β) lying on the left-side of the green curve, see Figure 1. From (2.24), we see that
(2.33)r4(0)=β(α-1)φ2(α,β)>0.
Combining (2.32) with (2.33) we get that r3(t) is strictly increasing on (0,∞).
Ifφ1(α,β)≥0, then 0<β<1 and r3(t)>0 follow from (2.22), which implies that r2(t) is strictly increasing in (0,∞). Thus we can obtain
(2.34)r2(t)<lim⁡t→∞r2(t)=αβ(αβ-1)<0.
Ifφ1(α,β)<0, then it follows from limt→∞r3(t)=+∞ or αβ(1+αβ)>0 that there exists σ1>0 such that r3(t)<0 for t∈(0,σ1) and r3(t)>0 for t∈(σ1,∞). Hence, r2(t) is strictly decreasing in (0,σ1) and strictly increasing in (σ1,∞). Then we can obtain
(2.35)r2(t)<max⁡{r2(0),lim⁡t→∞r2(t)}≤0.
Finally, we conclude that r(t)<0 for t∈(0,∞) follows from (2.17), (2.18) together with (2.34), (2.35).Case 2.
If(α,β)∈Ω4, then 1/2≤α<1 and 1≤β≤1/α. It follows from (2.21) that
(2.36)r2(t)=-αβ(1+αβ)e-t+e(α-1)βt[α(1+β)e-t+(1-α)(β-1)]+αβ(αβ-1)<α(1-αβ2)e-t+(1-α)(β-1)+αβ(αβ-1)≤α(1-αβ2)+(1-α)(β-1)+αβ(αβ-1)=(β-1)(1-2α)≤0.
Therefore,r(t)<0 for t∈(0,∞) follows from (2.17), (2.18) together with (2.36).Case 3.
If(α,β)∈Ω5, then 1/2≤α<1 and β-αβ-1≥0. From (2.26), we know that
(2.37)r4′(t)≥0.
In terms of the location of Ω3, we know that φ2(α,β)<0. From (2.24), we see that
(2.38)r4(0)=β(α-1)φ2(α,β)>0.
It follows from (2.37) and (2.38) that r3(t) is strictly increasing on (0,∞).
Ifφ1(α,β)≥0, then 1/2<α<1 and r3(t)>0 follow that from (2.22), which implies that r2(t) is strictly increasing on (0,∞). From (2.20) and (2.21), we see that
(2.39)r2(0)=(β-1)(1-2α)<0,lim⁡t→+∞r2(t)=αβ(αβ-1)>0.
Thus there exists σ2>0 such that r2(t)<0 for t∈(0,σ2) and r2(t)>0 for t∈(σ2,∞), which implies that r1(t) is strictly decreasing on (0,σ2) and strictly increasing on (σ2,∞). It follows from (2.18) and limt→∞r1(t)=α>0 that σ3>σ2 such that r1(t)<0 for t∈(0,σ3) and r1(t)>0 for t∈(σ3,∞), which implies that r(t) is strictly decreasing on (0,σ3) and strictly increasing on (σ3,∞). Therefore, it follows from (2.17) and limt→∞r(t)=α-1<0 that
(2.40)r(t)<max⁡{r(0),lim⁡t→∞r(t)}=0
for t∈(0,∞).
Ifφ1(α,β)<0, then there exists σ4>0 such that r3(t)<0 for t∈(0,σ4) and r3(t)>0 for t∈(σ4,∞) follows from limt→∞r3(t)=αβ(1+αβ)>0 or limt→∞r3(t)=β[(α-1/2)2+β(2α-1)+3/4]>0. This leads to r2(t) being strictly decreasing in (0,σ4) and strictly increasing in (σ4,∞). From (2.20), we clearly see that
(2.41)r2(0)≤0.
For special case ofαβ=1, that is, α=1/2 and β=2, it follows from (2.41) and (2.21) that
(2.42)r2(t)<max⁡{r2(0),lim⁡t→∞r2(t)}=0,
which implies that r(t)<0 for t∈(0,∞) follows from (2.17) and (2.18).
Forαβ>1, it follows from (2.38) and limt→∞r2(t)=αβ(αβ-1)>0 that there exists σ5>σ4>0 such that r2(t)<0 for t∈(0,σ5) and r2(t)>0 for t∈(σ5,∞). Making use of the same arguments as the case of φ1(α,β)≥0, then r(t)<0 for t∈(0,∞) follows from (2.17).
(3) If0<α<1/2, β>1 or 1/2<α<1, 0<β<1, then we have
(2.43)lim⁡t→∞r(t)=α-1<0.
From (2.20), we know that
(2.44)r2(0)=(β-1)(1-2α)>0.
It follows from (2.44) that there exists δ1>0 such that r2(t)>0 for t∈(0,δ1), which implies that r1(t) is strictly increasing on (0,δ1). Therefore, r(t)>0 for t∈(0,δ1) follows from (2.17) and (2.18).
From (2.43), we know that there exists δ2≫δ1>0 such that r(t)<0 for t∈(δ2,∞).
(4) Ifα>1, β>1, then we have
(2.45)lim⁡t→∞r(t)=α-1>0.
From (2.15), we know that
(2.46)r2(0)=(β-1)(1-2α)<0.
Making use of (2.45) and (2.46) together with the same arguments as in Lemma 2.2(3), we know that there exist δ4≫δ3>0 such that r2(t)<0 for t∈(0,δ3) and r(t)>0 for t∈(δ4,∞).


## 3. Proof of Theorem1.1

Proof of Theorem1.1.
From (2.15), we have
(3.1)(-1)n[log⁡fα,β(x)](n)=(-1)n[(-1)nα(n-1)!xn+βψ(n-1)(x+α)-βnψ(n-1)(βx)]=α∫0∞sn-1e-xsds+β∫0∞sn-11-e-se-(x+α)sds-βn∫0∞tn-11-e-te-βxtdt=αβn∫0∞tn-1e-βxtdt+βn+1∫0∞tn-11-e-βte-β(x+α)tdt-βn∫0∞tn-11-e-te-βxtdt=βn∫0∞tn-1e-βxt(1-e-t)(1-e-βt)r(t)dt,
where
(3.2)r(t)=(1-e-t)(βe-αβt-αe-βt)+e-βt-αe-t+α-1.
(1) If(α,β)∈Ω1∪Ω2, then from (3.1) and (3.2) together with Lemma 2.2(1) we clearly see that
(3.3)(-1)n[log⁡fα,β(x)](n)>0.
Therefore, fα,β(x) is strictly logarithmically completely monotonic on (0,∞) following from (3.3).
(2) If(α,β)∈Ω3∪Ω4∪Ω5, then from (3.1) we can get
(3.4)(-1)n{log⁡[fα,β(x)]-1}(n)=-βn∫0∞tn-1e-βxt(1-e-t)(1-e-βt)r(t)dt,
where r(t) is defined as (3.2).
Therefore,[fα,β(x)]-1 is strictly logarithmically completely monotonic on (0,∞) following from (3.4) and Lemma 2.2 (2).Remark 3.1.
Note that neitherfα,β(x) nor [fα,β(x)]-1 is strictly logarithmically completely monotonic on (0,∞) for (α,β)∈{(α,β):0<α<1/2,β>1}∪{(α,β):1/2<α<1,0<β<1}∪{(α,β):α>1,β>1} following from Lemma 2.2 (3) and (4), it is known that the logarithmically completely monotonicity properties of fα,β(x) and [fα,β(x)]-1 are not completely continuously depended on α and β.Remark 3.2.
Compared with Theorem 9 of [40], we can also extend Ω3 onto one component of its boundaries, which is
(3.5)Ω3→Ω̃3={(α,β):0≤α≤12,0<β≤1}∖{α=0,β=1}.
Then [fα,β(x)]-1 is strictly logarithmically completely monotonic on (0,∞) for (α,β)∈Ω̃3.



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*Source: 896483-2011-07-21.xml*