AAAAbstract and Applied Analysis1687-04091085-3375Hindawi Publishing Corporation89648310.1155/2011/896483896483Research ArticleLogarithmically Complete Monotonicity Properties Relating to the Gamma FunctionZhaoTie-Hong1ChuYu-Ming1WangHua2ApreuteseiNarcisa C.1Department of MathematicsHuzhou Teachers CollegeHuzhou 313000China2Department of MathematicsChangsha University of Science and TechnologyChangsha 410076Chinacsust.edu.cn20112172011201126032011180520112011Copyright © 2011 Tie-Hong Zhao et al.This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

We prove that the function fα,β(x)=Γβ(x+α)/xαΓ(βx) is strictly logarithmically completely monotonic on (0,) if (α,β){( α,β):1/αβ1, α1}{(α,β):0<β1,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<α1/2,0<β1}{(α,β):1β1/α2,α1}{(α,β):1/2α<1,β1/(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.

1. Introduction

It is well known that the classical Euler’s gamma function Γ(x) is defined for x>0 asΓ(x)=0tx-1e-tdt. The logarithmic derivative of Γ(x) defined byψ(x)=Γ(x)Γ(x) is called the psi or digamma function and ψi(x) for i 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 . Over the past half century, many authors have established inequalities and monotonicity for these functions (see ).

Recall that a real-valued function f:I is said to be completely monotonic on I if f has derivatives of all orders on I and(-1)nf(n)(x)0 for all xI and n0. Moreover, f is said to be strictly completely monotonic if inequality (1.3) is strict.

Recall also that a positive real-valued function f: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)k[logf(x)](k)0 for all xI and k. 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, 2337] and the references therein. In 1997, Merkle  proved that F(x)=Γ(2x)/Γ2(x) is strictly log-concave on (0,). Later, Chen  showed that [F(x)]-1=Γ2(x)/Γ(2x) is strictly logarithmically completely monotonic on (0,). In , 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  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 functionsfα,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: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, Ω1={(α,β):1αβ1,α1},Ω2={(α,β):0<β1,φ1(α,β)0,φ2(α,β)0},Ω3={(α,β):0<α12,0<β1},Ω4={(α,β):1β1α2,α1},Ω5={(α,β):12α<1,β11-α  }.

We summarize the result as follows.

Theorem 1.1.

Let α>0, β>0, and fα,β(x) be defined as (1.5); then the following statements are true:

fα,β(x) is strictly logarithmically completely monotonic on (0,) if (α,β)Ω1Ω2;

[fα,β(x)]-1 is strictly logarithmically completely monotonic on (0,) if (α,β)Ω3Ω4Ω5.

Note that fα,β(x) is the constant 1 for α=β=1 since Γ(x+1)=xΓ(x).

2. Lemmas

In order to prove our Theorem 1.1, we need two lemmas which we present in this section.

We consider φ1(x,y) and φ2(x,y) defined as (1.6) and discuss the properties for these functions, see Figure 1 more clearly.

The blue curve is the graph of the equation φ1(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.

The shading areas are respectively denoted by the subsets Ωi for i=1,2,,5. The function fα,β(x) is strictly logarithmically completely monotonic on (0,) if (α,β)Ω1Ω2, and [fα,β(x)]-1 is strictly logarithmically completely monotonic on (0,) if (α,β)Ω3Ω4Ω5.

2.1. The Properties of Function <inline-formula><mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" id="M90"><mml:msub><mml:mrow><mml:mi>φ</mml:mi></mml:mrow><mml:mrow><mml:mn>1</mml:mn></mml:mrow></mml:msub><mml:mo stretchy="false">(</mml:mo><mml:mi>x</mml:mi><mml:mo>,</mml:mo><mml:mi>y</mml:mi><mml:mo stretchy="false">)</mml:mo></mml:math></inline-formula>

The function φ1(x,y) can be interpreted as a quadric equation with respect to y. Let 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 Δ1(x)=b12(x)-4a1(x)c1(x)=4x4-8x3+17x2-10x+1.

If x=(5-1)/2, then it is easy to see that φ1(5-12,y)=11-552y-5-12<0 for y>0.

Let x1, x2 be two real roots of Δ1(x) with x1<x2; then we claim that 0<x1<x2<(5-1)/2. Indeed, Δ1(0)=1,limxΔ1(x)=+,Δ1(0)=-10,Δ1(x)=16x3-24x2+34x-10,Δ1′′(x)=48x2-48x+34>0. From (2.5)–(2.7), we know that Δ1(x) has only one root ξ, which is ξ=12+(-27+8715)1/3262/3-112[6(-27+8715)]1/30.365. 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.

If x1<x<x2, then φ1(x,y)<0 since Δ1(x)<0 and x2+x-1<0.

If x2<x<(5-1)/2, then a1(x)<0, b1(x)<0,c1(x)<0, which implies φ1(x,y)<0.

If 0<xx1 or x>(5-1)/2, then Δ1(x)0. We can solve two roots of the equation φ1(x,y)=0, which are 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<xx1, 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 Function <inline-formula><mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" id="M157"><mml:msub><mml:mrow><mml:mi>φ</mml:mi></mml:mrow><mml:mrow><mml:mn>2</mml:mn></mml:mrow></mml:msub><mml:mo stretchy="false">(</mml:mo><mml:mi>x</mml:mi><mml:mo>,</mml:mo><mml:mi>y</mml:mi><mml:mo stretchy="false">)</mml:mo></mml:math></inline-formula>

The function φ2(x,y) can also be interpreted as a quadric equation with respect to y. Let 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 Δ2(x)=b22(x)-4a2(x)c2(x)=4x4-20x3+33x2-16x.

If x=1, then we have φ2(1,y)=-y-1<0 for y>0.

If x<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.

If x>1, then we can solve the roots of the equation φ2(x,y)=0 but only one of the roots is positive, that is,y2(x)=-2x2+5x-2+4x4-20x3+33x2-16x2(x-1).

Therefore, we conclude that φ2(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 x1 and y2(x)0 as x+.

Finally, we calculate an intersection point of φ1(x,y)=0 and φ2(x,y)=0, that is, the point (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 ψ(x)=Γ(x)Γ(x)=-γ+0e-t-e-xt1-e-tdt,ψ(n)(x)=(-1)n+10tn1-e-te-xtdt for x>0 and n:={1,2,}, where γ=0.5772 is Euler’s constant.

Lemma 2.2.

Let (α,β)(0,)×(0,) and r(t)=(1-e-t)(βe-αβt-αe-βt)+e-βt-αe-t+α-1. Then the following statements are true:

if (α,β)Ω1Ω2, then r(t)>0 for t(0,);

if (α,β)Ω3Ω4Ω5, then r(t)<0 for t(0,);

if 0<α<1/2, β>1 or 1/2<α<1, 0<β<1, then there exist δ2δ1>0 such that r(t)>0 for t(0,δ1) and r(t)<0 for t(δ2,);

if α>1, β>1, then there exist δ4δ3>0 such that r(t)<0 for t(0,δ3) and r(t)>0 for t(δ4,).

Proof.

Let r1(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 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,r1(0)=r(0)=0,r1(t)=β(1+αβ)e-αβt-α(1+β)e-βt-αβ2e-(αβ-1)t+β(α-1)e-(β-1)t+α,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,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),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,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),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,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),r4(t)=β(α-1)2(β-1)(β-αβ-1)et.

(1) If (α,β)Ω1Ω2, then we divide the proof into two cases. Note that Ω1Ω2={(α,β):max{1/α,y2(α)}β1}, see Figure 2.

Case 1.

If (α,β)Ω1, then 1/αβ1, α1, and it follows from (2.21) that 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<β1, 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 r2(0)0,r3(0)=φ1(α,β)0,r4(0)=β(α-1)φ2(α,β)0,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)Ω3Ω4Ω5, then we divide the proof into three cases.

Case 1.

If (α,β)Ω3, then 0<α1/2 and 0<β1<1/(1-α). From (2.26), we clearly see that r4(t)0.

In terms of the properties of φ2(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 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,).

If φ1(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 r2(t)<limtr2(t)=αβ(αβ-1)<0.

If φ1(α,β)<0, then it follows from limtr3(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 r2(t)<max{r2(0),limtr2(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 1/2α<1 and 1β1/α. It follows from (2.21) that 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 1/2α<1 and β-αβ-10. From (2.26), we know that r4(t)0. In terms of the location of Ω3, we know that φ2(α,β)<0. From (2.24), we see that 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,).

If φ1(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 r2(0)=(β-1)(1-2α)<0,  limt+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 limtr1(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 limtr(t)=α-1<0 that r(t)<max{r(0),limtr(t)}=0 for t(0,).

If φ1(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 limtr3(t)=αβ(1+αβ)>0 or limtr3(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 r2(0)0.

For special case of αβ=1, that is, α=1/2 and β=2, it follows from (2.41) and (2.21) that r2(t)<max{r2(0),limtr2(t)}=0, which implies that r(t)<0 for t(0,) follows from (2.17) and (2.18).

For αβ>1, it follows from (2.38) and limtr2(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) If 0<α<1/2, β>1 or 1/2<α<1, 0<β<1, then we have limtr(t)=α-1<0.

From (2.20), we know that 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 δ2δ1>0 such that r(t)<0 for t(δ2,).

(4) If α>1, β>1, then we have limtr(t)=α-1>0.

From (2.15), we know that 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 δ4δ3>0 such that r2(t)<0 for t(0,δ3) and r(t)>0 for t(δ4,).

3. Proof of Theorem <xref ref-type="statement" rid="thm1.1">1.1</xref>Proof of Theorem <xref ref-type="statement" rid="thm1.1">1.1</xref>.

From (2.15), we have (-1)n[logfα,β(x)](n)=(-1)n[(-1)nα(n-1)!xn+βψ(n-1)(x+α)-βnψ(n-1)(βx)]=α0sn-1e-xsds+β0sn-11-e-se-(x+α)sds-βn0tn-11-e-te-βxtdt=αβn0tn-1e-βxtdt+βn+10tn-11-e-βte-β(x+α)tdt-βn0tn-11-e-te-βxtdt=βn0tn-1e-βxt(1-e-t)(1-e-βt)r(t)dt, where r(t)=(1-e-t)(βe-αβt-αe-βt)+e-βt-αe-t+α-1.

(1) If (α,β)Ω1Ω2, then from (3.1) and (3.2) together with Lemma 2.2(1) we clearly see that (-1)n[logfα,β(x)](n)>0. Therefore, fα,β(x) is strictly logarithmically completely monotonic on (0,) following from (3.3).

(2) If (α,β)Ω3Ω4Ω5, then from (3.1) we can get (-1)n{log[fα,β(x)]-1}(n)=-βn0tn-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 neither fα,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 , we can also extend Ω3 onto one component of its boundaries, which is Ω3Ω̃3={(α,β):0α12,0<β1}{α=0,β=1}. Then [fα,β(x)]-1 is strictly logarithmically completely monotonic on (0,) for (α,β)Ω̃3.

Acknowledgment

The first author is supported by the China-funded Postgraduates Studying Aboard Program for Building Top University.

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