Abstract

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/(1𝛼)}, where 𝜑1(𝛼,𝛽)=(𝛼2+𝛼1)𝛽2+(2𝛼23𝛼+1)𝛽𝛼 and 𝜑2(𝛼,𝛽)=(𝛼1)𝛽2+(2𝛼25𝛼+2)𝛽1.

1. Introduction

It is well known that the classical Euler’s gamma function Γ(𝑥) is defined for 𝑥>0 asΓ(𝑥)=0𝑡𝑥1𝑒𝑡𝑑𝑡.(1.1) The logarithmic derivative of Γ(𝑥) defined byΓ𝜓(𝑥)=(𝑥)Γ(𝑥)(1.2) 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 [1]. Over the past half century, many authors have established inequalities and monotonicity for these functions (see [222]).

Recall that a real-valued function 𝑓𝐼 is said to be completely monotonic on 𝐼 if 𝑓 has derivatives of all orders on 𝐼 and(1)𝑛𝑓(𝑛)(𝑥)0(1.3) 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 𝑓𝐼(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)𝑘log𝑓(𝑥)(𝑘)0(1.4) 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 [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 functions𝑓𝛼,𝛽Γ(𝑥)=𝛽(𝑥+𝛼)𝑥𝛼Γ(𝛽𝑥)(1.5) 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:𝜑1𝑥(𝑥,𝑦)=2𝑦+𝑥12+2𝑥2𝜑3𝑥+1𝑦𝑥,2(𝑥,𝑦)=(𝑥1)𝑦2+2𝑥25𝑥+2𝑦1.(1.6)

For convenience, we need to define five subsets of (0,)×(0,) and refer to Figure 2, 𝛀1=1(𝛼,𝛽)𝛼,𝛀𝛽1,𝛼12=(𝛼,𝛽)0<𝛽1,𝜑1(𝛼,𝛽)0,𝜑2,𝛀(𝛼,𝛽)03=1(𝛼,𝛽)0<𝛼2,𝛀,0<𝛽14=(1𝛼,𝛽)1𝛽𝛼,𝛀2,𝛼15=1(𝛼,𝛽)21𝛼<1,𝛽.1𝛼(1.7)

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: (1)𝑓𝛼,𝛽(𝑥) is strictly logarithmically completely monotonic on (0,) if (𝛼,𝛽)Ω1Ω2;(2)[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.

2.1. The Properties of Function 𝜑1(𝑥,𝑦)

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),(2.1) where 𝑎1(𝑥)=𝑥2+𝑥1, 𝑏1(𝑥)=2𝑥23𝑥+1,𝑐1(𝑥)=𝑥, and its discriminant function Δ1(𝑥)=𝑏21(𝑥)4𝑎1(𝑥)𝑐1(𝑥)=4𝑥48𝑥3+17𝑥210𝑥+1.(2.2)

If 𝑥=(51)/2, then it is easy to see that 𝜑1512=,𝑦11552𝑦512<0(2.3) for 𝑦>0.

Let 𝑥1, 𝑥2 be two real roots of Δ1(𝑥) with 𝑥1<𝑥2; then we claim that 0<𝑥1<𝑥2<(51)/2. Indeed, Δ1(0)=1,lim𝑥Δ1Δ(𝑥)=+,(2.4)1Δ(0)=10,(2.5)1(𝑥)=16𝑥324𝑥2Δ+34𝑥10,(2.6)1(𝑥)=48𝑥248𝑥+34>0.(2.7) From (2.5)–(2.7), we know that Δ1(𝑥) has only one root 𝜉, which is 1𝜉=2+27+87151/3262/3112627+87151/30.365.(2.8) 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<(51)/2, Δ1(𝜉)<0, and Δ1((51)/2)>0. Combining with (2.4), there exist two real roots 𝑥1,𝑥2 such that 0<𝑥1<𝑥2<(51)/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+𝑥1<0.

If 𝑥2<𝑥<(51)/2, then 𝑎1(𝑥)<0, 𝑏1(𝑥)<0,𝑐1(𝑥)<0, which implies 𝜑1(𝑥,𝑦)<0.

If 0<𝑥𝑥1 or 𝑥>(51)/2, then Δ1(𝑥)0. We can solve two roots of the equation 𝜑1(𝑥,𝑦)=0, which are ̃𝑦1(𝑥)=2𝑥2+3𝑥14𝑥48𝑥3+17𝑥210𝑥+12𝑥2,𝑦+𝑥11(𝑥)=2𝑥2+3𝑥1+4𝑥48𝑥3+17𝑥210𝑥+12𝑥2.+𝑥1(2.9) For 0<𝑥𝑥1, we know that 𝜑1(𝑥,𝑦)>0 for 𝑦1(𝑥)<𝑦<̃𝑦1(𝑥) and 𝜑1(𝑥,𝑦)<0 for 0<𝑦<𝑦1(𝑥) or 𝑦>̃𝑦1(𝑥). For 𝑥>(51)/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(51)/2 and 𝑦1(𝑥)0 as 𝑥+.

2.2. The Properties of Function 𝜑2(𝑥,𝑦)

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),(2.10) where 𝑎2(𝑥)=𝑥1, 𝑏2(𝑥)=2𝑥25𝑥+2,𝑐2(𝑥)=1, and its discriminant function Δ2(𝑥)=𝑏22(𝑥)4𝑎2(𝑥)𝑐2(𝑥)=4𝑥420𝑥3+33𝑥216𝑥.(2.11)

If 𝑥=1, then we have 𝜑2(1,𝑦)=𝑦1<0 for 𝑦>0.

If 𝑥<1, then a simple calculation leads to Δ2(𝑥)<0 for 0<𝑥<(1/6)[101/(53678)1/3(53678)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+5𝑥2+4𝑥420𝑥3+33𝑥216𝑥2.(𝑥1)(2.12)

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 2333,23.(2.13)

Lemma 2.1. The psi or digamma function, the logarithmic derivative of the gamma function, and the polygamma functions can be expressed as Γ𝜓(𝑥)=(𝑥)Γ(𝑥)=𝛾+0𝑒𝑡𝑒𝑥𝑡1𝑒𝑡𝜓𝑑𝑡,(2.14)(𝑛)(𝑥)=(1)𝑛+10𝑡𝑛1𝑒𝑡𝑒𝑥𝑡𝑑𝑡(2.15) 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)=1𝑒𝑡𝛽𝑒𝛼𝛽𝑡𝛼𝑒𝛽𝑡+𝑒𝛽𝑡𝛼𝑒𝑡+𝛼1.(2.16) Then the following statements are true: (1)if (𝛼,𝛽)Ω1Ω2, then 𝑟(𝑡)>0 for 𝑡(0,); (2)if (𝛼,𝛽)Ω3Ω4Ω5, then 𝑟(𝑡)<0 for 𝑡(0,); (3)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,); (4)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(𝛼𝛽1)𝑡𝑟1(𝑡),𝑟3(𝑡)=𝑒𝑡𝑟2(𝑡), and 𝑟4(𝑡)=𝑒(𝛽𝛼𝛽)𝑡𝑟3(𝑡). Then simple calculations lead to 𝑟𝑟(0)=0,(𝑡)=𝛽+𝛼𝛽2𝑒(𝛼𝛽+1)𝑡(𝛼+𝛼𝛽)𝑒(𝛽+1)𝑡𝛼𝛽2𝑒𝛼𝛽𝑡+(𝛼𝛽𝛽)𝑒𝛽𝑡+𝛼𝑒𝑡,𝑟(2.17)1(0)=𝑟𝑟(0)=0,(2.18)1(𝑡)=𝛽(1+𝛼𝛽)𝑒𝛼𝛽𝑡𝛼(1+𝛽)𝑒𝛽𝑡𝛼𝛽2𝑒(𝛼𝛽1)𝑡+𝛽(𝛼1)𝑒(𝛽1)𝑡𝑟+𝛼,(2.19)1(𝑡)=𝛼𝛽2(1+𝛼𝛽)𝑒𝛼𝛽𝑡+𝛼𝛽(1+𝛽)𝑒𝛽𝑡+𝛼𝛽2(𝛼𝛽1)𝑒(𝛼𝛽1)𝑡𝛽(𝛼1)(𝛽1)𝑒(𝛽1)𝑡,𝑟(2.20)21(0)=𝛽𝑟1𝑟(0)=(𝛽1)(12𝛼),2(𝑡)=𝛼𝛽(1+𝛼𝛽)𝑒𝑡+𝛼(1+𝛽)𝑒(𝛼𝛽𝛽1)𝑡(𝛼1)(𝛽1)𝑒(𝛼1)𝛽𝑡𝑟+𝛼𝛽(𝛼𝛽1),(2.21)2(𝑡)=𝛼𝛽(1+𝛼𝛽)𝑒𝑡+𝛼(1+𝛽)(𝛼𝛽𝛽1)𝑒(𝛼𝛽𝛽1)𝑡𝛽(𝛼1)2(𝛽1)𝑒(𝛼1)𝛽𝑡,𝑟(2.22)3(0)=𝑟2(0)=𝜑1𝑟(𝛼,𝛽),3(𝑡)=𝛼(𝛽+1)(𝛼𝛽𝛽1)𝑒(𝛼1)𝛽𝑡𝛽(𝛼1)2(𝛽1)𝑒(𝛼𝛽𝛽+1)𝑡𝑟+𝛼𝛽(1+𝛼𝛽),(2.23)3(𝑡)=𝛼𝛽(𝛼1)(𝛽+1)(𝛼𝛽𝛽1)𝑒(𝛼1)𝛽𝑡+𝛽(𝛼1)2(𝛽1)(𝛽𝛼𝛽1)𝑒(𝛼𝛽𝛽+1)𝑡,𝑟(2.24)4(0)=𝑟3(0)=𝛽(𝛼1)𝜑2𝑟(𝛼,𝛽),4(𝑡)=𝛽(𝛼1)2(𝛽1)(𝛽𝛼𝛽1)𝑒𝑡𝑟+𝛼𝛽(𝛼1)(𝛽+1)(𝛼𝛽𝛽1),(2.25)4(𝑡)=𝛽(𝛼1)2(𝛽1)(𝛽𝛼𝛽1)𝑒𝑡.(2.26)
(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)𝑒𝑡+𝑒(𝛼1)𝛽𝑡𝛼(1+𝛽)𝑒𝑡+(𝛼1)(1𝛽)+𝛼𝛽(𝛼𝛽1)>𝛼1𝛼𝛽2𝑒𝑡+(𝛼1)(1𝛽)+𝛼𝛽(𝛼𝛽1)𝛼1𝛼𝛽2+(𝛼1)(1𝛽)+𝛼𝛽(𝛼𝛽1)=(𝛽1)(12𝛼)0.(2.27)
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 𝑟2𝑟(0)0,(2.28)3(0)=𝜑1(𝑟𝛼,𝛽)0,(2.29)4(0)=𝛽(𝛼1)𝜑2𝑟(𝛼,𝛽)0,(2.30)4(𝑡)0.(2.31) 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/(33), 𝛽=23 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/(1𝛼). From (2.26), we clearly see that 𝑟4(𝑡)0.(2.32)
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(𝛼1)𝜑2(𝛼,𝛽)>0.(2.33) 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(𝛼𝛽1)<0.(2.34)
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 𝑟2𝑟(𝑡)<max2(0),lim𝑡𝑟2(𝑡)0.(2.35) 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)𝑒𝑡+𝑒(𝛼1)𝛽𝑡𝛼(1+𝛽)𝑒𝑡+(1𝛼)(𝛽1)+𝛼𝛽(𝛼𝛽1)<𝛼1𝛼𝛽2𝑒𝑡+(1𝛼)(𝛽1)+𝛼𝛽(𝛼𝛽1)𝛼1𝛼𝛽2+(1𝛼)(𝛽1)+𝛼𝛽(𝛼𝛽1)=(𝛽1)(12𝛼)0.(2.36)
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 𝛽𝛼𝛽10. From (2.26), we know that 𝑟4(𝑡)0.(2.37) In terms of the location of Ω3, we know that 𝜑2(𝛼,𝛽)<0. From (2.24), we see that 𝑟4(0)=𝛽(𝛼1)𝜑2(𝛼,𝛽)>0.(2.38) 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)=(𝛽1)(12𝛼)<0,lim𝑡+𝑟2(𝑡)=𝛼𝛽(𝛼𝛽1)>0.(2.39) 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)=𝛼1<0 that 𝑟(𝑡)<max𝑟(0),lim𝑡𝑟(𝑡)=0(2.40) 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[(𝛼1/2)2+𝛽(2𝛼1)+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.(2.41)
For special case of 𝛼𝛽=1, that is, 𝛼=1/2 and 𝛽=2, it follows from (2.41) and (2.21) that 𝑟2𝑟(𝑡)<max2(0),lim𝑡𝑟2(𝑡)=0,(2.42) 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(𝛼𝛽1)>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)=𝛼1<0.(2.43)
From (2.20), we know that 𝑟2(0)=(𝛽1)(12𝛼)>0.(2.44)
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)=𝛼1>0.(2.45)
From (2.15), we know that 𝑟2(0)=(𝛽1)(12𝛼)<0.(2.46)
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 1.1

Proof of Theorem 1.1. From (2.15), we have (1)𝑛log𝑓𝛼,𝛽(𝑥)(𝑛)=(1)𝑛(1)𝑛𝛼(𝑛1)!𝑥𝑛+𝛽𝜓(𝑛1)(𝑥+𝛼)𝛽𝑛𝜓(𝑛1)(𝛽𝑥)=𝛼0𝑠𝑛1𝑒𝑥𝑠𝑑𝑠+𝛽0𝑠𝑛11𝑒𝑠𝑒(𝑥+𝛼)𝑠𝑑𝑠𝛽𝑛0𝑡𝑛11𝑒𝑡𝑒𝛽𝑥𝑡𝑑𝑡=𝛼𝛽𝑛0𝑡𝑛1𝑒𝛽𝑥𝑡𝑑𝑡+𝛽𝑛+10𝑡𝑛11𝑒𝛽𝑡𝑒𝛽(𝑥+𝛼)𝑡𝑑𝑡𝛽𝑛0𝑡𝑛11𝑒𝑡𝑒𝛽𝑥𝑡𝑑𝑡=𝛽𝑛0𝑡𝑛1𝑒𝛽𝑥𝑡(1𝑒𝑡)1𝑒𝛽𝑡𝑟(𝑡)𝑑𝑡,(3.1) where 𝑟(𝑡)=1𝑒𝑡𝛽𝑒𝛼𝛽𝑡𝛼𝑒𝛽𝑡+𝑒𝛽𝑡𝛼𝑒𝑡+𝛼1.(3.2)
(1) If (𝛼,𝛽)Ω1Ω2, then from (3.1) and (3.2) together with Lemma 2.2(1) we clearly see that (1)𝑛log𝑓𝛼,𝛽(𝑥)(𝑛)>0.(3.3) 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)𝑛𝑓log𝛼,𝛽(𝑥)1(𝑛)=𝛽𝑛0𝑡𝑛1𝑒𝛽𝑥𝑡(1𝑒𝑡)1𝑒𝛽𝑡𝑟(𝑡)𝑑𝑡,(3.4) 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 [40], we can also extend Ω3 onto one component of its boundaries, which is 𝛀3𝛀3=1(𝛼,𝛽)0𝛼2,0<𝛽1{𝛼=0,𝛽=1}.(3.5) Then [𝑓𝛼,𝛽(𝑥)]1 is strictly logarithmically completely monotonic on (0,) for Ω(𝛼,𝛽)3.

Acknowledgment

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