Abstract

We prove that the function 𝑓𝛼,𝛽(π‘₯)=Γ𝛽(π‘₯+𝛼)/π‘₯𝛼Γ(𝛽π‘₯) is strictly logarithmically completely monotonic on (0,∞) if √(𝛼,𝛽)∈{(𝛼,𝛽)∢1/𝛼≀𝛽≀1,𝛼≠1}βˆͺ{(𝛼,𝛽)∢0<𝛽≀1,πœ‘1(𝛼,𝛽)β‰₯0,πœ‘2(𝛼,𝛽)β‰₯0} and [𝑓𝛼,𝛽(π‘₯)]βˆ’1 is strictly logarithmically completely monotonic on (0,∞) if √(𝛼,𝛽)∈{(𝛼,𝛽)∢0<𝛼≀1/2,0<𝛽≀1}βˆͺ{(𝛼,𝛽)∢1≀𝛽≀1/βˆšπ›Όβ‰€2,𝛼≠1}βˆͺ{(𝛼,𝛽)∢1/2≀𝛼<1,𝛽β‰₯1/(1βˆ’π›Ό)}, where πœ‘1(𝛼,𝛽)=(𝛼2+π›Όβˆ’1)𝛽2+(2𝛼2βˆ’3𝛼+1)π›½βˆ’π›Ό and πœ‘2(𝛼,𝛽)=(π›Όβˆ’1)𝛽2+(2𝛼2βˆ’5𝛼+2)π›½βˆ’1.

1. Introduction

It is well known that the classical Euler’s gamma function Ξ“(π‘₯) is defined for π‘₯>0 asξ€œΞ“(π‘₯)=∞0𝑑π‘₯βˆ’1π‘’βˆ’π‘‘π‘‘π‘‘.(1.1) The logarithmic derivative of Ξ“(π‘₯) defined byΞ“πœ“(π‘₯)=ξ…ž(π‘₯)Ξ“(π‘₯)(1.2) is called the psi or digamma function and πœ“π‘–(π‘₯) for π‘–βˆˆβ„• 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 function π‘“βˆΆπΌβ†’β„ is said to be completely monotonic on 𝐼 if 𝑓 has derivatives of all orders on 𝐼 and(βˆ’1)𝑛𝑓(𝑛)(π‘₯)β‰₯0(1.3) for all π‘₯∈𝐼 and 𝑛β‰₯0. Moreover, 𝑓 is said to be strictly completely monotonic if inequality (1.3) is strict.

Recall also that a positive real-valued function π‘“βˆΆπΌβ†’(0,∞) is said to be logarithmically completely monotonic on 𝐼 if 𝑓 has derivatives of all orders on 𝐼 and its logarithm log𝑓 satisfies(βˆ’1)π‘˜ξ€Ίξ€»log𝑓(π‘₯)(π‘˜)β‰₯0(1.4) for all π‘₯∈𝐼 and π‘˜βˆˆβ„•. 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(π‘₯)=Ξ“(2π‘₯)/Ξ“2(π‘₯) is strictly log-concave on (0,∞). Later, Chen [39] showed that [𝐹(π‘₯)]βˆ’1=Ξ“2(π‘₯)/Ξ“(2π‘₯) is strictly logarithmically completely monotonic on (0,∞). In [40], Li and Chen proved that 𝐹𝛽(π‘₯)=Γ𝛽(π‘₯)/Ξ“(𝛽π‘₯) is strictly logarithmically completely monotonic on (0,∞) for 𝛽>1, and [𝐹𝛽(π‘₯)]βˆ’1 is strictly logarithmically completely monotonic on (0,∞) for 0<𝛽<1. Qi et al. in their article [41] showed that 𝑓𝛼(π‘₯)=Ξ“(π‘₯+𝛼)/π‘₯𝛼Γ(π‘₯) is strictly logarithmically complete monotonic on (0,∞) for 𝛼>1, and [𝑓𝛼(π‘₯)]βˆ’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(π‘₯)=𝛽(π‘₯+𝛼)π‘₯𝛼Γ(𝛽π‘₯)(1.5) and [𝑓𝛼,𝛽(π‘₯)]βˆ’1 on (0,∞) where 𝛼>0 and 𝛽>0. The function 𝑓𝛼,𝛽(π‘₯) 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 (π‘₯,𝑦)∈(0,∞)Γ—(0,∞), we define two binary functions as follows:πœ‘1ξ€·π‘₯(π‘₯,𝑦)=2𝑦+π‘₯βˆ’12+ξ€·2π‘₯2ξ€Έπœ‘βˆ’3π‘₯+1π‘¦βˆ’π‘₯,2(π‘₯,𝑦)=(π‘₯βˆ’1)𝑦2+ξ€·2π‘₯2ξ€Έβˆ’5π‘₯+2π‘¦βˆ’1.(1.6)

For convenience, we need to define five subsets of (0,∞)Γ—(0,∞) and refer to Figure 2, 𝛀1=ξƒ―1(𝛼,𝛽)βˆΆβˆšπ›Όξƒ°,𝛀≀𝛽≀1,𝛼≠12=ξ€½(𝛼,𝛽)∢0<𝛽≀1,πœ‘1(𝛼,𝛽)β‰₯0,πœ‘2ξ€Ύ,𝛀(𝛼,𝛽)β‰₯03=1(𝛼,𝛽)∢0<𝛼≀2,𝛀,0<𝛽≀14=ξƒ―(1𝛼,𝛽)∢1β‰€π›½β‰€βˆšπ›Όβ‰€βˆšξƒ°,𝛀2,𝛼≠15=1(𝛼,𝛽)∢21≀𝛼<1,𝛽β‰₯.1βˆ’π›Ό(1.7)

We summarize the result as follows.

Theorem 1.1. Let 𝛼>0, 𝛽>0, and 𝑓𝛼,𝛽(π‘₯) be defined as (1.5); then the following statements are true: (1)𝑓𝛼,𝛽(π‘₯) is strictly logarithmically completely monotonic on (0,∞) if (𝛼,𝛽)∈Ω1βˆͺΞ©2;(2)[𝑓𝛼,𝛽(π‘₯)]βˆ’1 is strictly logarithmically completely monotonic on (0,∞) if (𝛼,𝛽)∈Ω3βˆͺΞ©4βˆͺΞ©5.
Note that 𝑓𝛼,𝛽(π‘₯) is the constant 1 for 𝛼=𝛽=1 since Ξ“(π‘₯+1)=π‘₯Ξ“(π‘₯).

2. Lemmas

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

We consider πœ‘1(π‘₯,𝑦) and πœ‘2(π‘₯,𝑦) defined as (1.6) and discuss the properties for these functions, see Figure 1 more clearly.

2.1. The Properties of Function πœ‘1(π‘₯,𝑦)

The function πœ‘1(π‘₯,𝑦) can be interpreted as a quadric equation with respect to 𝑦. Let πœ‘1(π‘₯,𝑦)=π‘Ž1(π‘₯)𝑦2+𝑏1(π‘₯)𝑦+𝑐1(π‘₯),(2.1) where π‘Ž1(π‘₯)=π‘₯2+π‘₯βˆ’1, 𝑏1(π‘₯)=2π‘₯2βˆ’3π‘₯+1,𝑐1(π‘₯)=βˆ’π‘₯, and its discriminant function Ξ”1(π‘₯)=𝑏21(π‘₯)βˆ’4π‘Ž1(π‘₯)𝑐1(π‘₯)=4π‘₯4βˆ’8π‘₯3+17π‘₯2βˆ’10π‘₯+1.(2.2)

If √π‘₯=(5βˆ’1)/2, then it is easy to see that πœ‘1ξƒ©βˆš5βˆ’12ξƒͺ=√,𝑦11βˆ’552βˆšπ‘¦βˆ’5βˆ’12<0(2.3) for 𝑦>0.

Let π‘₯1, π‘₯2 be two real roots of Ξ”1(π‘₯) with π‘₯1<π‘₯2; then we claim that 0<π‘₯1<π‘₯2√<(5βˆ’1)/2. Indeed, Ξ”1(0)=1,limπ‘₯β†’βˆžΞ”1Ξ”(π‘₯)=+∞,(2.4)ξ…ž1Ξ”(0)=βˆ’10,(2.5)ξ…ž1(π‘₯)=16π‘₯3βˆ’24π‘₯2Ξ”+34π‘₯βˆ’10,(2.6)1ξ…žξ…ž(π‘₯)=48π‘₯2βˆ’48π‘₯+34>0.(2.7) From (2.5)–(2.7), we know that Ξ”ξ…ž1(π‘₯) has only one root πœ‰, which is 1πœ‰=2+ξ‚€βˆšβˆ’27+87151/3262/3βˆ’1126ξ‚€βˆšβˆ’27+87151/3β‰ˆ0.365….(2.8) Moreover, Ξ”ξ…ž1(π‘₯)<0 for π‘₯∈(0,πœ‰) and Ξ”ξ…ž1(π‘₯)>0 for π‘₯∈(πœ‰,∞), which implies that Ξ”1(π‘₯) is strictly decreasing on (0,πœ‰) and strictly increasing on (πœ‰,∞). An easy computation shows that βˆšπœ‰<(5βˆ’1)/2, Ξ”1(πœ‰)<0, and Ξ”1√((5βˆ’1)/2)>0. Combining with (2.4), there exist two real roots π‘₯1,π‘₯2 such that 0<π‘₯1<π‘₯2√<(5βˆ’1)/2. Furthermore, we conclude that Ξ”1(π‘₯)>0 for 0<π‘₯<π‘₯1 or π‘₯>π‘₯2 and Ξ”1(π‘₯)<0 for π‘₯1<π‘₯<π‘₯2.

If π‘₯1<π‘₯<π‘₯2, then πœ‘1(π‘₯,𝑦)<0 since Ξ”1(π‘₯)<0 and π‘₯2+π‘₯βˆ’1<0.

If π‘₯2√<π‘₯<(5βˆ’1)/2, then π‘Ž1(π‘₯)<0, 𝑏1(π‘₯)<0,𝑐1(π‘₯)<0, which implies πœ‘1(π‘₯,𝑦)<0.

If 0<π‘₯≀π‘₯1 or √π‘₯>(5βˆ’1)/2, then Ξ”1(π‘₯)β‰₯0. We can solve two roots of the equation πœ‘1(π‘₯,𝑦)=0, which are ̃𝑦1(π‘₯)=βˆ’2π‘₯2√+3π‘₯βˆ’1βˆ’4π‘₯4βˆ’8π‘₯3+17π‘₯2βˆ’10π‘₯+12ξ€·π‘₯2ξ€Έ,𝑦+π‘₯βˆ’11(π‘₯)=βˆ’2π‘₯2√+3π‘₯βˆ’1+4π‘₯4βˆ’8π‘₯3+17π‘₯2βˆ’10π‘₯+12ξ€·π‘₯2ξ€Έ.+π‘₯βˆ’1(2.9) For 0<π‘₯≀π‘₯1, we know that πœ‘1(π‘₯,𝑦)>0 for 𝑦1(π‘₯)<𝑦<̃𝑦1(π‘₯) and πœ‘1(π‘₯,𝑦)<0 for 0<𝑦<𝑦1(π‘₯) or 𝑦>̃𝑦1(π‘₯). For √π‘₯>(5βˆ’1)/2, we know that πœ‘1(π‘₯,𝑦)<0 for 0<𝑦<𝑦1(π‘₯) and πœ‘1(π‘₯,𝑦)>0 for 𝑦>𝑦1(π‘₯). Moreover, we see that 𝑦1(π‘₯)β†’+∞ as √π‘₯β†’(5βˆ’1)/2 and 𝑦1(π‘₯)β†’0 as π‘₯β†’+∞.

2.2. The Properties of Function πœ‘2(π‘₯,𝑦)

The function πœ‘2(π‘₯,𝑦) can also be interpreted as a quadric equation with respect to 𝑦. Let πœ‘2(π‘₯,𝑦)=π‘Ž2(π‘₯)𝑦2+𝑏2(π‘₯)𝑦+𝑐2(π‘₯),(2.10) where π‘Ž2(π‘₯)=π‘₯βˆ’1, 𝑏2(π‘₯)=2π‘₯2βˆ’5π‘₯+2,𝑐2(π‘₯)=βˆ’1, and its discriminant function Ξ”2(π‘₯)=𝑏22(π‘₯)βˆ’4π‘Ž2(π‘₯)𝑐2(π‘₯)=4π‘₯4βˆ’20π‘₯3+33π‘₯2βˆ’16π‘₯.(2.11)

If π‘₯=1, then we have πœ‘2(1,𝑦)=βˆ’π‘¦βˆ’1<0 for 𝑦>0.

If π‘₯<1, then a simple calculation leads to Ξ”2(π‘₯)<0 for √0<π‘₯<(1/6)[10βˆ’1/(53βˆ’678)1/3βˆšβˆ’(53βˆ’678)1/3]β‰ˆ0.8427…. This implies that πœ‘2(π‘₯,𝑦)<0. Notice that π‘Ž2(π‘₯)<0, 𝑏2(π‘₯)<0, and 𝑐2(π‘₯)=βˆ’1; for 1/2<π‘₯<1, then we have πœ‘2(π‘₯,𝑦)<0.

If π‘₯>1, then we can solve the roots of the equation πœ‘2(π‘₯,𝑦)=0 but only one of the roots is positive, that is,𝑦2(π‘₯)=βˆ’2π‘₯2√+5π‘₯βˆ’2+4π‘₯4βˆ’20π‘₯3+33π‘₯2βˆ’16π‘₯2.(π‘₯βˆ’1)(2.12)

Therefore, we conclude that πœ‘2(π‘₯,𝑦)<0 for 0<𝑦<𝑦2(π‘₯) and πœ‘2(π‘₯,𝑦)>0 for 𝑦>𝑦2(π‘₯). Moreover, it is easy to see that 𝑦2(π‘₯)β†’+∞ as π‘₯β†’1 and 𝑦2(π‘₯)β†’0 as π‘₯β†’+∞.

Finally, we calculate an intersection point of πœ‘1(π‘₯,𝑦)=0 and πœ‘2(π‘₯,𝑦)=0, that is, the point 2√3√3βˆ’3√,2βˆ’3ξƒͺ.(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 Ξ“πœ“(π‘₯)=ξ…ž(π‘₯)ξ€œΞ“(π‘₯)=βˆ’π›Ύ+∞0π‘’βˆ’π‘‘βˆ’π‘’βˆ’π‘₯𝑑1βˆ’π‘’βˆ’π‘‘πœ“π‘‘π‘‘,(2.14)(𝑛)(π‘₯)=(βˆ’1)𝑛+1ξ€œβˆž0𝑑𝑛1βˆ’π‘’βˆ’π‘‘π‘’βˆ’π‘₯𝑑𝑑𝑑(2.15) for π‘₯>0 and π‘›βˆˆβ„•βˆΆ={1,2,…}, where 𝛾=0.5772… is Euler’s constant.

Lemma 2.2. Let (𝛼,𝛽)∈(0,∞)Γ—(0,∞) and ξ€·π‘Ÿ(𝑑)=1βˆ’π‘’βˆ’π‘‘ξ€Έξ€·π›½π‘’βˆ’π›Όπ›½π‘‘βˆ’π›Όπ‘’βˆ’π›½π‘‘ξ€Έ+π‘’βˆ’π›½π‘‘βˆ’π›Όπ‘’βˆ’π‘‘+π›Όβˆ’1.(2.16) Then the following statements are true: (1)if (𝛼,𝛽)∈Ω1βˆͺΞ©2, then π‘Ÿ(𝑑)>0 for π‘‘βˆˆ(0,∞); (2)if (𝛼,𝛽)∈Ω3βˆͺΞ©4βˆͺΞ©5, then π‘Ÿ(𝑑)<0 for π‘‘βˆˆ(0,∞); (3)if 0<𝛼<1/2, 𝛽>1 or 1/2<𝛼<1, 0<𝛽<1, then there exist 𝛿2≫𝛿1>0 such that π‘Ÿ(𝑑)>0 for π‘‘βˆˆ(0,𝛿1) and π‘Ÿ(𝑑)<0 for π‘‘βˆˆ(𝛿2,∞); (4)if 𝛼>1, 𝛽>1, then there exist 𝛿4≫𝛿3>0 such that π‘Ÿ(𝑑)<0 for π‘‘βˆˆ(0,𝛿3) and π‘Ÿ(𝑑)>0 for π‘‘βˆˆ(𝛿4,∞).

Proof. Let π‘Ÿ1(𝑑)=π‘’π‘‘π‘Ÿξ…ž(𝑑),π‘Ÿ2(𝑑)=(1/𝛽)𝑒(π›Όπ›½βˆ’1)π‘‘π‘Ÿξ…ž1(𝑑),π‘Ÿ3(𝑑)=π‘’π‘‘π‘Ÿξ…ž2(𝑑), and π‘Ÿ4(𝑑)=𝑒(π›½βˆ’π›Όπ›½)π‘‘π‘Ÿξ…ž3(𝑑). Then simple calculations lead to π‘Ÿπ‘Ÿ(0)=0,ξ…žξ€·(𝑑)=𝛽+𝛼𝛽2ξ€Έπ‘’βˆ’(𝛼𝛽+1)π‘‘βˆ’(𝛼+𝛼𝛽)π‘’βˆ’(𝛽+1)π‘‘βˆ’π›Όπ›½2π‘’βˆ’π›Όπ›½π‘‘+(π›Όπ›½βˆ’π›½)π‘’βˆ’π›½π‘‘+π›Όπ‘’βˆ’π‘‘,π‘Ÿ(2.17)1(0)=π‘Ÿξ…žπ‘Ÿ(0)=0,(2.18)1(𝑑)=𝛽(1+𝛼𝛽)π‘’βˆ’π›Όπ›½π‘‘βˆ’π›Ό(1+𝛽)π‘’βˆ’π›½π‘‘βˆ’π›Όπ›½2π‘’βˆ’(π›Όπ›½βˆ’1)𝑑+𝛽(π›Όβˆ’1)π‘’βˆ’(π›½βˆ’1)π‘‘π‘Ÿ+𝛼,(2.19)ξ…ž1(𝑑)=βˆ’π›Όπ›½2(1+𝛼𝛽)π‘’βˆ’π›Όπ›½π‘‘+𝛼𝛽(1+𝛽)π‘’βˆ’π›½π‘‘+𝛼𝛽2(π›Όπ›½βˆ’1)π‘’βˆ’(π›Όπ›½βˆ’1)π‘‘βˆ’π›½(π›Όβˆ’1)(π›½βˆ’1)π‘’βˆ’(π›½βˆ’1)𝑑,π‘Ÿ(2.20)21(0)=π›½π‘Ÿξ…ž1π‘Ÿ(0)=(π›½βˆ’1)(1βˆ’2𝛼),2(𝑑)=βˆ’π›Όπ›½(1+𝛼𝛽)π‘’βˆ’π‘‘+𝛼(1+𝛽)𝑒(π›Όπ›½βˆ’π›½βˆ’1)π‘‘βˆ’(π›Όβˆ’1)(π›½βˆ’1)𝑒(π›Όβˆ’1)π›½π‘‘π‘Ÿ+𝛼𝛽(π›Όπ›½βˆ’1),(2.21)ξ…ž2(𝑑)=𝛼𝛽(1+𝛼𝛽)π‘’βˆ’π‘‘+𝛼(1+𝛽)(π›Όπ›½βˆ’π›½βˆ’1)𝑒(π›Όπ›½βˆ’π›½βˆ’1)π‘‘βˆ’π›½(π›Όβˆ’1)2(π›½βˆ’1)𝑒(π›Όβˆ’1)𝛽𝑑,π‘Ÿ(2.22)3(0)=π‘Ÿξ…ž2(0)=πœ‘1π‘Ÿ(𝛼,𝛽),3(𝑑)=𝛼(𝛽+1)(π›Όπ›½βˆ’π›½βˆ’1)𝑒(π›Όβˆ’1)π›½π‘‘βˆ’π›½(π›Όβˆ’1)2(π›½βˆ’1)𝑒(π›Όπ›½βˆ’π›½+1)π‘‘π‘Ÿ+𝛼𝛽(1+𝛼𝛽),(2.23)ξ…ž3(𝑑)=𝛼𝛽(π›Όβˆ’1)(𝛽+1)(π›Όπ›½βˆ’π›½βˆ’1)𝑒(π›Όβˆ’1)𝛽𝑑+𝛽(π›Όβˆ’1)2(π›½βˆ’1)(π›½βˆ’π›Όπ›½βˆ’1)𝑒(π›Όπ›½βˆ’π›½+1)𝑑,π‘Ÿ(2.24)4(0)=π‘Ÿξ…ž3(0)=𝛽(π›Όβˆ’1)πœ‘2π‘Ÿ(𝛼,𝛽),4(𝑑)=𝛽(π›Όβˆ’1)2(π›½βˆ’1)(π›½βˆ’π›Όπ›½βˆ’1)π‘’π‘‘π‘Ÿ+𝛼𝛽(π›Όβˆ’1)(𝛽+1)(π›Όπ›½βˆ’π›½βˆ’1),(2.25)ξ…ž4(𝑑)=𝛽(π›Όβˆ’1)2(π›½βˆ’1)(π›½βˆ’π›Όπ›½βˆ’1)𝑒𝑑.(2.26)
(1) If (𝛼,𝛽)∈Ω1βˆͺΞ©2, then we divide the proof into two cases. Note that Ξ©1∩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 π‘Ÿ2(𝑑)=βˆ’π›Όπ›½(1+𝛼𝛽)π‘’βˆ’π‘‘+𝑒(π›Όβˆ’1)𝛽𝑑𝛼(1+𝛽)π‘’βˆ’π‘‘ξ€»ξ€·+(π›Όβˆ’1)(1βˆ’π›½)+𝛼𝛽(π›Όπ›½βˆ’1)>𝛼1βˆ’π›Όπ›½2ξ€Έπ‘’βˆ’π‘‘ξ€·+(π›Όβˆ’1)(1βˆ’π›½)+𝛼𝛽(π›Όπ›½βˆ’1)β‰₯𝛼1βˆ’π›Όπ›½2ξ€Έ+(π›Όβˆ’1)(1βˆ’π›½)+𝛼𝛽(π›Όπ›½βˆ’1)=(π›½βˆ’1)(1βˆ’2𝛼)β‰₯0.(2.27)
Therefore, π‘Ÿ(𝑑)>0 for π‘‘βˆˆ(0,∞) follows from (2.17), (2.18) together with (2.27).
Case 2. If (𝛼,𝛽)∈Ω2, then 0<𝛽≀1, πœ‘1(𝛼,𝛽)β‰₯0, and πœ‘2(𝛼,𝛽)β‰₯0. It follows from πœ‘2(𝛼,𝛽)β‰₯0 that 𝛼>1 and then (2.20) and (2.22) together with (2.24) lead to π‘Ÿ2π‘Ÿ(0)β‰₯0,(2.28)3(0)=πœ‘1(π‘Ÿπ›Ό,𝛽)β‰₯0,(2.29)4(0)=𝛽(π›Όβˆ’1)πœ‘2π‘Ÿ(𝛼,𝛽)β‰₯0,(2.30)ξ…ž4(𝑑)β‰₯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 βˆšπ›Ό=2√3/(3βˆ’3), βˆšπ›½=2βˆ’3 while the qualities of (2.29) and (2.30) hold only for 𝛽=1.
Therefore, π‘Ÿ(𝑑)>0 for π‘‘βˆˆ(0,∞) follows from (2.17) and (2.18) together with (2.28)–(2.31).

(2) If (𝛼,𝛽)∈Ω3βˆͺΞ©4βˆͺΞ©5, then we divide the proof into three cases.
Case 1. If (𝛼,𝛽)∈Ω3, then 0<𝛼≀1/2 and 0<𝛽≀1<1/(1βˆ’π›Ό). From (2.26), we clearly see that π‘Ÿξ…ž4(𝑑)β‰₯0.(2.32)
In terms of the properties of πœ‘2(π‘₯,𝑦), we know that πœ‘2(𝛼,𝛽)<0 for (𝛼,𝛽) lying on the left-side of the green curve, see Figure 1. From (2.24), we see that π‘Ÿ4(0)=𝛽(π›Όβˆ’1)πœ‘2(𝛼,𝛽)>0.(2.33) Combining (2.32) with (2.33) we get that π‘Ÿ3(𝑑) is strictly increasing on (0,∞).
If πœ‘1(𝛼,𝛽)β‰₯0, then 0<𝛽<1 and π‘Ÿ3(𝑑)>0 follow from (2.22), which implies that π‘Ÿ2(𝑑) is strictly increasing in (0,∞). Thus we can obtain π‘Ÿ2(𝑑)<limπ‘‘β†’βˆžπ‘Ÿ2(𝑑)=𝛼𝛽(π›Όπ›½βˆ’1)<0.(2.34)
If πœ‘1(𝛼,𝛽)<0, then it follows from limπ‘‘β†’βˆžπ‘Ÿ3(𝑑)=+∞ or 𝛼𝛽(1+𝛼𝛽)>0 that there exists 𝜎1>0 such that π‘Ÿ3(𝑑)<0 for π‘‘βˆˆ(0,𝜎1) and π‘Ÿ3(𝑑)>0 for π‘‘βˆˆ(𝜎1,∞). Hence, π‘Ÿ2(𝑑) is strictly decreasing in (0,𝜎1) and strictly increasing in (𝜎1,∞). Then we can obtain π‘Ÿ2ξ‚»π‘Ÿ(𝑑)<max2(0),limπ‘‘β†’βˆžπ‘Ÿ2ξ‚Ό(𝑑)≀0.(2.35) Finally, we conclude that π‘Ÿ(𝑑)<0 for π‘‘βˆˆ(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 π‘Ÿ2(𝑑)=βˆ’π›Όπ›½(1+𝛼𝛽)π‘’βˆ’π‘‘+𝑒(π›Όβˆ’1)𝛽𝑑𝛼(1+𝛽)π‘’βˆ’π‘‘ξ€»ξ€·+(1βˆ’π›Ό)(π›½βˆ’1)+𝛼𝛽(π›Όπ›½βˆ’1)<𝛼1βˆ’π›Όπ›½2ξ€Έπ‘’βˆ’π‘‘ξ€·+(1βˆ’π›Ό)(π›½βˆ’1)+𝛼𝛽(π›Όπ›½βˆ’1)≀𝛼1βˆ’π›Όπ›½2ξ€Έ+(1βˆ’π›Ό)(π›½βˆ’1)+𝛼𝛽(π›Όπ›½βˆ’1)=(π›½βˆ’1)(1βˆ’2𝛼)≀0.(2.36)
Therefore, π‘Ÿ(𝑑)<0 for π‘‘βˆˆ(0,∞) follows from (2.17), (2.18) together with (2.36).
Case 3. If (𝛼,𝛽)∈Ω5, then 1/2≀𝛼<1 and π›½βˆ’π›Όπ›½βˆ’1β‰₯0. From (2.26), we know that π‘Ÿξ…ž4(𝑑)β‰₯0.(2.37) In terms of the location of Ξ©3, we know that πœ‘2(𝛼,𝛽)<0. From (2.24), we see that π‘Ÿ4(0)=𝛽(π›Όβˆ’1)πœ‘2(𝛼,𝛽)>0.(2.38) It follows from (2.37) and (2.38) that π‘Ÿ3(𝑑) is strictly increasing on (0,∞).
If πœ‘1(𝛼,𝛽)β‰₯0, then 1/2<𝛼<1 and π‘Ÿ3(𝑑)>0 follow that from (2.22), which implies that π‘Ÿ2(𝑑) is strictly increasing on (0,∞). From (2.20) and (2.21), we see that π‘Ÿ2(0)=(π›½βˆ’1)(1βˆ’2𝛼)<0,lim𝑑→+βˆžπ‘Ÿ2(𝑑)=𝛼𝛽(π›Όπ›½βˆ’1)>0.(2.39) Thus there exists 𝜎2>0 such that π‘Ÿ2(𝑑)<0 for π‘‘βˆˆ(0,𝜎2) and π‘Ÿ2(𝑑)>0 for π‘‘βˆˆ(𝜎2,∞), which implies that π‘Ÿ1(𝑑) is strictly decreasing on (0,𝜎2) and strictly increasing on (𝜎2,∞). It follows from (2.18) and limπ‘‘β†’βˆžπ‘Ÿ1(𝑑)=𝛼>0 that 𝜎3>𝜎2 such that π‘Ÿ1(𝑑)<0 for π‘‘βˆˆ(0,𝜎3) and π‘Ÿ1(𝑑)>0 for π‘‘βˆˆ(𝜎3,∞), which implies that π‘Ÿ(𝑑) is strictly decreasing on (0,𝜎3) and strictly increasing on (𝜎3,∞). Therefore, it follows from (2.17) and limπ‘‘β†’βˆžπ‘Ÿ(𝑑)=π›Όβˆ’1<0 that ξ‚»π‘Ÿ(𝑑)<maxπ‘Ÿ(0),limπ‘‘β†’βˆžξ‚Όπ‘Ÿ(𝑑)=0(2.40) for π‘‘βˆˆ(0,∞).
If πœ‘1(𝛼,𝛽)<0, then there exists 𝜎4>0 such that π‘Ÿ3(𝑑)<0 for π‘‘βˆˆ(0,𝜎4) and π‘Ÿ3(𝑑)>0 for π‘‘βˆˆ(𝜎4,∞) follows from limπ‘‘β†’βˆžπ‘Ÿ3(𝑑)=𝛼𝛽(1+𝛼𝛽)>0 or limπ‘‘β†’βˆžπ‘Ÿ3(𝑑)=𝛽[(π›Όβˆ’1/2)2+𝛽(2π›Όβˆ’1)+3/4]>0. This leads to π‘Ÿ2(𝑑) being strictly decreasing in (0,𝜎4) and strictly increasing in (𝜎4,∞). From (2.20), we clearly see that π‘Ÿ2(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 π‘Ÿ2ξ‚»π‘Ÿ(𝑑)<max2(0),limπ‘‘β†’βˆžπ‘Ÿ2ξ‚Ό(𝑑)=0,(2.42) which implies that π‘Ÿ(𝑑)<0 for π‘‘βˆˆ(0,∞) follows from (2.17) and (2.18).
For 𝛼𝛽>1, it follows from (2.38) and limπ‘‘β†’βˆžπ‘Ÿ2(𝑑)=𝛼𝛽(π›Όπ›½βˆ’1)>0 that there exists 𝜎5>𝜎4>0 such that π‘Ÿ2(𝑑)<0 for π‘‘βˆˆ(0,𝜎5) and π‘Ÿ2(𝑑)>0 for π‘‘βˆˆ(𝜎5,∞). Making use of the same arguments as the case of πœ‘1(𝛼,𝛽)β‰₯0, then π‘Ÿ(𝑑)<0 for π‘‘βˆˆ(0,∞) follows from (2.17).

(3) If 0<𝛼<1/2, 𝛽>1 or 1/2<𝛼<1, 0<𝛽<1, then we have limπ‘‘β†’βˆžπ‘Ÿ(𝑑)=π›Όβˆ’1<0.(2.43)
From (2.20), we know that π‘Ÿ2(0)=(π›½βˆ’1)(1βˆ’2𝛼)>0.(2.44)
It follows from (2.44) that there exists 𝛿1>0 such that π‘Ÿ2(𝑑)>0 for π‘‘βˆˆ(0,𝛿1), which implies that π‘Ÿ1(𝑑) is strictly increasing on (0,𝛿1). Therefore, π‘Ÿ(𝑑)>0 for π‘‘βˆˆ(0,𝛿1) follows from (2.17) and (2.18).
From (2.43), we know that there exists 𝛿2≫𝛿1>0 such that π‘Ÿ(𝑑)<0 for π‘‘βˆˆ(𝛿2,∞).
(4) If 𝛼>1, 𝛽>1, then we have limπ‘‘β†’βˆžπ‘Ÿ(𝑑)=π›Όβˆ’1>0.(2.45)
From (2.15), we know that π‘Ÿ2(0)=(π›½βˆ’1)(1βˆ’2𝛼)<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 𝛿4≫𝛿3>0 such that π‘Ÿ2(𝑑)<0 for π‘‘βˆˆ(0,𝛿3) and π‘Ÿ(𝑑)>0 for π‘‘βˆˆ(𝛿4,∞).

3. Proof of Theorem 1.1

Proof of Theorem 1.1. From (2.15), we have (βˆ’1)𝑛log𝑓𝛼,𝛽(π‘₯)(𝑛)=(βˆ’1)𝑛(βˆ’1)𝑛𝛼(π‘›βˆ’1)!π‘₯𝑛+π›½πœ“(π‘›βˆ’1)(π‘₯+𝛼)βˆ’π›½π‘›πœ“(π‘›βˆ’1)ξ‚Ήξ€œ(𝛽π‘₯)=π›Όβˆž0π‘ π‘›βˆ’1π‘’βˆ’π‘₯π‘ ξ€œπ‘‘π‘ +π›½βˆž0π‘ π‘›βˆ’11βˆ’π‘’βˆ’π‘ π‘’βˆ’(π‘₯+𝛼)π‘ π‘‘π‘ βˆ’π›½π‘›ξ€œβˆž0π‘‘π‘›βˆ’11βˆ’π‘’βˆ’π‘‘π‘’βˆ’π›½π‘₯𝑑𝑑𝑑=π›Όπ›½π‘›ξ€œβˆž0π‘‘π‘›βˆ’1π‘’βˆ’π›½π‘₯𝑑𝑑𝑑+𝛽𝑛+1ξ€œβˆž0π‘‘π‘›βˆ’11βˆ’π‘’βˆ’π›½π‘‘π‘’βˆ’π›½(π‘₯+𝛼)π‘‘π‘‘π‘‘βˆ’π›½π‘›ξ€œβˆž0π‘‘π‘›βˆ’11βˆ’π‘’βˆ’π‘‘π‘’βˆ’π›½π‘₯𝑑𝑑𝑑=π›½π‘›ξ€œβˆž0π‘‘π‘›βˆ’1π‘’βˆ’π›½π‘₯𝑑(1βˆ’π‘’βˆ’π‘‘)ξ€·1βˆ’π‘’βˆ’π›½π‘‘ξ€Έπ‘Ÿ(𝑑)𝑑𝑑,(3.1) where ξ€·π‘Ÿ(𝑑)=1βˆ’π‘’βˆ’π‘‘ξ€Έξ€·π›½π‘’βˆ’π›Όπ›½π‘‘βˆ’π›Όπ‘’βˆ’π›½π‘‘ξ€Έ+π‘’βˆ’π›½π‘‘βˆ’π›Όπ‘’βˆ’π‘‘+π›Όβˆ’1.(3.2)
(1) If (𝛼,𝛽)∈Ω1βˆͺΞ©2, then from (3.1) and (3.2) together with Lemma 2.2(1) we clearly see that (βˆ’1)𝑛log𝑓𝛼,𝛽(π‘₯)(𝑛)>0.(3.3) Therefore, 𝑓𝛼,𝛽(π‘₯) is strictly logarithmically completely monotonic on (0,∞) following from (3.3).
(2) If (𝛼,𝛽)∈Ω3βˆͺΞ©4βˆͺΞ©5, then from (3.1) we can get (βˆ’1)𝑛𝑓log𝛼,𝛽(π‘₯)βˆ’1(𝑛)=βˆ’π›½π‘›ξ€œβˆž0π‘‘π‘›βˆ’1π‘’βˆ’π›½π‘₯𝑑(1βˆ’π‘’βˆ’π‘‘)ξ€·1βˆ’π‘’βˆ’π›½π‘‘ξ€Έπ‘Ÿ(𝑑)𝑑𝑑,(3.4) where π‘Ÿ(𝑑) is defined as (3.2).
Therefore, [𝑓𝛼,𝛽(π‘₯)]βˆ’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(π‘₯) nor [𝑓𝛼,𝛽(π‘₯)]βˆ’1 is strictly logarithmically completely monotonic on (0,∞) for (𝛼,𝛽)∈{(𝛼,𝛽)∢0<𝛼<1/2,𝛽>1}βˆͺ{(𝛼,𝛽)∢1/2<𝛼<1,0<𝛽<1}βˆͺ{(𝛼,𝛽)βˆΆπ›Ό>1,𝛽>1} following from Lemma 2.2 (3) and (4), it is known that the logarithmically completely monotonicity properties of 𝑓𝛼,𝛽(π‘₯) and [𝑓𝛼,𝛽(π‘₯)]βˆ’1 are not completely continuously depended on 𝛼 and 𝛽.

Remark 3.2. Compared with Theorem 9 of [40], we can also extend Ξ©3 onto one component of its boundaries, which is 𝛀3→𝛀3=1(𝛼,𝛽)∢0≀𝛼≀2,0<𝛽≀1⧡{𝛼=0,𝛽=1}.(3.5) Then [𝑓𝛼,𝛽(π‘₯)]βˆ’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.