Journal of Inequalities and Applications
Volume 2009 (2009), Article ID 728612, 13 pages
doi:10.1155/2009/728612
Research Article

Monotonic and Logarithmically Convex Properties of a Function Involving Gamma Functions

Tie-Hong Zhao,1 Yu-Ming Chu,2 and Yue-Ping Jiang3

1Institut de Mathématiques, Université Pierre et Marie Curie, 4 Place Jussieu, 75252 Paris, France
2Department of Mathematics, Huzhou Teachers College, Huzhou 313000, Zhejiang, China
3College of Mathematics and Econometrics, Hunan University, Changsha 410082, Hunan, China

Received 14 October 2008; Accepted 27 February 2009

Academic Editor: Sever Dragomir

Copyright © 2009 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.

Abstract

Using the series-expansion of digamma functions and other techniques, some monotonicity and logarithmical concavity involving the ratio of gamma function are obtained, which is to give a partially affirmative answer to an open problem posed by B.-N.Guo and F.Qi. Several inequalities for the geometric means of natural numbers are established.

1. Introduction

For real and positive values of 𝑥 the Euler gamma function Γ and its logarithmic derivative 𝜓 , the so-called digamma function, are defined as Γ ( 𝑥 ) = 0 + ∞ 𝑡 𝑥 − 1 𝑒 − 𝑡 Γ 𝑑 𝑡 , 𝜓 ( 𝑥 ) = ( 𝑥 ) Γ ( 𝑥 ) . ( 1 . 1 ) For extension of these functions to complex variables and for basic properties see [1].

In recent years, many monotonicity results and inequalities involving the Gamma and incomplete Gamma functions have been established. This article is stimulated by an open problem posed by Guo and Qi in [2]. The extensions and generalizations of this problem can be found in [3–5] and some references therein.

Using Stirling formula, for all nonnegative integers 𝑘 , natural numbers 𝑛 and 𝑚 , an upper bound of the quotient of two geometrical means of natural numbers was established in [4] as follows: ∏ 𝑛 + 𝑘 𝑖 = 𝑘 + 1 𝑖 1 / 𝑛 ∏ 𝑛 + 𝑚 + 𝑘 𝑖 = 𝑘 + 1 𝑖 1 / ( 𝑛 + 𝑚 ) ≤ 𝑛 + 𝑘 𝑛 + 𝑚 + 𝑘 , ( 1 . 2 ) and the following lower bound was appeared in [2, 5]: 𝑛 + 𝑘 + 1 < 𝑛 + 𝑚 + 𝑘 + 1 𝑛 √ ( 𝑛 + 𝑘 ) ! / 𝑘 ! 𝑛 + 𝑚 √ ( 𝑛 + 𝑚 + 𝑘 ) ! / 𝑘 ! , ( 1 . 3 )

Since Γ ( 𝑛 + 1 ) = 𝑛 ! , as a generalization of inequality (1.3), the following monotonicity result was obtained by Guo and Qi in [2]. The function [ Γ ( 𝑥 + 𝑦 + 1 ) / Γ ( 𝑦 + 1 ) ] 1 / 𝑥 𝑥 + 𝑦 + 1 ( 1 . 4 ) is decreasing with respect to 𝑥 on [ 1 , ∞ ) for fixed 𝑦 ≥ 0 . Hence, for positive real numbers 𝑥 and 𝑦 , we have 𝑥 + 𝑦 + 1 ≤ 𝑥 + 𝑦 + 2 Γ ( 𝑥 + 𝑦 + 1 ) / Γ ( 𝑦 + 1 ) 1 / 𝑥 Γ ( 𝑥 + 𝑦 + 2 ) / Γ ( 𝑦 + 1 ) 1 / ( 𝑥 + 1 ) . ( 1 . 5 )

Recently, in [6], Qi and Sun proved that the function Γ ( 𝑥 + 𝑦 + 1 ) / Γ ( 𝑦 + 1 ) 1 / 𝑥 √ 𝑥 + 𝑦 ( 1 . 6 ) is strictly increasing with respect to 𝑥 ∈ [ 𝑦 + 1 , ∞ ) for all 𝑦 ≥ 𝑦 0 .

Now, we generalize the function in (1.4) as follows: for positive real numbers 𝑥 and 𝑦 , 𝛼 ≥ 0 , let 𝐹 𝛼 ( 𝑥 , 𝑦 ) = Γ ( 𝑥 + 𝑦 + 1 ) / Γ ( 𝑦 + 1 ) 1 / 𝑥 ( 𝑥 + 𝑦 + 1 ) 𝛼 . ( 1 . 7 )

The aim of this paper is to discuss the monotonicity and logarithmical convexity of the function 𝐹 𝛼 ( 𝑥 , 𝑦 ) with respect to parameter 𝛼 .

For convenience of the readers, we recall the definitions and basic knowledge of convex function and logarithmically convex function.

Definition 1.1. Let 𝐷 ⊂ 𝑅 2 be a convex set, 𝑓 ∶ 𝐷 → 𝑅 is called a convex function on 𝐷 if 𝑓 𝐱 + 𝐲 2 ≤ 𝑓 ( 𝐱 ) + 𝑓 ( 𝐲 ) 2 ( 1 . 8 ) for all 𝐱 , 𝐲 ∈ 𝐷 , and 𝑓 is called concave if − 𝑓 is convex.

Definition 1.2. Let 𝐷 ⊂ 𝑅 2 be a convex set, 𝑓 ∶ 𝐷 → ( 0 , ∞ ) is called a logarithmically convex function on 𝐷 if l n 𝑓 is convex on 𝐷 , and 𝑓 is called logarithmically concave if l n 𝑓 is concave.
The following criterion for convexity of function was established by Fichtenholz in [7].

Proposition 1.3. Let 𝐷 ⊂ 𝑅 2 be a convex set, if 𝑓 ∶ 𝐷 → 𝑅 have continuous second partial derivatives, then 𝑓 is a convex (or concave) function on 𝐷 if and only if 𝐿 ( 𝐱 ) is a positive (or negative) semidefinite matrix for all 𝐱 ∈ 𝐷 , where 𝑓 𝐿 ( 𝐱 ) = 1 1 𝑓 1 2 𝑓 2 1 𝑓 2 2 ( 1 . 9 ) and 𝑓 𝑖 𝑗 = 𝜕 2 𝑓 ( 𝑥 1 , 𝑥 2 ) / 𝜕 𝑥 𝑖 𝜕 𝑥 𝑗 for 𝐱 = ( 𝑥 1 , 𝑥 2 ) , 𝑖 , 𝑗 = 1 , 2 .

Notation. In Definitions 1.1, 1.2 and Proposition 1.3, we denote 𝐱 , 𝐲 by the points (or vectors) of 𝑅 2 , and denote 𝑥 , 𝑦 by the real variables in the later.
Our main results are Theorems 1.4 and 1.5.

Theorem 1.4. (1) For any fixed 𝑦 ≥ 0 , 𝐹 𝛼 ( 𝑥 , 𝑦 ) is strictly increasing (or decreasing, resp.) with respect to 𝑥 on ( 0 , ∞ ) if and only if 0 ≤ 𝛼 ≤ 1 / 2 (or 𝛼 ≥ 1 , resp.);
(2) For any fixed 𝑥 > 0 , 𝐹 𝛼 ( 𝑥 , 𝑦 ) is strictly increasing with respect to 𝑦 on [ 0 , ∞ ) if and only if 0 ≤ 𝛼 ≤ 1 .

Theorem 1.5. (1) If 0 ≤ 𝛼 ≤ 1 / 4 , then 𝐹 𝛼 ( 𝑥 , 𝑦 ) is logarithmically concave with respect to ( 𝑥 , 𝑦 ) ∈ ( 0 , ∞ ) × ( 0 , ∞ ) ;
(2) If 𝐸 ⊂ ( 0 , ∞ ) × ( 0 , ∞ ) is a convex set with nonempty interior and 𝛼 ≥ 1 , then 𝐹 𝛼 ( 𝑥 , 𝑦 ) is neither logarithmically convex nor logarithmically concave with respect to ( 𝑥 , 𝑦 ) on 𝐸 .

The following two corollaries can be derived from Theorems 1.4 and 1.5 immediately.

Corollary 1.6. If ( 𝑥 , 𝑦 ) ∈ ( 0 , ∞ ) × ( 0 , ∞ ) , then 𝑥 + 𝑦 + 1 < 𝑥 + 𝑦 + 2 Γ ( 𝑥 + 𝑦 + 1 ) / Γ ( 𝑦 + 1 ) 1 / 𝑥 Γ ( 𝑥 + 𝑦 + 2 ) / Γ ( 𝑦 + 1 ) 1 / ( 𝑥 + 1 ) < 𝑥 + 𝑦 + 1 𝑥 + 𝑦 + 2 . ( 1 . 1 0 )

Remark 1.7. Inequality (1.3) can be derived from Corollary 1.6 if we take 𝑥 , 𝑦 ∈ ℕ . Although we cannot get the inequality (1.2) exactly from Corollary 1.6, but we can get the following inequality which is close to inequality (1.2): ∏ 𝑛 + 𝑘 𝑖 = 𝑘 + 1 𝑖 1 / 𝑛 ∏ 𝑛 + 𝑚 + 𝑘 𝑖 = 𝑘 + 1 𝑖 1 / ( 𝑛 + 𝑚 ) ≤ 𝑛 + 𝑘 + 1 𝑛 + 𝑚 + 𝑘 + 1 . ( 1 . 1 1 )

Corollary 1.8. If ( 𝑥 1 , 𝑦 1 ) , ( 𝑥 2 , 𝑦 2 ) ∈ ( 0 , ∞ ) × ( 0 , ∞ ) , then Γ 𝑥 1 + 𝑦 1 𝑦 + 1 / Γ 1 + 1 1 / 𝑥 1 ⋅ Γ 𝑥 2 + 𝑦 2 𝑦 + 1 / Γ 2 + 1 1 / 𝑥 2 Γ 𝑥 1 + 𝑥 2 + 𝑦 1 + 𝑦 2 𝑦 ) / 2 + 1 / Γ 1 + 𝑦 2 ) / 2 + 1 4 / ( 𝑥 1 + 𝑥 2 ) ≤ √ 2 𝑥 1 + 𝑦 1 𝑥 + 1 2 + 𝑦 2 + 1 1 / 4 √ 𝑥 1 + 𝑦 1 + 𝑥 2 + 𝑦 2 . + 2 ( 1 . 1 2 )

Remark 1.9. We conjecture that the inequality (1.2) can be improved if we can choose two pairs of integers ( 𝑥 1 , 𝑦 1 ) and ( 𝑥 2 , 𝑦 2 ) properly.

2. Lemmas

It is well known that the Bernoulli numbers 𝐵 𝑛 is defined [8] in general by 1 𝑒 𝑡 + 1 − 1 2 − 1 𝑡 = ∞ 𝑛 = 1 ( − 1 ) 𝑛 − 1 𝑡 2 𝑛 𝐵 ( 2 𝑛 ) ! 𝑛 . ( 2 . 1 ) In particular, we have 𝐵 1 = 1 6 , 𝐵 2 = 1 3 0 , 𝐵 3 = 1 4 2 , 𝐵 4 = 1 3 0 . ( 2 . 2 ) In [9], the following summation formula is given: ∞ 𝑛 = 0 ( − 1 ) 𝑛 ( 2 𝑛 + 1 ) 2 𝑘 + 1 = 𝜋 2 𝑘 + 1 𝐸 𝑘 2 2 𝑘 + 2 ( 2 𝑘 ) ! ( 2 . 3 ) for nonnegative integer 𝑘 , where 𝐸 𝑘 denotes the Euler number, which implies 𝐵 𝑛 = 2 ( 2 𝑛 ) ! ( 2 𝜋 ) ∞ 2 𝑛 𝑚 = 1 1 𝑚 2 𝑛 , 𝑛 ∈ ℕ . ( 2 . 4 )

Recently, the Bernoulli and Euler numbers and polynomials are generalized in [10–13]. The following two Lemmas were established by Qi and Guo in [3, 14].

Lemma 2.1 (see [3]). For real number 𝑥 > 0 and natural number 𝑚 , one has 1 l n Γ ( 𝑥 ) = 2 1 l n ( 2 𝜋 ) + 𝑥 − 2 l n 𝑥 − 𝑥 + 𝑚 𝑛 = 1 ( − 1 ) 𝑛 − 1 𝐵 𝑛 ⋅ 1 2 ( 2 𝑛 − 1 ) 𝑛 𝑥 2 𝑛 − 1 + ( − 1 ) 𝑚 𝜃 1 𝐵 𝑚 + 1 ⋅ 1 ( 2 𝑚 + 1 ) ( 2 𝑚 + 2 ) 𝑥 2 𝑚 + 1 , 0 < 𝜃 1 1 < 1 ; ( 2 . 5 ) 𝜓 ( 𝑥 ) = l n 𝑥 − + 2 𝑥 𝑚 𝑛 = 1 ( − 1 ) 𝑛 𝐵 𝑛 ⋅ 1 2 𝑛 𝑥 2 𝑛 + ( − 1 ) 𝑚 + 1 𝜃 2 𝐵 𝑚 + 1 ⋅ 1 ( 2 𝑚 + 2 ) 𝑥 2 𝑚 + 2 , 0 < 𝜃 2 𝜓 < 1 ; ( 2 . 6 ) 1 ( 𝑥 ) = 𝑥 + 1 2 𝑥 2 + 𝑚 𝑛 = 1 ( − 1 ) 𝑛 − 1 𝐵 𝑛 𝑥 2 𝑛 + 1 + ( − 1 ) 𝑚 𝜃 3 ⋅ 𝐵 𝑚 + 1 𝑥 2 𝑚 + 3 , 0 < 𝜃 3 𝜓 < 1 ; ( 2 . 7 ) 1 ( 𝑥 ) = − 𝑥 2 − 1 𝑥 3 + 𝑚 𝑛 = 1 ( − 1 ) 𝑛 𝐵 ( 2 𝑛 + 1 ) 𝑛 𝑥 2 𝑛 + 2 + ( − 1 ) 𝑚 + 1 ( 2 𝑚 + 3 ) 𝜃 4 ⋅ 𝐵 𝑚 + 1 𝑥 2 𝑚 + 4 , 0 < 𝜃 4 < 1 . ( 2 . 8 )

Lemma 2.2 (see [14]). Inequalities 1 l n 𝑥 − 𝑥 1 ≤ 𝜓 ( 𝑥 ) ≤ l n 𝑥 − , 2 𝑥 ( 2 . 9 ) ( 𝑘 − 1 ) ! 𝑥 𝑘 + 𝑘 ! 2 𝑥 𝑘 + 1 ≤ ( − 1 ) 𝑘 + 1 𝜓 ( 𝑘 ) ( 𝑥 ) ≤ ( 𝑘 − 1 ) ! 𝑥 𝑘 + 𝑘 ! 𝑥 𝑘 + 1 ( 2 . 9 ) hold in ( 0 , ∞ ) for 𝑘 ∈ ℕ .

Lemma 2.3. Let 𝑟 ( 𝑥 , 𝑦 ) = 𝜓 ( 𝑥 + 𝑦 + 1 ) − 𝜓 ( 𝑦 + 1 ) − 𝛼 𝑥 / ( 𝑥 + 𝑦 + 1 ) , then the following statements are true:
(1) if 0 ≤ 𝛼 ≤ 1 , then 𝑟 ( 𝑥 , 𝑦 ) ≥ 0 for ( 𝑥 , 𝑦 ) ∈ ( 0 , ∞ ) × [ 0 , ∞ ) ; (2) if 𝛼 > 1 , then 𝑟 ( 𝛼 , 𝑦 ) < 0 for 𝑦 ∈ ( 2 / ( 𝛼 − 1 ) , ∞ ) .

Proof. (1) Making use of (2.6) we get l i m 𝑦 → ∞ 𝑟 ( 𝑥 , 𝑦 ) = l i m 𝑦 → ∞ l n ( 𝑥 + 𝑦 + 1 ) − l n ( 𝑦 + 1 ) = 0 ( 2 . 1 1 ) for any fixed 𝑥 > 0 .
Since 𝜓 ( 𝑥 + 1 ) = 1 / 𝑥 + 𝜓 ( 𝑥 ) and 0 ≤ 𝛼 ≤ 1 , we have 𝑥 𝑟 ( 𝑥 , 𝑦 ) − 𝑟 ( 𝑥 , 𝑦 + 1 ) = ( 1 − 𝛼 ) 𝑦 + 𝑥 + 2 − 𝛼 ( 𝑦 + 1 ) ( 𝑥 + 𝑦 + 1 ) ( 𝑥 + 𝑦 + 2 ) > 0 ( 2 . 1 2 ) for all ( 𝑥 , 𝑦 ) ∈ ( 0 , ∞ ) × [ 0 , ∞ ) .
Therefore, Lemma 2.3(1) follows from (2.11) and (2.12).
(2) If 𝛼 > 1 , then (2.12) leads to 𝑟 ( 𝛼 , 𝑦 ) − 𝑟 ( 𝛼 , 𝑦 + 1 ) < 0 ( 2 . 1 3 ) for 𝑦 ∈ ( 2 / ( 𝛼 − 1 ) , ∞ ) .
Therefore, Lemma 2.3(2) follows from (2.11) and (2.13).

Lemma 2.4. If 𝑔 ( 𝑥 , 𝑦 ) = 2 𝑥 𝜓 ( 𝑦 + 1 ) − 2 [ l n Γ ( 𝑥 + 𝑦 + 1 ) − l n Γ ( 𝑦 + 1 ) ] + 𝑥 2 𝜓 ( 𝑦 + 1 ) , then 𝑔 ( 𝑥 , 𝑦 ) > 0 for ( 𝑥 , 𝑦 ) ∈ ( 0 , ∞ ) × ( 0 , ∞ ) .

Proof. It is easy to see that 𝑔 ( 0 , 𝑦 ) = 0 ( 2 . 1 4 ) for all 𝑦 ∈ ( 0 , ∞ ) .
Let 𝑔 1 ( 𝑥 , 𝑦 ) = 𝜕 𝑔 ( 𝑥 , 𝑦 ) / 𝜕 𝑥 , then 𝑔 1 ( 𝑥 , 𝑦 ) = 2 𝑥 𝜓 𝑔 ( 𝑦 + 1 ) − 𝜓 ( 𝑥 + 𝑦 + 1 ) + 𝜓 ( 𝑦 + 1 ) , ( 2 . 1 5 ) 1 ( 0 , 𝑦 ) = 0 , ( 2 . 1 6 ) 𝜕 𝑔 1 ( 𝑥 , 𝑦 ) 𝜓 𝜕 𝑥 = 2 ( 𝑦 + 1 ) − 𝜓 ( 𝑥 + 𝑦 + 1 ) > 0 ( 2 . 1 7 ) for 𝑥 > 0 . On the other hand, from (2.10) we know that 𝜓 ( 𝑥 ) is strictly decreasing on ( 0 , ∞ ) .
Therefore, Lemma 2.4 follows from (2.14)–(2.17).

Remark 2.5. Let 2 𝑎 ( 𝑥 , 𝑦 ) = 𝑥 3 − 2 l n Γ ( 𝑥 + 𝑦 + 1 ) − l n Γ ( 𝑦 + 1 ) 𝑥 2 1 𝜓 ( 𝑥 + 𝑦 + 1 ) , 𝑏 ( 𝑥 , 𝑦 ) = − 𝑥 2 , 1 𝜓 ( 𝑥 + 𝑦 + 1 ) − 𝜓 ( 𝑦 + 1 ) 𝑐 ( 𝑥 , 𝑦 ) = − 𝑥 𝜓 ( 𝑦 + 1 ) . ( 2 . 1 8 ) Then simple computation shows that 𝑔 ( 𝑥 , 𝑦 ) = 𝑥 3 2 𝑏 ( 𝑥 , 𝑦 ) − 𝑎 ( 𝑥 , 𝑦 ) − 𝑐 ( 𝑥 , 𝑦 ) . ( 2 . 1 9 )

Lemma 2.6. Let 𝑑 ( 𝑥 , 𝑦 ) = ( 1 / 𝑥 ) 𝜓 ( 𝑥 + 𝑦 + 1 ) + 𝛼 / ( 𝑥 + 𝑦 + 1 ) 2 , then the following statements are true:
(1) if 0 ≤ 𝛼 ≤ 1 / 4 , then > 𝑎 ( 𝑥 , 𝑦 ) + 𝑑 ( 𝑥 , 𝑦 ) 𝑐 ( 𝑥 , 𝑦 ) + 𝑑 ( 𝑥 , 𝑦 ) 𝑏 ( 𝑥 , 𝑦 ) + 𝑑 ( 𝑥 , 𝑦 ) 2 ( 2 . 2 0 ) for ( 𝑥 , 𝑦 ) ∈ ( 0 , ∞ ) × ( 0 , ∞ ) ; (2) if 𝛼 ≥ 1 , then < 𝑎 ( 𝑥 , 𝑦 ) + 𝑑 ( 𝑥 , 𝑦 ) 𝑐 ( 𝑥 , 𝑦 ) + 𝑑 ( 𝑥 , 𝑦 ) 𝑏 ( 𝑥 , 𝑦 ) + 𝑑 ( 𝑥 , 𝑦 ) 2 ( 2 . 2 1 ) for ( 𝑥 , 𝑦 ) ∈ ( 0 , ∞ ) × ( 0 , ∞ ) .

Proof. Let 𝑓 ( 𝑥 , 𝑦 ) = 2 𝜓 − ( 𝑦 + 1 ) 𝑥 𝜓 ( 𝑥 + 𝑦 + 1 ) − l n Γ ( 𝑥 + 𝑦 + 1 ) + l n Γ ( 𝑦 + 1 ) 𝜓 ( 𝑥 + 𝑦 + 1 ) − 𝜓 ( 𝑦 + 1 ) 2 , 𝜓 𝑝 ( 𝑥 , 𝑦 ) = 𝑓 ( 𝑥 , 𝑦 ) − 𝑔 ( 𝑥 , 𝑦 ) ( 𝑥 + 𝑦 + 1 ) + 𝛼 𝑥 ( 𝑥 + 𝑦 + 1 ) 2 . ( 2 . 2 2 ) Then it is not difficult to verify 𝑝 ( 0 , 𝑦 ) = 0 , ( 2 . 2 3 ) 𝑝 ( 𝑥 , 𝑦 ) = 𝑥 4 − 𝑎 ( 𝑥 , 𝑦 ) + 𝑑 ( 𝑥 , 𝑦 ) 𝑐 ( 𝑥 , 𝑦 ) + 𝑑 ( 𝑥 , 𝑦 ) 𝑏 ( 𝑥 , 𝑦 ) + 𝑑 ( 𝑥 , 𝑦 ) 2 , ( 2 . 2 4 ) 𝜕 𝑝 ( 𝑥 , 𝑦 ) 𝜕 𝑥 = − 𝛼 𝑥 ( 𝑥 + 𝑦 + 1 ) 2 𝜕 𝑔 ( 𝑥 , 𝑦 ) 𝜓 𝜕 𝑥 − 𝑔 ( 𝑥 , 𝑦 ) 𝛼 ( 𝑥 + 𝑦 + 1 ) + ( 𝑥 + 𝑦 + 1 ) 2 − 2 𝛼 𝑥 ( 𝑥 + 𝑦 + 1 ) 3 . ( 2 . 2 5 )
(1) If 0 ≤ 𝛼 ≤ 1 / 4 , then making use of Lemmas 2.2, 2.4 and (2.25) we get 𝜕 𝑝 ( 𝑥 , 𝑦 ) 𝜕 𝑥 > − 𝛼 𝑥 ( 𝑥 + 𝑦 + 1 ) 2 𝜕 𝑔 ( 𝑥 , 𝑦 ) 1 𝜕 𝑥 + 𝑔 ( 𝑥 , 𝑦 ) ( 𝑥 + 𝑦 + 1 ) 2 + 1 ( 𝑥 + 𝑦 + 1 ) 3 − 𝛼 ( 𝑥 + 𝑦 + 1 ) 2 + 2 𝛼 𝑥 ( 𝑥 + 𝑦 + 1 ) 3 > 1 ( 𝑥 + 𝑦 + 1 ) 2 ( 1 − 𝛼 ) 𝑔 ( 𝑥 , 𝑦 ) − 𝛼 𝑥 𝜕 𝑔 ( 𝑥 , 𝑦 ) 𝜕 𝑥 , ( 2 . 2 6 ) for ( 𝑥 , 𝑦 ) ∈ ( 0 , ∞ ) × ( 0 , ∞ ) .
Let 𝑔 𝑖 ( 𝑥 , 𝑦 ) = 𝜕 𝑖 𝑔 ( 𝑥 , 𝑦 ) / 𝜕 𝑥 𝑖 , 𝑖 = 1 , 2 , 3 , 4 , 𝑞 ( 𝑥 , 𝑦 ) = ( 1 − 𝛼 ) 𝑔 ( 𝑥 , 𝑦 ) − 𝛼 𝑥 ( 𝜕 𝑔 ( 𝑥 , 𝑦 ) / 𝜕 𝑥 ) , and 𝑞 𝑗 ( 𝑥 , 𝑦 ) = 𝜕 𝑗 𝑞 ( 𝑥 , 𝑦 ) / 𝜕 𝑥 𝑗 , 𝑗 = 1 , 2 . Then simple computation leads to 𝑔 3 ( 𝑥 , 𝑦 ) = − 2 𝜓 𝑔 ( 𝑥 + 𝑦 + 1 ) , ( 2 . 2 7 ) 4 ( 𝑥 , 𝑦 ) = − 2 𝜓 ( 𝑥 + 𝑦 + 1 ) , ( 2 . 2 8 ) 𝜕 𝑞 2 ( 𝑥 , 𝑦 ) 𝜕 𝑥 = ( 1 − 4 𝛼 ) 𝑔 3 ( 𝑥 , 𝑦 ) − 𝛼 𝑥 𝑔 4 𝑞 ( 𝑥 , 𝑦 ) , ( 2 . 2 9 ) 2 ( 0 , 𝑦 ) = 𝑞 1 ( 0 , 𝑦 ) = 𝑞 ( 0 , 𝑦 ) = 0 ( 2 . 3 0 ) for all 𝑦 ∈ ( 0 , ∞ ) .
It is well known that ∑ l n Γ ( 𝑥 ) = − 𝑐 𝑥 + ∞ 𝑘 = 1 [ 𝑥 / 𝑘 − l n ( 1 + 𝑥 / 𝑘 ) ] − l n 𝑥 , where 𝑐 = 0 . 5 7 7 2 1 5 ⋯ is the Euler's constant. From this we get 𝜓 ( 𝑛 ) = ( − 1 ) 𝑛 + 1 𝑛 ! ∞ 𝑘 = 0 1 ( 𝑘 + 𝑥 ) 𝑛 + 1 . ( 2 . 3 1 )
From Lemma 2.2, (2.27)–(2.29), (2.31) and the assumption 0 ≤ 𝛼 ≤ 1 / 4 , we conclude that 𝜕 𝑞 2 ( 𝑥 , 𝑦 ) 𝜕 𝑥 > 0 . ( 2 . 3 2 )
Therefore, Lemma 2.6(1) follows from (2.23)–(2.26), (2.30), and (2.32).
(2) If 𝛼 ≥ 1 , then making use of (2.8), Lemma 2.4 and (2.25) we obtain 𝜕 𝑝 ( 𝑥 , 𝑦 ) 𝜕 𝑥 < − 𝛼 𝑥 ( 𝑥 + 𝑦 + 1 ) 2 𝜕 𝑔 ( 𝑥 , 𝑦 ) 1 𝜕 𝑥 + 𝑔 ( 𝑥 , 𝑦 ) ( 𝑥 + 𝑦 + 1 ) 3 + 1 2 ( 𝑥 + 𝑦 + 1 ) 4 + 2 𝛼 𝑥 ( 𝑥 + 𝑦 + 1 ) 3 < − 𝛼 𝑥 ( 𝑥 + 𝑦 + 1 ) 2 𝜕 𝑔 ( 𝑥 , 𝑦 ) 𝜕 𝑥 + 𝑔 ( 𝑥 , 𝑦 ) 2 𝛼 ( 𝑥 + 1 ) ( 𝑥 + 𝑦 + 1 ) 3 < 𝛼 ( 𝑥 + 1 ) ( 𝑥 + 𝑦 + 1 ) 3 2 𝑔 ( 𝑥 , 𝑦 ) − 𝑥 𝜕 𝑔 ( 𝑥 , 𝑦 ) . 𝜕 𝑥 ( 2 . 3 3 ) Let 𝑣 ( 𝑥 , 𝑦 ) = 2 𝑔 ( 𝑥 , 𝑦 ) − 𝑥 𝜕 𝑔 ( 𝑥 , 𝑦 ) 𝜕 𝑥 , 𝑣 𝑖 𝜕 ( 𝑥 , 𝑦 ) = 𝑖 𝑣 ( 𝑥 , 𝑦 ) 𝜕 𝑥 𝑖 , 𝑖 = 1 , 2 . ( 2 . 3 4 ) Then 𝑣 2 ( 𝑥 , 𝑦 ) = 2 𝑥 𝜓 ( 𝑥 + 𝑦 + 1 ) < 0 ( 2 . 3 5 ) for ( 𝑥 , 𝑦 ) ∈ ( 0 , ∞ ) × ( 0 , ∞ ) by Lemma 2.2, and 𝑣 ( 0 , 𝑦 ) = 𝑣 1 ( 0 , 𝑦 ) = 0 ( 2 . 3 6 ) for 𝑦 ∈ ( 0 , ∞ ) .
Therefore, Lemma 2.6(2) follows from (2.23)–(2.25) and (2.33)–(2.36).

3. Proofs of Theorems 1.4 and 1.5

Proof of Theorem 1.4. (1) Let 𝐺 ( 𝑥 , 𝑦 ) = l n 𝐹 𝛼 ( 𝑥 , 𝑦 ) and 𝐺 1 ( 𝑥 , 𝑦 ) = 𝑥 2 ( 𝜕 𝐺 ( 𝑥 , 𝑦 ) / 𝜕 𝑥 ) , then 𝐺 1 ( 𝑥 , 𝑦 ) = − l n Γ ( 𝑥 + 𝑦 + 1 ) − l n Γ ( 𝑦 + 1 ) + 𝑥 𝜓 ( 𝑥 + 𝑦 + 1 ) − 𝛼 𝑥 2 𝑥 + 𝑦 + 1 . ( 3 . 1 ) The following three cases will complete the proof of Theorem 1.4(1).
Case 1. If 0 ≤ 𝛼 ≤ 1 / 2 , then (3.1) and Lemma 2.2 imply 𝜕 𝐺 1 ( 𝑥 , 𝑦 ) 𝜓 𝜕 𝑥 = 𝑥 ( 𝑥 + 𝑦 + 1 ) − 𝛼 ( 𝑥 + 2 𝑦 + 2 ) ( 𝑥 + 𝑦 + 1 ) 2 1 > 𝑥 + 1 𝑥 + 𝑦 + 1 2 ( 𝑥 + 𝑦 + 1 ) 2 − 𝛼 ( 𝑥 + 2 𝑦 + 2 ) ( 𝑥 + 𝑦 + 1 ) 2 = 𝑥 2 ( 𝑥 + 𝑦 + 1 ) 2 ( 2 − 2 𝛼 ) 𝑥 + ( 2 − 4 𝛼 ) 𝑦 + 3 − 4 𝛼 > 0 ( 3 . 2 ) for ( 𝑥 , 𝑦 ) ∈ ( 0 , ∞ ) × [ 0 , ∞ ) .
From (3.2) and the fact that 𝐺 1 ( 0 , 𝑦 ) = 0 for all 𝑦 ∈ [ 0 , ∞ ) we know that 𝐹 𝛼 ( 𝑥 , 𝑦 ) is strictly increasing with respect to 𝑥 on ( 0 , ∞ ) for any fixed 𝑦 ∈ [ 0 , ∞ ) .Case 2. If 𝛼 ≥ 1 , then (3.1) and (2.7) imply 𝜕 𝐺 1 ( 𝑥 , 𝑦 ) 1 𝜕 𝑥 < 𝑥 + 1 𝑥 + 𝑦 + 1 2 ( 𝑥 + 𝑦 + 1 ) 2 + 1 6 ( 𝑥 + 𝑦 + 1 ) 3 − 𝛼 ( 𝑥 + 2 𝑦 + 2 ) ( 𝑥 + 𝑦 + 1 ) 2 = 𝑥 6 ( 𝑥 + 𝑦 + 1 ) 3 ( 6 − 6 𝛼 ) 𝑥 2 + 𝜆 1 ( 𝑦 ) 𝑥 + 𝜆 2 ( 𝑦 ) < 0 ( 3 . 3 ) for ( 𝑥 , 𝑦 ) ∈ ( 0 , ∞ ) × [ 0 , ∞ ) , where 𝜆 1 ( 𝑦 ) = ( 1 2 − 1 8 𝛼 ) 𝑦 + 1 5 − 1 8 𝛼 < 0 and 𝜆 2 ( 𝑦 ) = 6 ( 1 − 2 𝛼 ) 𝑦 2 + ( 1 5 − 2 4 𝛼 ) 𝑦 + 1 0 − 1 2 𝛼 < 0 .
From (3.3) and the fact that 𝐺 1 ( 0 , 𝑦 ) = 0 for all 𝑦 ∈ [ 0 , ∞ ) we know that 𝐹 𝛼 ( 𝑥 , 𝑦 ) is strictly decreasing with respect to 𝑥 on ( 0 , ∞ ) for any fixed 𝑦 ∈ [ 0 , ∞ ) .Case 3. If 1 / 2 < 𝛼 < 1 , let 𝐺 2 ( 𝑥 , 𝑦 ) = 𝜓 ( 𝑥 + 𝑦 + 1 ) − 𝛼 ( 𝑥 + 2 𝑦 + 2 ) ( 𝑥 + 𝑦 + 1 ) 2 . ( 3 . 4 ) Then 𝜕 𝐺 1 ( 𝑥 , 𝑦 ) 𝜕 𝑥 = 𝑥 𝐺 2 ( 𝑥 , 𝑦 ) , ( 3 . 5 )
𝐺 2 1 ( 0 , 𝑦 ) < + 1 𝑦 + 1 2 ( 𝑦 + 1 ) 2 + 1 6 ( 𝑦 + 1 ) 3 − 2 𝛼 = 1 𝑦 + 1 6 ( 𝑦 + 1 ) 3 6 ( 1 − 2 𝛼 ) 𝑦 2 + ( 1 5 − 2 4 𝛼 ) 𝑦 + 1 0 − 1 2 𝛼 < 0 ( 3 . 6 ) for √ 𝑦 ≥ ( 1 5 − 2 4 𝛼 + 4 8 𝛼 − 1 5 ) / ( 2 4 𝛼 − 1 2 ) .
It is obvious that (3.6) implies 𝐺 2 √ 0 , 1 5 + 4 8 𝛼 − 1 5 2 4 𝛼 − 1 2 < 0 . ( 3 . 7 ) The continuity of 𝐺 2 ( 𝑥 , 𝑦 ) with respect to 𝑥 ∈ ( 0 , ∞ ) for any fixed 𝑦 ∈ [ 0 , ∞ ) and (3.7) imply that there exists 𝛿 = 𝛿 ( 𝛼 ) > 0 such that 𝐺 2 √ 𝑥 , 1 5 + 4 8 𝛼 − 1 5 2 4 𝛼 − 1 2 < 0 ( 3 . 8 ) for 𝑥 ∈ ( 0 , 𝛿 ) .
From (3.5), (3.8) and 𝐺 1 √ ( 0 , ( 1 5 + 4 8 𝛼 − 1 5 ) / ( 2 4 𝛼 − 1 2 ) ) = 0 we know that 𝐹 𝛼 ( 𝑥 , 𝑦 ) is strictly decreasing with respect to 𝑥 on ( 0 , 𝛿 ) for √ 𝑦 = ( 1 5 + 4 8 𝛼 − 1 5 ) / ( 2 4 𝛼 − 1 2 ) .
On the other hand, making use of (2.5) and (2.6) we have l i m 𝑥 → + ∞ 𝐺 1 ( 𝑥 , 𝑦 ) = l i m 𝑥 → + ∞ 𝑥 1 1 − 𝑦 + 2 l n ( 𝑥 + 𝑦 + 1 ) 𝑥 − 𝛼 𝑥 𝑥 + 𝑦 + 1 + 𝐶 𝑦 , 𝜃 1 = l i m 𝑥 → + ∞ ( 1 − 𝛼 ) 𝑥 + 𝐶 𝑦 , 𝜃 1 = + ∞ , ( 3 . 9 ) where 𝐶 𝑦 , 𝜃 1 = 1 𝑦 + 2 1 l n ( 𝑦 + 1 ) + − 1 1 2 ( 𝑦 + 1 ) 2 − 𝜃 1 3 6 0 ( 𝑦 + 1 ) 3 ( 3 . 1 0 ) for 𝑦 ∈ [ 0 , ∞ ) and 0 < 𝜃 1 < 1 .
Equation (3.9) implies that there exists 𝑀 = 𝑀 ( 𝛼 ) > 𝛿 ( 𝛼 ) such that 𝐺 1 √ 𝑥 , 1 5 + 4 8 𝛼 − 1 5 2 4 𝛼 − 1 2 > 0 ( 3 . 1 1 ) for 𝑥 ∈ ( 𝑀 , ∞ ) .
Hence, from (3.11) we know that 𝐹 𝛼 ( 𝑥 , 𝑦 ) is strictly increasing with respect to 𝑥 on ( 𝑀 , ∞ ) for √ 𝑦 = ( 1 5 + 4 8 𝛼 − 1 5 ) / ( 2 4 𝛼 − 1 2 ) .
(2) Since 𝑥 𝜕 𝐺 ( 𝑥 , 𝑦 ) 𝜕 𝑦 = 𝜓 ( 𝑥 + 𝑦 + 1 ) − 𝜓 ( 𝑦 + 1 ) − 𝛼 𝑥 𝑥 + 𝑦 + 1 = 𝑟 ( 𝑥 , 𝑦 ) , ( 3 . 1 2 ) then, Theorem 1.4(2) follows from (3.12) and Lemma 2.3.

Proof of Theorem 1.5. Let 𝐺 ( 𝑥 , 𝑦 ) = l n 𝐹 𝛼 ( 𝑥 , 𝑦 ) , 𝐺 1 1 ( 𝑥 , 𝑦 ) = 𝜕 2 𝐺 ( 𝑥 , 𝑦 ) / 𝜕 𝑥 2 , 𝐺 1 2 = 𝜕 2 𝐺 ( 𝑥 , 𝑦 ) / 𝜕 𝑥 𝜕 𝑦 and 𝐺 2 2 ( 𝑥 , 𝑦 ) = 𝜕 2 𝐺 ( 𝑥 , 𝑦 ) / 𝜕 𝑦 2 , then simple calculation yields 𝐺 1 1 2 ( 𝑥 , 𝑦 ) = 𝑥 3 − 2 l n Γ ( 𝑥 + 𝑦 + 1 ) − l n Γ ( 𝑦 + 1 ) 𝑥 2 + 1 𝜓 ( 𝑥 + 𝑦 + 1 ) 𝑥 𝜓 𝛼 ( 𝑥 + 𝑦 + 1 ) + ( 𝑥 + 𝑦 + 1 ) 2 𝐺 = 𝑎 ( 𝑥 , 𝑦 ) + 𝑑 ( 𝑥 , 𝑦 ) , ( 3 . 1 3 ) 1 2 1 ( 𝑥 , 𝑦 ) = − 𝑥 2 + 1 𝜓 ( 𝑥 + 𝑦 + 1 ) − 𝜓 ( 𝑦 + 1 ) 𝑥 𝜓 𝛼 ( 𝑥 + 𝑦 + 1 ) + ( 𝑥 + 𝑦 + 1 ) 2 𝐺 = 𝑏 ( 𝑥 , 𝑦 ) + 𝑑 ( 𝑥 , 𝑦 ) , ( 3 . 1 4 ) 2 2 1 ( 𝑥 , 𝑦 ) = 𝑥 𝜓 ( 𝑥 + 𝑦 + 1 ) − 𝜓 + 𝛼 ( 𝑦 + 1 ) ( 𝑥 + 𝑦 + 1 ) 2 = 𝑐 ( 𝑥 , 𝑦 ) + 𝑑 (