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Necessary and sufficient conditions for a class of functions and their reciprocals to be logarithmically completely monotonic

Journal of Inequalities and Applications20112011:36

https://doi.org/10.1186/1029-242X-2011-36

©  Lv et al; licensee Springer. 2011

Received: 9 February 2011

Accepted: 22 August 2011

Published: 22 August 2011

Abstract

We prove that the function F α,β (x) = x α Γ β (x)/Γ(βx) is strictly logarithmically completely monotonic on (0, ∞) if and only if (α, β) ∈ {(α, β) : β > 0, β ≥ 2α + 1, β ≥ α + 1}\{(α, β) : α = 0, β = 1} and that [F α,β (x)]-1 is strictly logarithmically completely monotonic on (0, ∞) if and only if (α, β) ∈ {(α, β ) : β > 0, β ≤ 2α + 1, β ≤ α + 1}\{(α, β ) : α = 0, β = 1}.

2010 Mathematics Subject Classification: 33B15; 26A48.

Keywords

completely monotonic logarithmically completely monotonic gamma function

1 Introduction

For real and positive values of x the Euler gamma function Γ and its logarithmic derivative ψ, the so-called digamma functions are defined by
Γ ( x ) = ∫ 0 ∞ t x - 1 e - t d t ,
(1.1)
ψ ( x ) = Γ ′ ( x ) Γ ( x ) = - γ + ∫ 0 ∞ e - t - e - x t 1 - e - t d t ,
(1.2)

where γ = 0.5772 ··· is the Euler's constant.

For extension of these functions to complex variable and for basic properties see [1]. Over the last half century, many authors have established inequalities and monotonicity for these functions [2–22].

We know that a real-valued function f : I → ℝ is said to be completely monotonic on I if f has derivatives of all orders on I and
( - 1 ) n f ( n ) ( x ) ≥ 0
(1.3)

for all x ∈ I and n ≥ 0. Moreover, f is said to be strictly completely monotonic if inequalities (1.3) are strict.

We also know 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 log f satisfies
( - 1 ) k [ l o g f ( x ) ] ( k ) ≥ 0
(1.4)

for all x ∈ I and k ∈ ℕ. Moreover, f is said to be strictly logarithmically completely monotonic if inequalities (1.4) are strict.

Recently, the completely monotonic or logarithmically completely monotonic functions have been the subject of intensive research. In particular, many complete monotonicity and logarithmically complete monotonicity properties related to the gamma function, psi function, and polygamma function can be found in the literature [17, 18, 23–37]. In 1997, Merkle [38] proved that F ( x ) = Γ ( 2 x ) Γ 2 ( x ) is strictly log-concave on (0, ∞). Later, Chen [39] showed that [ F ( x ) ] - 1 = Γ 2 ( x ) Γ ( 2 x ) is strictly logarithmically completely monotonic on (0, ∞). In [40], Li and Chen proved that F β ( x ) = Γ β ( x ) Γ ( β x ) is strictly logarithmically completely monotonic on (0, ∞) for β > 1, and that [F β (x)]-1 is strictly logarithmically completely monotonic on (0, ∞) for 0 < β < 1. The purpose of this article is to generalize Li and Chen's result. Our main result is as follows.

Theorem 1.1 Let α ∈ ℝ, β > 0 and F α,β (x) = x α Γ β (x)/Γ(βx), then
  1. (1)

    F α,β (x) is strictly logarithmically completely monotonic on (0, ∞) if and only if (α, β) ∈ {(α, β) : β > 0, β ≥ 2α + 1, β ≥ α + 1}\{(α, β) : α = 0, β = 1};

     
  2. (2)

    [F α,β (x)]-1 is strictly logarithmically completely monotonic on (0, ∞) if and only if (α, β) ∈ {(α, β) : β > 0, β ≤ 2α + 1, β ≤ α + 1}\{(α, β) : α = 0, β = 1}.

     

2 Lemma

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

Lemma 2.1 Let α ∈ ℝ, β ∈ (0, 1) ∪ (1, ∞) and
h ( t ) = - α e - ( β + 1 ) t + ( α + 1 ) e - β t + ( α - β ) e - t + β - α - 1 .
Then the following statements are true:
  1. (1)

    If β ≤ α + 1 and β ≤ 2α + 1, then h(t) < 0 for t ∈ (0, ∞);

     
  2. (2)

    If α + 1 < β < 2α + 1, then there exists λ 1 ∈ (0, ∞) such that h(t) < 0 for t ∈ (0, λ 1) and h(t) > 0 for t ∈ (λ 1, ∞);

     
  3. (3)

    If β ≥ α + 1 and β ≥ 2α + 1, then h(t) > 0 for t ∈ (0, ∞);

     
  4. (4)

    If 2α + 1 < β < α + 1, then there exists λ 2 ∈ (0, ∞) such that h(t) > 0 for t ∈ (0, λ 2) and h(t) < 0 for t ∈ (λ 2, ∞).

     
Proof Let h1(t) = e(β + 1)th'(t) and h 2 ( t ) = e - t h 1 ′ ( t ) . Then simple computations lead to
h ( 0 ) = 0 ,
(2.1)
h ′ ( t ) = α ( β + 1 ) e - ( β + 1 ) t - β ( α + 1 ) e - β t - ( α - β ) e - t , h 1 ( 0 ) = h ′ ( 0 ) = 0 ,
(2.2)
h 1 ( t ) = α ( β + 1 ) - β ( α + 1 ) e t - ( α - β ) e β t , h ′ 1 ( t ) = - β ( α + 1 ) e t - β ( α - β ) e β t ,
(2.3)
h 2 ( 0 ) = h 1 ′ ( 0 ) = β ( β - 2 α - 1 ) ,
(2.4)
h 2 ( t ) = - β ( α + 1 ) - β ( α - β ) e ( β - 1 ) t
(2.5)
and
h 2 ′ ( t ) = β ( β - 1 ) ( β - α ) e ( β - 1 ) t .
(2.6)
  1. (1)

    If β ≤ α + 1 and β ≤ 2α + 1, then we divide the proof into four cases.

     
Case 1 If 0 < β < 1 and α < β ≤ 2α + 1, then from (2.4) and (2.6) we clearly see that
h 2 ( 0 ) ≤ 0 ,
(2.7)
h 2 ′ ( t ) < 0 .
(2.8)

Therefore, h(t) < 0 for t ∈ (0, ∞), which follows from (2.7) and (2.8) together with (2.1) and (2.2).

Case 2 If 0 < β < 1 and β ≤ α, then (2.5) and (2.6) lead to
lim t → + ∞ h 2 ( t ) = - β ( α + 1 ) < 0 ,
(2.9)
h 2 ′ ( t ) ≥ 0 .
(2.10)

Therefore, h(t) < 0 for t ∈ (0, ∞), which follows from (2.9) and (2.10) together with (2.1) and (2.2).

Case 3 If 1 < β ≤ α, then (2.4) and (2.6) lead to
h 2 ( 0 ) < 0 ,
(2.11)
h 2 ′ ( t ) ≤ 0 .
(2.12)

From equations (2.1) and (2.2) together with inequalities (2.11) and (2.12), we clearly see that h(t) < 0 for t ∈ (0, ∞).

Case 4 If β > 1 and α < β ≤ α + 1, then we clearly see that
lim t → + ∞ h ( t ) = β - α - 1 ≤ 0 .
(2.13)
From (2.3)-(2.6), we know that
lim t → + ∞ h 1 ( t ) = + ∞ ,
(2.14)
h 2 ( 0 ) < 0 ,
(2.15)
lim t → + ∞ h 2 ( t ) = + ∞ ,
(2.16)
h 2 ′ ( t ) > 0 .
(2.17)

From (2.15)-(2.17), we clearly see that there exists t1> 0 such that h2(t) < 0 for t ∈ (0, t1) and h2(t) > 0 for t ∈ (t1, ∞). Hence, h1(t) is strictly decreasing in [0, t1] and strictly increasing in [t1, ∞).

From (2.2) and (2.14) together with the monotonicity of h1(t), we know that there exists t2> 0 such that h1(t) < 0 for t ∈ (0, t2) and h1(t) > 0 for t ∈ (t2, ∞). Hence, h(t) is strictly decreasing in [0, t2] and strictly increasing in [t2, ∞).

Therefore, h(t) < 0 for t ∈ (0, ∞) follows from (2.1) and (2.13) together with the monotonicity of h(t).

(2) If α + 1 < β < 2α + 1, then we clearly see that
lim t → + ∞ h ( t ) = β - α - 1 > 0
(2.18)

and (2.14)-(2.17) hold again. From the proof of Case 4 in Lemma 2.1(1), we know that there exists λ > 0 such that h(t) is strictly decreasing in [0, λ] and strictly increasing in [λ, ∞).

Therefore, Lemma 2.1(2) follows from (2.1) and (2.18) together with the monotonicity of h(t).

(3) If β ≥ α + 1 and β ≥ 2α + 1, then we divide the proof into three cases.

Case I If β > 1 and β ≥ 2α + 1, then
β > α
(2.19)
and it follows from (2.4) that
h 2 ( 0 ) ≥ 0 .
(2.20)
Equation (2.6) and inequality (2.19) lead to
h 2 ′ ( t ) > 0 .
(2.21)

Therefore, h(t) > 0 for t ∈ (0, ∞) follows from (2.1), (2.2), (2.20), and (2.21).

Case II If 0 < β < 1 and α ≤ -1, then from (2.5) and (2.6) we clearly see that
lim t → + ∞ h 2 ( t ) = - β ( α + 1 ) ≥ 0 ,
(2.22)
h 2 ′ ( t ) < 0 .
(2.23)
Inequalities (2.22) and (2.23) imply that
h 2 ( t ) > 0
(2.24)

for t ∈ (0, ∞).

Therefore, h(t) > 0 for t ∈ (0, ∞) follows from (2.1) and (2.2) together with (2.24).

Case III If 0 < α + 1 ≤ β < 1, then we clearly see that
lim t → + ∞ h ( t ) = β - α - 1 ≥ 0 .
(2.25)
It follows from (2.3)-(2.6) that
lim t → + ∞ h 1 ( t ) = - ∞ ,
(2.26)
h 2 ( 0 ) = β ( β - 2 α - 1 ) > 0 ,
(2.27)
lim t → + ∞ h 2 ( t ) = - β ( α + 1 ) < 0 ,
(2.28)
h 2 ′ ( t ) < 0 .
(2.29)

Inequalities (2.27)-(2.29) imply that there exists t3> 0 such that h2(t) > 0 for t ∈ (0, t3) and h2(t) < 0 for t ∈ (t3, ∞). Hence, h1(t) is strictly increasing in [0, t3] and strictly decreasing in [t3, ∞).

It follows from (2.2) and (2.26) together with the monotonicity of h1(t) that there exists t4> 0 such that h1(t) > 0 for t ∈ (0, t4) and h1(t) < 0 for t ∈ (t4, ∞). Hence, h(t) is strictly increasing in [0, t4] and strictly decreasing in [t4, ∞).

Therefore, h(t) > 0 for t ∈ (0, ∞) follows from (2.1) and (2.25) together with the monotonicity of h(t).

(4) If 2α + 1 < β < α + 1, then we clearly see that
lim t → + ∞ h ( t ) = β - α - 1 < 0
(2.30)

and (2.26)-(2.29) hold again.

From the proof of Case III in Lemma 2.1(3) we know that there exists μ > 0 such that h(t) is strictly increasing in [0, μ] and strictly decreasing in [μ, ∞).

Therefore, Lemma 2.1(4) follows from (2.1) and (2.30) together with the monotonicity of h(t).

3 Proof of Theorem 1.1

Proof of Theorem 1.1 Let E1 = {(α, β) : 0 < β < 1, β ≥ α + 1}, E2 = {(α, β) : β > 1, β ≥ 2α +1}, E3 = {(α, β) : α < 0, β = 1}, E4 = {(α, β) : α = 0, β = 1}, E5 = {(α, β) : α + 1 < β < 2α + 1}, E6 = {(α, β) : β > 0, 2α + 1 < β < α + 1}, E7 = {(α, β) : 0 < β < 1, β ≤ 2α + 1}, E8 = {(α, β) : β > 1, β ≤ α + 1} and E9 = {(α, β) : α > 0, β = 1}. Then
{ ( α , β ) : α ∈ ℝ , β > 0 } = ∪ i = 1 9 E i , { ( α , β ) : β > 0 , β ≥ 2 α + 1 , β ≥ α + 1 } \ { ( α , β ) : α = 0 , β = 1 } = E 1 ∪ E 2 ∪ E 3 , { ( α , β ) : β > 0 , β ≤ 2 α + 1 , β ≤ α + 1 } \ { ( α , β ) : α = 0 , β = 1 } = E 7 ∪ E 8 ∪ E 9 .
  1. (1)

    We divide the proof of Theorem 1.1(1) into five cases.

     
Case 1.1 (α, β) ∈ E1 ∪ E2. From (1.1), (1.2), and applying
ψ m ( x ) = ( − 1 ) m + 1 ∫ 0 ∞ t m 1 − e − t e − x t d t ( x > 0, m = 1,2,... ),
we obtain for n ≥ 1,
( − 1 ) n [ log F α , β ( x ) ] ( n ) = ( − 1 ) n [ ( − 1 ) n − 1 α ( n − 1 ) ! x n + β ψ ( n − 1 ) ( x ) − β n ψ ( n − 1 ) ( β x ) ] = − α ∫ 0 ∞ s n − 1 e − x s d s + β ∫ 0 ∞ s n − 1 1 − e − s e − x s d s − β n ∫ 0 ∞ t n − 1 1 − e − t e − β x t d t = − α β n ∫ 0 ∞ t n − 1 e − β x t d t + β n + 1 ∫ 0 ∞ t n − 1 1 − e − β t e − β x t d t − β n ∫ 0 ∞ t n − 1 1 − e − t e − β x t d t = β n ∫ 0 ∞ t n − 1 e − β x t ( 1 − e − t ) ( 1 − e − β t ) h ( t ) d t ,
(3.1)
where
h ( t ) = - α e - ( β + 1 ) t + ( α + 1 ) e - β t + ( α - β ) e - t - α + β - 1 .
(3.2)

Therefore, F α,β (x) is strictly logarithmically completely monotonic on (0, ∞), which follows from (3.1) and (3.2) together with Lemma 2.1(3).

Case 1.2 (α, β) ∈ E3. Then we clearly see that
( − 1 ) n [ l o g F α , β ( x ) ] ( n ) = ( − 1 ) n α ( n − 1 ) ! ( − 1 ) n − 1 x n = − α ( n − 1 ) ! x n > 0
(3.3)

for all x > 0.

Therefore, F α,β (x) is strictly logarithmically completely monotonic on (0, ∞), which follows from (3.3).

Case 1.3 (α,β) ∈ E4. Then F α,β (x) = 1 and
( - 1 ) n [ l o g F α , β ( x ) ] ( n ) = 0 .
(3.4)

Therefore, F α,β (x) is not strictly logarithmically completely monotonic on (0, ∞), which follows from (3.4).

Case 1.4 (α, β) ∈ E5∪E6∪E7∪E8. Then F α,β (x) is not strictly logarithmically completely monotonic on (0, ∞), which follows from Lemmas 2.1(2), 2.1(4), 2.1(1), and equations (3.1) and (3.2).

Case 1.5 (α, β) ∈ E9. Then
( - 1 ) n [ l o g F α , β ( x ) ] ( n ) = - α ( n - 1 ) ! x n < 0
(3.5)

for all x > 0.

Therefore, F α,β (x) is not strictly logarithmically completely monotonic on (0, ∞), which follows from (3.5).
  1. (2)

    We divide the proof of Theorem 1.1(2) into five cases.

     
Case 2.1 (α, β) ∈ E7 ∪ E8. Then from (3.1) we get
( - 1 ) n { l o g [ F α , β ( x ) ] - 1 } ( n ) = - β n ∫ 0 ∞ t n - 1 e - β x t ( 1 - e - t ) ( 1 - e - β t ) h ( t ) d t ,
(3.6)
where
h ( t ) = - α e - ( β + 1 ) t + ( α + 1 ) e - β t + ( α - β ) e - t - α + β - 1 .
(3.7)

Therefore, [F α,β (x)]-1 is strictly logarithmically completely monotonic on (0, ∞), which follows from (3.6) and (3.7) together with Lemma 2.1(1).

Case 2.2 (α, β) ∈ E9. Then
( - 1 ) n { l o g [ F α , β ( x ) ] - 1 } ( n ) = α ( n - 1 ) ! x n > 0
(3.8)

for all x > 0.

Therefore, [F α,β (x)]-1 is strictly logarithmically completely monotonic on (0, ∞), which follows from (3.8).

Case 2.3 (α, β) ∈ E4. Then [F α,β (x)]-1 = 1 and
( - 1 ) n { l o g [ F α , β ( x ) ] - 1 } ( n ) = 0 .
(3.9)

Therefore, [F α,β (x)]-1 is not strictly logarithmically completely monotonic on (0, ∞), which follows from (3.9).

Case 2.4 (α, β) ∈ E1 ∪ E2 ∪ E5 ∪ E6. Then [F α,β (x)]-1 is not strictly logarithmically completely monotonic on (0, ∞), which follows from equations (3.6) and (3.7) and Lemmas 2.1(3), 2.1(2) and 2.1(4).

Case 2.5 (α, β) ∈ E3. Then
( - 1 ) n { l o g [ F α , β ( x ) ] - 1 } ( n ) = α ( n - 1 ) ! x n < 0
(3.10)

for all x > 0.

Therefore, [F α,β (x)]-1 is not strictly logarithmically completely monotonic on (0, ∞), which follows from (3.10). □

Declarations

Acknowledgements

This study is partly supported by the Natural Science Foundation of China (Grant no. 11071069), the Innovation Team Foundation of the Department of Education of Zhejiang Province (Grant no. T200924), and the Natural Science Foundation of the Department of Education of Zhejiang Province (Grant no. Y200805602).

Authors’ Affiliations

(1)
Department of Mathematics, Huzhou Teachers College

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© Lv et al; licensee Springer. 2011

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