Volume 2009, Article ID 728612, 13 pagesdoi:10.1155/2009/728612 Research Article Monotonic and Logarithmically Convex Properties of a Function Involving Gamma Functions Tie-Hong Zhao,1 Y
Trang 1Volume 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
1 Institut de Math´ematiques, Universit´e Pierre et Marie Curie, 4 Place Jussieu, 75252 Paris, France
2 Department of Mathematics, Huzhou Teachers College, Huzhou 313000, Zhejiang, China
3 College of Mathematics and Econometrics, Hunan University, Changsha 410082, Hunan, China
Correspondence should be addressed to Yu-Ming Chu,chuyuming2005@yahoo.com.cn
Received 14 October 2008; Accepted 27 February 2009
Recommended by Sever Dragomir
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
Copyrightq 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
1 Introduction
For real and positive values of x the Euler gamma functionΓ and its logarithmic derivative
ψ, the so-called digamma function, are defined as
Γx
∞
0
t x−1e −t dt, ψ x Γx
For extension of these functions to complex variables and for basic properties see1
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 in2 The extensions and generalizations of this problem can
be found in3 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
Trang 22 Journal of Inequalities and Applications
in4 as follows:
n k
i k1 i1/n
n mk
i k1 i1/nm ≤
n k
and the following lower bound was appeared in2,5 :
n k 1
n m k 1 <
n
n k!/k!
nm
SinceΓn 1 n!, as a generalization of inequality 1.3, the following monotonicity result was obtained by Guo and Qi in2 The function
Γx y 1/Γy 1 1/x
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 ≤
Γx y 1/Γy 1 1/x
Γx y 2/Γy 1 1/x1 . 1.5
Recently, in6 , Qi and Sun proved that the function
Γx y 1/Γy 1 1/x
is strictly increasing with respect to x ∈ y 1, ∞ for all y ≥ y0.
Now, we generalize the function in1.4 as follows: for positive real numbers x and y,
α≥ 0, let
F α x, y
Γx y 1/Γy 1 1/x
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
Trang 3Definition 1.1 Let D ⊂ R2be a convex set, f : D → R is called a convex function on D if
f
2
≤ f x fy
for allx, y ∈ D, and f is called concave if −f is convex.
Definition 1.2 Let D ⊂ R2 be a convex set, f : D → 0, ∞ is called a logarithmically convex function on D if ln f is convex on D, and f is called logarithmically concave if ln f
is concave
The following criterion for convexity of function was established by Fichtenholz in7
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
Lx
f11 f12
and f ij ∂2f x1, x2/∂x i ∂x j for x x1, x2, i, j 1, 2.
Notation 1 In Definitions1.1,1.2andProposition 1.3, we denotex, y by the points or vectors
of R2, and denote x, y by the real variables in the later.
Our main results are Theorems1.4and1.5
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 Theorems1.4and1.5immediately
Corollary 1.6 If x, y ∈ 0, ∞ × 0, ∞, then
x y 1
x y 2 <
Γx y 1/Γy 1 1/x
Γx y 2/Γy 1 1/x1 <
x y 1
Trang 44 Journal of Inequalities and Applications
Remark 1.7 Inequality1.3 can be derived fromCorollary 1.6if we take x, y ∈ N Although
we cannot get the inequality1.2 exactly fromCorollary 1.6, but we can get the following inequality which is close to inequality1.2:
n k
i k1 i1/n
n mk
i k1 i1/nm ≤
n k 1
Corollary 1.8 If x1, y1, x2, y2 ∈ 0, ∞ × 0, ∞, then
Γx1 y1 1/Γy1 1 1/x1·Γx2 y2 1/Γy2 1 1/x2
Γx1 x2 y1 y2/2 1/Γy1 y2/2 1 4/x1x2
≤
√
2
x1 y1 1x2 y2 1 1/4
x1 y1 x2 y2 2 .
1.12
Remark 1.9 We conjecture that the inequality1.2 can be improved if we can choose two pairs of integersx1, y1 and x2, y2 properly
2 Lemmas
It is well known that the Bernoulli numbers B nis defined8 in general by
1
e t− 1
1
2 −1
t ∞
n1
−1n−1 t 2n
In particular, we have
B1 1
6, B2 1
30, B3 1
42, B4 1
In9 , the following summation formula is given:
∞
n0
−1n
2n 1 2k1 π 2k1 E k
for nonnegative integer k, where E kdenotes the Euler number, which implies
B n 22n!
2π 2n
∞
m1
1
Recently, the Bernoulli and Euler numbers and polynomials are generalized in10–13 The following two Lemmas were established by Qi and Guo in3,14
Trang 5Lemma 2.1 see 3 For real number x > 0 and natural number m, one has
lnΓx 1
2ln2π x− 1
2
ln x − x m
n1
−1n−1 B n
22n − 1n·
1
x 2n−1
−1m θ1 B m1
2m 12m 2·
1
x 2m1 , 0 < θ1< 1;
2.5
ψ x ln x − 1
2xm
n1
−1n B n
2n· 1
x 2n −1m1θ2 B m1
2m 2·
1
x 2m2 , 0 < θ2< 1; 2.6
ψx 1
x 1
2x2 m
n1
−1n−1 B n
x 2n1 −1m θ3·B m1
x 2m3 , 0 < θ3< 1; 2.7
ψx −1
x2 − 1
x3 m
n1
−1n 2n 1 B n
x 2n2 −1m12m 3θ4·B m1
x 2m4 , 0 < θ4< 1. 2.8
Lemma 2.2 see 14 Inequalities
ln x− 1
x ≤ ψx ≤ ln x − 1
k − 1!
x k k!
2x k1 ≤ −1k1ψ k x ≤ k − 1!
x k k!
hold in 0, ∞ for k ∈ N.
Lemma 2.3 Let rx, y ψx y 1 − ψy 1 − αx/x y 1, then the following statements
are true:
1 if 0 ≤ α ≤ 1, then rx, y ≥ 0 for x, y ∈ 0, ∞ × 0, ∞;
2 if α > 1, then rα, y < 0 for y ∈ 2/α − 1, ∞.
Proof. 1 Making use of 2.6 we get
lim
y→ ∞r x, y lim
y→ ∞
lnx y 1 − lny 1 0 2.11
for any fixed x > 0.
Since ψx 1 1/x ψx and 0 ≤ α ≤ 1, we have
r x, y − rx, y 1 x
1 − αy x 2 − α
y 1x y 1x y 2 > 0 2.12
for allx, y ∈ 0, ∞ × 0, ∞.
Trang 66 Journal of Inequalities and Applications Therefore,Lemma 2.31 follows from 2.11 and 2.12
2 If α > 1, then 2.12 leads to
for y ∈ 2/α − 1, ∞.
Therefore,Lemma 2.32 follows from 2.11 and 2.13
for x, y ∈ 0, ∞ × 0, ∞.
Proof It is easy to see that
for all y ∈ 0, ∞.
Let g1x, y ∂gx, y/∂x, then
g1x, y 2xψy 1 − ψx y 1 ψy 1 , 2.15
∂g1x, y
∂x 2ψy 1 − ψx y 1 > 0 2.17
for x > 0 On the other hand, from2.10 we know that ψx is strictly decreasing on 0, ∞.
Therefore,Lemma 2.4follows from2.14–2.17
Remark 2.5 Let
a x, y 2
x3
lnΓx y 1 − ln Γy 1 − 2
x2ψ x y 1,
b x, y − 1
x2
ψ x y 1 − ψy 1 ,
c x, y −1
x ψ
y 1.
2.18
Then simple computation shows that
g x, y x3
2bx, y − ax, y − cx, y . 2.19
Lemma 2.6 Let dx, y 1/x ψx y 1 α/x y 12, then the following statements are true:
1 if 0 ≤ α ≤ 1/4, then
a x, y dx, y c x, y dx, y >
b x, y dx, y 2 2.20
for x, y ∈ 0, ∞ × 0, ∞;
Trang 72 if α ≥ 1, then
a x, y dx, y c x, y dx, y <
b x, y dx, y 2 2.21
for x, y ∈ 0, ∞ × 0, ∞.
Proof Let
f x, y 2ψy 1xψ x y 1 − ln Γx y 1 ln Γy 1 −ψ x y 1 − ψy 1 2
,
p x, y fx, y − gx, y
ψx y 1 αx
x y 12
.
2.22 Then it is not difficult to verify
p x, y x4
a x, y dx, y c x, y dx, y −b x, y dx, y 2
∂p x, y
x y 12
∂g x, y
∂x − gx, y
ψx y 1 α
x y 12 − 2αx
x y 13
.
2.25
1 If 0 ≤ α ≤ 1/4, then making use of Lemmas2.2,2.4and2.25 we get
∂p x, y
∂x >− αx
x y 12
∂g x, y
∂x
gx, y
1
x y 12 1
x y 13 − α
x y 12 2αx
x y 13
> 1
x y 12
1 − αgx, y − αx ∂g x, y
∂x
, 2.26
forx, y ∈ 0, ∞ × 0, ∞.
Let g i x, y ∂ i g x, y/∂x i , i 1, 2, 3, 4, qx, y 1 − αgx, y − αx∂gx, y/∂x, and q j x, y ∂ j q x, y/∂x j , j 1, 2 Then simple computation leads to
∂q2x, y
∂x 1 − 4αg3x, y − αxg4x, y, 2.29
for all y ∈ 0, ∞.
Trang 88 Journal of Inequalities and Applications
It is well known that lnΓx −cx ∞k1x/k − ln1 x/k − ln x, where c 0.577215· · · is the Euler’s constant From this we get
ψ n −1n1n!
∞
k0
1
FromLemma 2.2,2.27–2.29, 2.31 and the assumption 0 ≤ α ≤ 1/4, we conclude
that
∂q2x, y
Therefore,Lemma 2.61 follows from 2.23–2.26, 2.30, and 2.32
2 If α ≥ 1, then making use of 2.8,Lemma 2.4and2.25 we obtain
∂p x, y
∂x <− αx
x y 12
∂g x, y
∂x gx, y
1
x y 13 1
2x y 14 2αx
x y 13
<− αx
x y 12
∂g x, y
∂x gx, y 2αx 1
x y 13
< α x 1
x y 13
2gx, y − x ∂g x, y
∂x
.
2.33 Let
v x, y 2gx, y − x ∂g x, y
∂x , v i x, y ∂ i v x, y
∂x i , i 1, 2. 2.34 Then
v2x, y 2xψx y 1 < 0 2.35
forx, y ∈ 0, ∞ × 0, ∞ byLemma 2.2, and
for y ∈ 0, ∞.
Therefore,Lemma 2.62 follows from 2.23–2.25 and 2.33–2.36
Trang 93 Proofs of Theorems 1.4 and 1.5
Proof of Theorem 1.4 1 Let Gx, y ln F α x, y and G1x, y x2∂Gx, y/∂x, then
G1x, y −lnΓx y 1 − ln Γy 1 xψx y 1 − αx2
x y 1 . 3.1
The following three cases will complete the proof ofTheorem 1.41
Case 1 If 0 ≤ α ≤ 1/2, then 3.1 andLemma 2.2imply
∂G1x, y
∂x x
ψx y 1 − α x 2y 2
x y 12
> x
1
x y 1
1 2x y 12 −α x 2y 2
x y 12
2x y 12
2 − 2αx 2 − 4αy 3 − 4α
> 0
3.2
forx, y ∈ 0, ∞ × 0, ∞.
From3.2 and the fact that G10, 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
∂G1x, y
∂x < x
1
x y 1
1 2x y 12 1
6x y 13 −α x 2y 2
x y 12
6x y 13
6 − 6αx2 λ1yx λ2y
< 0
3.3
forx, y ∈ 0, ∞ × 0, ∞, where λ1y 12 − 18αy 15 − 18α < 0 and λ2y 61 − 2αy2
15 − 24αy 10 − 12α < 0.
From3.3 and the fact that G10, 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
G2x, y ψx y 1 − α x 2y 2
Trang 1010 Journal of Inequalities and Applications Then
∂G1x, y
G20, y < 1
y 1
1 2y 12 1
6y 13 − 2α
y 1
6y 13
61 − 2αy2 15 − 24αy 10 − 12α < 0
3.6
for y ≥ 15 − 24α √48α − 15/24α − 12.
It is obvious that3.6 implies
G2 0,15√48α− 15
24α− 12
The continuity of G2x, y with respect to x ∈ 0, ∞ for any fixed y ∈ 0, ∞ and 3.7 imply
that there exists δ δα > 0 such that
G2 x,15√48α− 15
24α− 12
for x ∈ 0, δ.
From3.5, 3.8 and G10, 15 √48α − 15/24α − 12 0 we know that F α x, y is strictly decreasing with respect to x on 0, δ for y 15 √48α − 15/24α − 12.
On the other hand, making use of2.5 and 2.6 we have
lim
x→ ∞G1x, y lim
x→ ∞x
1− y 1
2
lnx y 1
x y 1
Cy, θ1
lim
x→ ∞1 − αx Cy, θ1
∞,
3.9
where
C
y, θ1
y1
2
lny 1 12y 11 − 1
for y ∈ 0, ∞ and 0 < θ1 < 1.
Equation3.9 implies that there exists M Mα > δα such that
G1 x,15√48α− 15
24α− 12
for x ∈ M, ∞.
Trang 11Hence, from3.11 we know that F α x, y is strictly increasing with respect to x on
M, ∞ for y 15 √48α − 15/24α − 12.
2 Since
x ∂G x, y
∂y ψx y 1 − ψy 1 − αx
x y 1 rx, y, 3.12
then,Theorem 1.42 follows from 3.12 andLemma 2.3
Proof of Theorem 1.5 Let G x, y ln F α x, y, G
11x, y ∂2G x, y/∂x2, G12 ∂2G x, y/
∂x∂y and G22x, y ∂2G x, y/∂y2, then simple calculation yields
G11x, y 2
x3
lnΓx y 1 − ln Γy 1 − 2
x2ψ x y 1
1
x ψ
x y 1 α
x y 12
ax, y dx, y,
3.13
G12x, y −1
x2
ψ x y 1 − ψy 1 1
x ψ
x y 1 α
x y 12
bx, y dx, y,
3.14
G22x, y 1
x
ψx y 1 − ψy 1 α
x y 12
cx, y dx, y,
3.15
where ax, y, bx, y, cx, y, and dx, y are defined inRemark 2.5andLemma 2.6
According to the Definition 1.2 and Proposition 1.3, to prove Theorem 1.5 we need only to show that
G11x, yG
22x, y −G12x, y 2≥ 0 3.17
for 0≤ α ≤ 1/4 and x, y ∈ 0, ∞ × 0, ∞, and
G11x, yG
22x, y −G12x, y 2
for α ≥ 1 and x, y ∈ 0, ∞ × 0, ∞.
Trang 1212 Journal of Inequalities and Applications
Next, let wx, y x3G11x, y, then
w x, y 2lnΓx y 1 − ln Γy 1 − 2xψx y 1 x2ψx y 1 αx3
x y 12,
w 0, y 0,
3.19
∂w x, y
∂x x2
ψx y 1 α x 3y 3
x y 13
< x2
α x 3y 3
x y 13 − 1
x y 12 − 1
x y 13
x y 13
α − 1x 3α − 1y 3α − 2
< 0
3.20
forx, y ∈ 0, ∞ × 0, ∞ byLemma 2.2and 0≤ α ≤ 1/4.
Therefore, 3.16 follows from 3.19 and 3.20, and 3.17 and 3.18 follow from
Lemma 2.6 The proof ofTheorem 1.5is completed
Acknowledgments
This research is partly supported by 973 Project of China under grant 2006CB708304, N S Foundation of China under Grant 10771195, and N S Foundation Zhejiang Province under Grant Y607128
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...for all y ∈ 0, ∞.
Trang 88 Journal of Inequalities and Applications
It is well known that...
for allx, y ∈ 0, ∞ × 0, ∞.
Trang 66 Journal of Inequalities and Applications Therefore,Lemma... class="text_page_counter">Trang 4
4 Journal of Inequalities and Applications
Remark 1.7 Inequality1.3 can be derived fromCorollary 1.6if we take