Hindawi Publishing Corporation
Journal of Inequalities and Applications
Volume 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 Mathématiques, Université Pierre et Marie Curie, 4 Place Jussieu, 75252 Paris, France
Department of Mathematics, Huzhou Teachers College, Huzhou 313000, Zhejiang, China
3
College of Mathematics and Econometrics, Hunan University, Changsha 410082, Hunan, China
2
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.
Copyright q 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 function Γ and its logarithmic derivative
ψ, the so-called digamma function, are defined as
Γx ∞
0
tx−1 e−t dt,
ψx Γ x
.
Γx
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
2
Journal of Inequalities and Applications
in 4 as follows:
nk
ik1 i
nmk
ik1 i
1/n
1/nm ≤
nk
,
nmk
1.2
and the following lower bound was appeared in 2, 5:
nk1
<
nmk1
n
nm
n k!/k!
n m k!/k!
,
1.3
Since Γn 1 n!, as a generalization of inequality 1.3, the following monotonicity
result was obtained by Guo and Qi in 2. The function
Γx y 1/Γy 11/x
xy1
1.4
is decreasing with respect to x on 1, ∞ for fixed y ≥ 0. Hence, for positive real numbers x
and y, we have
1/x
Γx y 1/Γy 1
xy1
≤
.
x y 2 Γx y 2/Γy 11/x1
1.5
Recently, in 6, Qi and Sun proved that the function
1/x
Γx y 1/Γy 1
√
xy
1.6
is strictly increasing with respect to x ∈ y 1, ∞ for all y ≥ y0 .
Now, we generalize the function in 1.4 as follows: for positive real numbers x and y,
α ≥ 0, let
Fα x, y 1/x
Γx y 1/Γy 1
x y 1α
.
1.7
The aim of this paper is to discuss the monotonicity and logarithmical convexity of the
function Fα 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.
Journal of Inequalities and Applications
3
Definition 1.1. Let D ⊂ R2 be a convex set, f : D → R is called a convex function on D if
f
xy
2
fx fy
2
≤
1.8
for all x, 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 in 7.
Proposition 1.3. Let D ⊂ R2 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 Lx is a positive (or negative)
semidefinite matrix for all x ∈ D, where
Lx f12
f11
f21
f22
1.9
and fij ∂2 fx1 , x2 /∂xi ∂xj for x x1 , x2 , i, j 1, 2.
Notation 1. In Definitions 1.1, 1.2 and Proposition 1.3, we denote x, y by the points or vectors
of R2 , 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, Fα 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, Fα 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 Fα 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 Fα 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
1/x
Γx y 1/Γy 1
xy1
xy1
<
.
<
1/x1
xy2
xy2
Γx y 2/Γy 1
1.10
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Journal of Inequalities and Applications
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:
nk
ik1 i
nmk
ik1 i
1/n
1/nm ≤
nk1
.
nmk1
1.11
Corollary 1.8. If x1 , y1 , x2 , y2 ∈ 0, ∞ × 0, ∞, then
1/x1 1/x2
· Γ x2 y2 1 /Γ y2 1
Γ x1 y1 1 /Γ y1 1
4/x1 x2 Γ x1 x2 y1 y2 /2 1 /Γ y1 y2 /2 1
√ 1/4
2 x1 y1 1 x2 y2 1
≤
.
x1 y1 x2 y2 2
1.12
Remark 1.9. We conjecture that the inequality 1.2 can be improved if we can choose two
pairs of integers x1 , y1 and x2 , y2 properly.
2. Lemmas
It is well known that the Bernoulli numbers Bn is defined 8 in general by
∞
1
t2n
1 1 −
Bn .
−1n−1
t
2n!
e − 1 2 t n1
2.1
In particular, we have
B1 1
,
6
B2 1
,
30
B3 1
,
42
B4 1
.
30
2.2
In 9, the following summation formula is given:
∞
−1n
n0 2n
12k1
π 2k1 Ek
22k2 2k!
2.3
for nonnegative integer k, where Ek denotes the Euler number, which implies
Bn ∞
1
22n! ,
2n
m
2π m1 2n
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.
Journal of Inequalities and Applications
5
Lemma 2.1 see 3. For real number x > 0 and natural number m, one has
ln Γx m
1
1
Bn
1
ln2π x −
· 2n−1
ln x − x −1n−1
2
2
22n
−
1n
x
n1
1
Bm1
·
−1 θ1
,
2m 12m 2 x2m1
m
ψx ln x −
ψ x 0 < θ1 < 1;
m
Bn 1
Bm1
1
1 −1n · 2n −1m1 θ2
·
,
2x n1
2n x
2m 2 x2m2
m
Bn
1
1
Bm1
2 −1n−1 2n1 −1m θ3 · 2m3 ,
x 2x
x
x
n1
ψ x −
2.5
0 < θ2 < 1;
2.7
0 < θ3 < 1;
m
1
1 Bn
Bm1
− 3 −1n 2n 1 2n2 −1m1 2m 3θ4 · 2m4 ,
2
x
x
x
x
n1
2.6
0 < θ4 < 1.
2.8
Lemma 2.2 see 14. Inequalities
ln x −
1
1
≤ ψx ≤ ln x −
,
x
2x
k − 1!
k!
k − 1!
k!
k1 ≤ −1k1 ψ k x ≤
k1
k
k
x
2x
x
x
2.9
2.10
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 rx, y lim lnx y 1 − lny 1 0
y→∞
y→∞
2.11
for any fixed x > 0.
Since ψx 1 1/x ψx and 0 ≤ α ≤ 1, we have
x 1 − αy x 2 − α
rx, y − rx, y 1 >0
y 1x y 1x y 2
for all x, y ∈ 0, ∞ × 0, ∞.
2.12
6
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
rα, y − rα, y 1 < 0
2.13
for y ∈ 2/α − 1, ∞.
Therefore, Lemma 2.32 follows from 2.11 and 2.13.
Lemma 2.4. If gx, y 2xψy 1 − 2ln Γx y 1 − ln Γy 1 x2 ψ y 1, then gx, y > 0
for x, y ∈ 0, ∞ × 0, ∞.
Proof. It is easy to see that
g0, y 0
2.14
g1 x, y 2 xψ y 1 − ψx y 1 ψy 1 ,
2.15
g1 0, y 0,
2.16
∂g1 x, y
2 ψ y 1 − ψ x y 1 > 0
∂x
2.17
for all y ∈ 0, ∞.
Let g1 x, y ∂gx, y/∂x, then
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
ax, y 2
2
ln Γx y 1 − ln Γy 1 − 2 ψx y 1,
x3
x
1
bx, y − 2 ψx y 1 − ψy 1 ,
x
1
cx, y − ψ y 1.
x
2.18
Then simple computation shows that
gx, y x3 2bx, y − ax, y − cx, y .
2.19
Lemma 2.6. Let dx, y 1/x ψ x y 1 α/x y 12 , then the following statements are
true:
1 if 0 ≤ α ≤ 1/4, then
2
ax, y dx, y cx, y dx, y > bx, y dx, y
for x, y ∈ 0, ∞ × 0, ∞;
2.20
Journal of Inequalities and Applications
7
2 if α ≥ 1, then
2
ax, y dx, y cx, y dx, y < bx, y dx, y
2.21
for x, y ∈ 0, ∞ × 0, ∞.
Proof. Let
2
fx, y 2ψ y 1 xψx y 1 − ln Γx y 1 ln Γy 1 − ψx y 1 − ψy 1 ,
αx
px, y fx, y − gx, y ψ x y 1 .
x y 12
2.22
Then it is not difficult to verify
2.23
2 2.24
px, y x ax, y dx, y cx, y dx, y − bx, y dx, y ,
∂px, y
∂gx, y
αx
α
2αx
−
−
gx,
y
ψ
x
y
1
−
.
∂x
∂x
x y 12
x y 12 x y 13
2.25
4
p0, y 0,
1 If 0 ≤ α ≤ 1/4, then making use of Lemmas 2.2, 2.4 and 2.25 we get
∂px, y
∂gx, y
αx
>−
2
∂x
∂x
x y 1
1
1
α
2αx
gx, y
−
,
x y 12 x y 13 x y 12 x y 13
∂gx, y
1
>
1 − αgx, y − αx
∂x
x y 12
2.26
for x, y ∈ 0, ∞ × 0, ∞.
Let gi x, y ∂i gx, y/∂xi , i 1, 2, 3, 4, qx, y 1 − αgx, y − αx∂gx, y/∂x,
and qj x, y ∂j qx, y/∂xj , j 1, 2. Then simple computation leads to
for all y ∈ 0, ∞.
g3 x, y −2ψ x y 1,
2.27
g4 x, y −2ψ x y 1,
2.28
∂q2 x, y
1 − 4αg3 x, y − αxg4 x, y,
∂x
2.29
q2 0, y q1 0, y q0, y 0
2.30
8
Journal of Inequalities and Applications
It is well known that ln Γx −cx ∞
k1 x/k − ln1 x/k − ln x, where c 0.577215 · · · is the Euler’s constant. From this we get
ψ n −1n1 n!
∞
k0 k
1
xn1
2.31
.
From Lemma 2.2, 2.27–2.29, 2.31 and the assumption 0 ≤ α ≤ 1/4, we conclude
that
∂q2 x, y
> 0.
∂x
2.32
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 2.4 and 2.25 we obtain
∂px, y
∂gx, y
αx
1
1
2αx
<−
gx, y
∂x
∂x
x y 12
x y 13 2x y 14 x y 13
∂gx, y
2αx 1
gx, y
∂x
x y 1
x y 13
∂gx, y
αx 1
.
<
2gx, y − x
∂x
x y 13
<−
αx
2
2.33
Let
vx, y 2gx, y − x
∂gx, y
,
∂x
vi x, y ∂i vx, y
,
∂xi
i 1, 2.
2.34
Then
v2 x, y 2xψ x y 1 < 0
2.35
for x, y ∈ 0, ∞ × 0, ∞ by Lemma 2.2, and
v0, y v1 0, y 0
for y ∈ 0, ∞.
Therefore, Lemma 2.62 follows from 2.23–2.25 and 2.33–2.36.
2.36
Journal of Inequalities and Applications
9
3. Proofs of Theorems 1.4 and 1.5
Proof of Theorem 1.4. 1 Let Gx, y ln Fα x, y and G1 x, y x2 ∂Gx, y/∂x, then
G1 x, y − ln Γx y 1 − ln Γy 1 xψx y 1 −
αx2
.
xy1
3.1
The following three cases will complete the proof of Theorem 1.41.
Case 1. If 0 ≤ α ≤ 1/2, then 3.1 and Lemma 2.2 imply
∂G1 x, y
αx 2y 2
x ψ x y 1 −
∂x
x y 12
αx 2y 2
1
1
>x
−
x y 1 2x y 12
x y 12
x
2 − 2αx 2 − 4αy 3 − 4α
2
2x y 1
3.2
>0
for x, y ∈ 0, ∞ × 0, ∞.
From 3.2 and the fact that G1 0, y 0 for all y ∈ 0, ∞ we know that Fα 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
αx 2y 2
∂G1 x, y
1
1
1
<x
−
∂x
x y 1 2x y 12 6x y 13
x y 12
x
6 − 6αx2 λ1 yx λ2 y
3
6x y 1
3.3
<0
for x, y ∈ 0, ∞ × 0, ∞, where λ1 y 12 − 18αy 15 − 18α < 0 and λ2 y 61 − 2αy2 15 − 24αy 10 − 12α < 0.
From 3.3 and the fact that G1 0, y 0 for all y ∈ 0, ∞ we know that Fα x, y is
strictly decreasing with respect to x on 0, ∞ for any fixed y ∈ 0, ∞.
Case 3. If 1/2 < α < 1, let
G2 x, y ψ x y 1 −
αx 2y 2
x y 12
.
3.4
10
Journal of Inequalities and Applications
Then
∂G1 x, y
xG2 x, y,
∂x
1
1
1
2α
−
G2 0, y <
y 1 2y 12 6y 13 y 1
for y ≥ 15 − 24α √
1
6y 1
61 − 2αy2 15 − 24αy 10 − 12α < 0
3
3.5
3.6
48α − 15/24α − 12.
It is obvious that 3.6 implies
√
15 48α − 15
G2 0,
< 0.
24α − 12
3.7
The continuity of G2 x, y with respect to x ∈ 0, ∞ for any fixed y ∈ 0, ∞ and 3.7 imply
that there exists δ δα > 0 such that
√
15 48α − 15
G2 x,
<0
24α − 12
3.8
for x ∈ 0, δ.
√
From 3.5, 3.8 and G1 0, 15 48α − 15/24α −√12 0 we know that Fα 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 of 2.5 and 2.6 we have
αx
1 lnx y 1
−
lim G1 x, y lim x 1 − y C y, θ1
x → ∞
x → ∞
2
x
xy1
lim 1 − αx C y, θ1
3.9
x → ∞
∞,
where
C y, θ1
1
θ1
1
1
− −
y
lny 1 2
12y 1 2 360y 13
3.10
for y ∈ 0, ∞ and 0 < θ1 < 1.
Equation 3.9 implies that there exists M Mα > δα such that
√
15 48α − 15
G1 x,
>0
24α − 12
for x ∈ M, ∞.
3.11
Journal of Inequalities and Applications
11
Hence, from 3.11
we know that Fα x, y is strictly increasing with respect to x on
√
M, ∞ for y 15 48α − 15/24α − 12.
2 Since
x
∂Gx, y
αx
ψx y 1 − ψy 1 −
rx, y,
∂y
xy1
3.12
then, Theorem 1.42 follows from 3.12 and Lemma 2.3.
Proof of Theorem 1.5. Let Gx, y ln Fα x, y, G11 x, y ∂2 Gx, y/∂x2 , G12 ∂2 Gx, y/
∂x∂y and G22 x, y ∂2 Gx, y/∂y2 , then simple calculation yields
G11 x, y 2
2
ln Γx y 1 − ln Γy 1 − 2 ψx y 1
x3
x
1 α
ψ x y 1 x
x y 12
3.13
ax, y dx, y,
G12 x, y −
1
1
α
ψx y 1 − ψy 1 ψ x y 1 2
x
x
x y 12
3.14
bx, y dx, y,
G22 x, y 1 α
ψ x y 1 − ψ y 1 x
x y 12
3.15
cx, y dx, y,
where ax, y, bx, y, cx, y, and dx, y are defined in Remark 2.5 and Lemma 2.6.
According to the Definition 1.2 and Proposition 1.3, to prove Theorem 1.5 we need
only to show that
G11 x, y ≤ 0,
2
G11 x, yG22 x, y − G12 x, y ≥ 0
3.16
3.17
for 0 ≤ α ≤ 1/4 and x, y ∈ 0, ∞ × 0, ∞, and
2
G11 x, yG22 x, y − G12 x, y < 0
for α ≥ 1 and x, y ∈ 0, ∞ × 0, ∞.
3.18
12
Journal of Inequalities and Applications
Next, let wx, y x3 G11 x, y, then
wx, y 2 ln Γx y 1 − ln Γy 1 − 2xψx y 1 x2 ψ x y 1 w0, y 0,
αx3
x y 12
,
3.19
∂wx, y
αx 3y 3
2
x ψ x y 1 ∂x
x y 13
αx 3y 3
1
1
< x2
−
−
x y 13
x y 12 x y 13
x
2
x y 1
3
3.20
α − 1x 3α − 1y 3α − 2
<0
for x, y ∈ 0, ∞ × 0, ∞ by Lemma 2.2 and 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 of Theorem 1.5 is 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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