International Journal of Mathematical Archive-2(3), Mar. - 2011,
Page: 351-355
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International Journal of Mathematical Archive- 2 (3), Mar. – 2011 351
SOME INEQUALITIES CONCERNING Q-GAMMA FUNCTIONS
W. T. Sulaiman*
Department of Computer Engineering, College of Engineering University of Mosul, Iraq.
E-mail: [email protected]
(Received on: 26-02-11; Accepted on: 05-03-11)
---
ABSTRACT
Several new inequalities concerning q-gamma functions are proved.
Mathematics subject classification (2010): 33B15, 33D05, 36D07, 24A48.
Keywords: q-gamma function, inequality, monotonic function.
1. INTRODUCTION:
The Euler gamma function
Γ
(
x
)
is defined forx
>
0
by
(
)
.
0 1
dt
e
t
x
x −t∞ −
=
Γ
(1.1)The psi or digamma function, the logarithmic derivative of the gamma function is defined by
,
0
.
)
(
)
(
)
(
>
Γ
Γ′
=
x
x
x
x
ψ
(1.2)The q-analogue of
Γ
(
x
)
is called q-gamma function, was introduced by Jackson in 1904 and defined for0
>
x
and0
<
q
<
1
by
∏
∞
=
+ + −
−
−
−
=
Γ
0
1 1
1
1
)
1
(
)
(
n
x n n x
q
q
q
q
x
(1.3)The q-gamma function satisfies the following
lim
(
)
(
)
1
1 q
x
x
q→ −
Γ
=
Γ
(1.4)and
(
),
(
1
)
1
.
1
1
)
1
(
Γ
Γ
=
−
−
=
+
Γ
q qx
q
x
q
q
x
(1.5)The q-analogue of
ψ
(
x
)
is called the q-psi function defined by
)
(
)
(
)
(
x
x
x
q q q
Γ
Γ′
=
ψ
. (1.6)---
From (1.3) and (1.6) it follows that
∞
= ∞
=
+ +
−
+
−
−
=
−
+
−
−
=
0
0
1
log
)
1
log(
1
log
)
1
log(
)
(
n
n nx
n
x n x n q
q
q
q
q
q
q
q
q
x
ψ
. (1.7)It is well-known that
ψ
q′
is strictly completely monotonic on(
0
,
∞
)
that is (see [1,page260] )
(
−
1
)
n(
′
q(
)
)
(n)>
0
x
ψ
forx
>
0
,
n
≥
0
.
(1.8)Concerning q-gamma functions, the following results were achieved
Theorem: 1.1[3]. Let
x
∈
[
0
,
1
],
q
∈
(
0
,
1
),
a
≥
b
>
0
,
c
,
d
positive real numbers withbc
>
ad
>
0
and.
0
)
(
b
+
ax
>
q
ψ
Then
.
)
(
)
(
)
(
)
(
)
(
)
(
d q
c q
d q
c q
d q
c q
b
a
b
a
ax
b
bx
a
b
a
+
Γ
+
Γ
≤
+
Γ
+
Γ
≤
Γ
Γ
(1.9)
Theorem: 1.2[2]. Let
0
<
q
<
1
,
A
≤
0
,
andb
≥
0
.
Then the function
f
(
x
)
=
x
A[
Γ
(
1
+
bx)
]
x (1.10)decreases with respect to
x
>
0
.
The object of the present paper is to give several new inequalities concerning the q-gamma functions.
2. RESULTS:
The following generalizes theorem 1.2.
Theorem: 2.1. Let f be a non-negative real function such that
f
′
<
0
and
f
′′
≥
0
. Let0
<
q
<
1
,
(
)
2)
(
)
(
)
(
,
0
)
(
)
(
lim
f
x
f
x
f
x
x
f
x
f
x
′
≥
′′
=
′
∞→ and
b
≥
0
.
Then the function
F
(
x
)
=
f
(
x
)
[
Γ
q(
1
+
bx)
]
x (2.1)decreases with respect to
x
>
0
.
Proof: We have,
log
(
)
log
(
)
log
(
1
bx)
q
x
x
f
x
F
=
+
Γ
+
,
(
1
)
log
(
1
)
(
)
(
)
(
)
)
(
)
(
)
(
F
x
g
1F
x
x
b
x
f
x
f
x
F
bx xq x
b
q
+
+
Γ
+
=
−
′
=
′
ψ
.By setting
x
=
1
/
y
,
we have)
1
(
log
)
1
(
)
(
)
(
)
(
11
by
by
by
y
f
y
f
y
g
=
′
−−
+
+
Γ
q+
−
*W. T. Sulaiman / Some inequalities concerning Q-gamma functions / IJMA- 2(3), Mar.-2011, Page: 351-355
© 2010, IJMA. All Rights Reserved 353
(
)
(
1
)
0
)
(
)
(
)
(
)
(
)
(
2 11 1
2 1
<
+
′
−
′′
−
′
=
′
−
− −
−
by
by
y
f
y
y
f
y
f
y
f
y
g
ψ
.Therefore
g
is non-increasing. Asg
(
0
)
=
0
,
theng
(
y
)
and henceg
(
x
−1)
<
0
,
which implies
F
′
(
x
)
<
0
. That is F is decreasing.Remark: 1 It may be mentioned that theorem 1.2 follows from theorem 2.1 by putting
f
(
x
)
=
x
A,
A
≤
0
.
In a similar way as the Beta function, we define the q-beta function by
,
,
0
.
)
(
)
(
)
(
)
,
(
>
+
Γ
Γ
Γ
=
x
y
y
x
y
x
y
x
B
q q q
q (2.2)
Lemma: 2.2 Let
0
<
q
<
1
,
0
<
s
≤
t
,
0
<
t
+
m
≤
s
+
n
,
thenn
t m s
q
q
q
q
−
≥
−
1
1
.Proof: We have
q
s+
q
t+m≥
q
t+
q
s+n,which implies
q
s−
q
s+n≥
q
t−
q
t+m,
q
s(
1
−
q
n)
≥
q
t(
1
−
q
m)
, hence the result.
Theorem: 2.3 Let
x
,
a
,
b
,
α
>
0
.
Then the functionB
q(
α
a
+
bx
,
α
b
+
ax
)
is non-increasing in x.Proof: Let
f
(
x
)
=
B
q(
α
a
+
bx
,
α
b
+
ax
)
, then, we have
log
f
(
x
)
=
log
Γ
q(
α
a
+
bx
)
+
log
Γ
(
α
b
+
ax
)
−
log
Γ
(
(
a
+
b
)
(
α
+
x
)
)
,
and
(
)
(
)
(
)
(
(
)(
)
)
)
(
)
(
x
b
a
b
a
ax
b
a
bx
a
b
x
f
x
f
+
+
+
−
+
+
+
=
′
α
ψ
α
ψ
α
ψ
−
+
−
−
+
−
=
∞
=
∞
=
∞
= + + + + + + +
+ + + +
+ + +
0 0 0
) ( ) (
) ( ) (
1
)
(
1
1
log
i i i
i x b a
i x b a i
ax b
i ax b i
bx a
i bx a
q
q
b
a
q
q
b
q
q
a
q
αα
α α
α α
−
−
−
=
∞
= + + +
+ + + +
+ + +
0
) ( ) (
) ( ) (
1
1
log
i
i i x b a
i x b a i bx a
i bx a
q
q
q
q
q
q
a
αα
α α
+
−
−
−
∞
=
+ + +
+ + + +
+ + +
0
) ( ) (
) ( ) (
1
1
log
i
i i x b a
i x b a i ax b
i ax b
q
q
q
q
q
q
b
αα
α α
.
Now, making use of lemma 2. With
and then exchang a and b, we get
f
′
(
x
)
/
f
(
x
)
≤
0
,
which impliesf
′
(
x
)
≤
0
.Therefore
f
(
x
)
is non-increasing. The proof is complete.Theorem: 2.4 Let
x
>
0
,
c
,
d
,
M
,
N
>
0
,
Mb
2<
Nd
2,
a
>
c
,
b
>
d
,
0
<
q
<
1
,
f
>
0
and0
<
′′
f
(that isf
′
is decreasing). Then the function
(
(
)
)
(
)
(
)
Nxx b q
x M x a q
x
f
x
L
+
Γ
+
Γ
=
1
1
)
(
)
(
is decreasing for
x
>
0
.
Proof: We have
L
x
f
x
Mx
(
(
ax)
)
Nx
(
q(
b
)
)
q
+
−
Γ
+
Γ
+
=
log
(
)
log
1
log
1
)
(
log
,
(
)
(
)
(
)
x
b
N
M
x
M
b
x
f
x
f
x
L
x
L
x a q x
a q
x a q
+
+
Γ
+
+
Γ
+
Γ′
−
′
=
′
1
1
1
)
(
)
(
)
(
)
(
(
)
(
)
(
x)
b q x
b q
x b q
N
Γ
+
−
+
Γ
+
Γ′
1
1
1
q
(
ax)
q(
ax)
q(
bx)
N
q(
bx)
x
b
N
M
x
M
b
x
f
x
f
+
Γ
−
+
+
+
Γ
+
+
−
′
=
1
1
1
1
)
(
)
(
ψ
ψ
.
)
(
)
(
x
f
x
G
=
Now,
(
)
q(
ax)
q(
bx)
x
b
N
x
a
M
x
f
x
f
x
f
x
f
x
G
′
=
′′
−
′
+
′
1
+
−
′
1
+
)
(
)
(
)
(
)
(
)
(
32 3
2 2
2
ψ
ψ
(
)
(
q(
xa)
q(
xb)
)
x
a
M
x
f
x
f
x
f
x
f
+
′
−
+
′
+
′
−
′′
≤
1
1
)
(
)
(
)
(
)
(
3 2 2
2
ψ
ψ
Since
ψ
q′′
(
x
)
<
0
,
thenψ
q′
is decreasing. ThereforeG
′
(
x
)
≤
0
, which impliesL
′
(
x
)
≤
0
,
and hence)
(
x
L
is decreasing.Theorem: 2.5 Let
x
,
y
>
0
,
0
<
q
<
1
,
c
,
d
>
0
,
a
≥
c
,
b
≥
d
,
Γ
q(
a
)
≥
1
,
ψ
q(
c
+
dx
)
≥
0
. Then the function
(
)
(
)
yx d
x y b
c
a
x
H
+
Γ
+
Γ
=
)
(
is non- decreasing for
x
>
0
.
Proof: Since
ψ
′
q>
0
,
thenψ
q is increasing and thereforeψ
(
a
+
bx
)
≥
ψ
(
c
+
dx
)
.Then we have
(
)
(
dx)
q y
b
q
a
y
c
x
x
H
(
)
=
log
Γ
+
−
log
Γ
+
log
,
(
)
(
)
)
(
)
(
log
)
(
)
(
2
x
F
x
g
c
x
yd
a
x
H
x
H
x d q y
b
+
+
=
+
Γ
=
′
ψ
.Now,
g
(
z
)
=
log
Γ
q(
a
+
bz
)
+
yz
2d
ψ
q(
c
+
dz
),
x
−1=
z
.
*W. T. Sulaiman / Some inequalities concerning Q-gamma functions / IJMA- 2(3), Mar.-2011, Page: 351-355
© 2010, IJMA. All Rights Reserved 355
≥
(
b
+
2
yz
)
ψ
q(
c
+
dz
)
+
ydz
2ψ
q′
(
c
+
dz
)
≥
0
.Therefore
g
is non-decreasing. Sinceg
(
0
)
≥
0
,
theng
(
x
)
≥
0
,
and henceH
′
(
x
)
≥
0
.
Theorem: 2.6 Let
f
(
x
)
≥
g
(
x
),
a
f
′
(
x
)
≥
b
g
′
(
x
)
>
0
,
ψ
(
g
(
x
)
)
>
0
,
a
,
b
>
0
,
then the function
(
)
(
)
bq a q
x
g
x
f
x
h
)
(
)
(
)
(
Γ
Γ
=
is non-decreasing in
x
.
Proof: We have
log
h
(
x
)
=
a
log
Γ
q(
f
(
x
)
)
−
b
log
Γ
q(
g
(
x
)
)
,
(
)
(
)
(
)
(
(
)
)
)
(
)
(
)
(
)
(
)
(
)
(
)
(
x
g
x
g
x
g
b
x
f
x
f
x
f
a
x
h
x
h
q q q
q
Γ
Γ′
′
−
Γ
Γ′
′
=
′
=
a
f
′
(
x
)
ψ
q(
f
(
x
)
)
−
b
g
′
(
x
)
ψ
q(
g
(
x
)
)
.Making use of (1.7), we have
(
)
(
)
(
)
(
1
)(
1
)
0
,
log
)
(
)
(
0
) ( )
( ) ( ) (
≥
−
−
−
=
−
∞
= + +
n
x g n x
f n
n x g x f
q
q
q
q
q
x
g
x
f
ψ
ψ
which implies
(
)
(
(
(
)
)
(
(
)
)
)
0
.
)
(
)
(
≥
−
′
≥
′
x
g
x
f
x
g
b
x
h
x
h
ψ
ψ
Therefore
h
′
(
x
)
≥
0
and h is non-decreasing.Remark: 2 Theorem 1.1 follows from theorem 2.6 by putting
f
(
x
)
=
a
+
bx
,
g
(
x
)
=
b
+
ax
,
0
≤
x
≤
1
,
and replacinga
,
b
byc
,
d
respectively. REFERENCES:[1] M. Abramowitz and I. A. Stegun, Handbook of Mathematical Functions with Formulas and Mathematical Tables, Dover, New york, 1965.
[2] C. G. Kokologiannaki, Monotonicity of of functions involving q-gamma functi-ons, Math. Ineq. Appl., Preprint. [3] T. Mansour, Some inequalities for the q-gamma functions, J. Ineq. Pure. Appl. Math. Volume 9 (2008), Issue 1, Article 1.
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