Applied Mathematics, 2011, 2, 465-469 doi:10.4236/am.2011.24059 Published Online April 2011 (http://www.SciRP.org/journal/am)
A Study on New q-Integral Inequalities Waad T. Sulaiman Department of Computer Engineering, College of Engineering, University of Mosul, Mosul, Iraq E-mail:
[email protected] Received February 12, 2011; revised March 4, 2011; accepted March 7, 2011
Abstract A q-analog, also called a q-extension or q-generalization is a mathematical expression parameterized by a quantity q that generalized a known expression and reduces to the known expression in the limit q 1 . There are q-analogs for the fractional, binomial coefficient, derivative, Integral, Fibonacci numbers and so on. In this paper, we give several results, some of them are new and others are generalizations of the main results of [1]. As well as we give a generalization to the key lemma ([2], Lemma 1.3). Keywords: q-Integral Inequlalities, Integral Inequalities b
1. Introduction
f x dq x F b F a .
For 0 q 1 , the q-analog of the derivative of the function f x , denoted by Dq f x is defined (see [5]) by
a
Dq f x
f x f qx
, x0
1 q x then Dq f 0 f 0 .
1
k k f x d q x 1 q k0 f q q ,
(1.2)
0
1
which reduces to f x dx as q 1 . The q-Jackson integral from 0 to a , for a more general case, can be defined (see [2,3]) by
k k f x d q x a 1 q k0 f aq q ,
(1.3)
0
provided the sum converges absolutely. The q-Jackson integral on a general interval may be defined (see [2,3]) by b
b
a
a
0
0
f x dq x f x dq x f x dq x .
(1.4)
The q-Jackson integral and q-derivative are related by the “fundamental theorem of quantum calculus” which can be restated ([3, p.73]) as follows: If F is an anti q-derivative of the function f, that is Dq F f , continuous at x a , then Copyright © 2011 SciRes.
x Dq f t d q t f x . a
(1.6)
It is not difficult to check that the q-analog of Leibniz’s rule is
Dq f x g x f x Dq g x g qx Dq f x . (1.7) For b 0 and a bq n with n , we denote
a, bq bq k : 0 k n and a, bq aq 1 , b q . (1.8) Some applications of q-integrals:
0
a
For any function f, we have
(1.1)
The q-deriIf f 0 exists, vative reduces to the usual derivative as q 1 . The q-analog of integration may be given (see [6]) by
(1.5)
1
x dx x 0
1 , 1 q
1
n x dq x 0
1 q 2 a 1 x 1 x dq x a q 0
q 1 1 q ln q , ln x d q x , n 1 q q 1 0
2
, where q a is a doubly pe-
riodic function. If q 1 , the integral reduces to a 1 x 1 x dx sin a . 0 In [4], the authors proved the following results: Theorem 1.1. If f x is a non-negative and increasing function on a, b q and satisfies
1 f 2 qx Dq f x 1 f 1 x x a
2
(1.9) AM
W. T. SULAIMAN
466
for 1 and 1 , then
Also,
f x dq x f x dq x . a
b
b
a
(1.10)
Theorem 1.2. If f x is a non-negative and increasing function on bq n m , b for m, n and satisfies q
1 Dq f x 1 f
1
q x x a m
2
Dq f
p
(1.11)
b f x dq x f qm x dq x . a
a
(1.12)
Theorem 1.3. If f x is a non-negative function on and satisfies
0, bq
b
b
f t dqt t dqt
x
(1.13)
x
for x 0, b q and 0 , then the inequality b
f
b
t d q t t
0
f t dqt
(1.14)
0
holds for all positive numbers and . In the coming section, we start with Lemma 2.1, which represent a generalization of ([1], Lemma 1.3). Theorems 2.3 and 2.4 are generalizations of Theorems 1.1 and 1.2 respectively, while Theorem 2.5 gives a generalization for more than one direction to Theorem 1.3. Other new results are also given.
Lemma 2.1. Let f x , f x , g x be nonnegative, nondecreasing functions defined on a, b q , and let p 1. Then
g qx f g qx Dq g x Dq f g x p 1 pf g x f g x Dq g x . p 1
p
g x
x p f 1 q x qx
f g x f g qx
.
(2.2)
Dq f g x Dq f , g Dq g x .
(2.3)
g x g qx
Then
Proof. It is follows as Dq f g x
f g x f g qx
1 q x f g x f g qx g x g qx g x g qx 1 q x Dq f , g Dq g x . Theorem 2.3. Let f x , f x , g x be nonnegative, nondecreasing functions defined on a, b q satis-
fies
f 1 g qx f g qx Dq g x
g x x a 1 Dq2 g x 2 1 1 f g x x a Dq g x
f
1
(2.4)
b
a
p 1
b f g x dq x f g x dq x . a
(2.5)
Proof. Let M x
(2.1)
x
a
x f g t dqt f g t dqt , a
and
Proof. We have
pf
g qx f g qx x g
t dt 1 q x qx pf p 1 g qx f g qx Dq g x .
Dq f , g
p
Dq f
p 1
for 1 , 2 , then
2. Results
pf
pf
Lemma 2.2. Define
on a, b q and for , 1 , then b
g x
f
p 1
p
1 q x
g t f g t g t dt
g x f g x x g
t dt 1 q x qx pf p 1 g x f g x g x g qx 1 q x pf p 1 g x f g x Dq g x .
Copyright © 2011 SciRes.
h g x
g x f p g qx
x
f g t dqt . a
Then via Lemma 2.1, we have Dq M x f g x Dq h g x
f g x h 1 g x h g x Dq g x f g x h 1 g x f g x Dq g x
f g x f g x h 1 g x Dq g x
g x F g x . h g x f g x x a ,
: f
As
then AM
W. T. SULAIMAN
467
Dq f g x Dq f g x Dq h 1 g x Dq g x
g x h g x D g x Dq h g x Dq g qx f 1 g qx f g qx Dq g x h 1 g x Dq2 g x 1 h 2 g x h g x Dq g x f 1 g qx f g qx Dq g x h 1 g x Dq2 g x 1 h 2 g x h g x Dq g x 1 1 f 1 g qx f g qx Dq g x f g x x a Dq2 g x 2 1 1 f g x x a Dq g x Dq f
1
1
2 q
The above implies Dq F g x 0 , by (2.4), which implies F g x 0 and hence Dq M x 0, so that M x 0 . Theorem 2.4. Let f x , f x , g x be nonnegative, nondecreasing functions defined on bq n m , b satisfies
b m f g x dq x f g q t dqt a a Proof. Let N x
q
f 1 g qx f g qx Dq g x
g q x x a D g x 1 f g q x x a D g x ,
f
1
1
m
1
m
2 q
b
x
a
(2.6)
(2.7)
x f g t dqt f g qmt dqt , a
and h g x
2
q
x
f
g q t d t . m
q
a
for 1, 2, then
Then via Lemma 2.1, we have
Dq N x f g x Dq h g x f g x h 1 g x h g x Dq g x
f g x h 1 g x f g q m x Dq g x f g x h 1 g x f g x Dq g x
g q x x a ,
f g x f g x h 1 g x Dq g x : f g x F g x .
As h g x f
m
then
Dq f g x Dq f g x Dq h 1 g x Dq g x
g x h g x D g x Dq h g x Dq g qx f 1 g qx f g qx Dq g x h 1 g x Dq2 g x 1 h 2 g x h g x Dq g x f 1 g qx f g qx Dq g x h 1 g x Dq2 g x 1 h 2 g x h g q m x Dq g x 1 1 f 1 g qx f g qx Dq g x f g q m x x a Dq2 g x Dq f
1 f
1
1
1
2 q
g q x x a m
2
The above implies Dq F g x 0 , by (2.4), which implies F g x 0 and hence Dq N x 0 , so that N x 0 . Theorem 2.5. Let f , g , h be nonnegative functions, h defined on a, b q and f , g are defined on R h , g, h are nondecreasing with g a h a 0 . If Copyright © 2011 SciRes.
Dq g x b
b
x
s
f h t dqt g h t dqt,
t a, b q ,
then b
a
f h x dq x
b
f h x g h x dq x
(2.8)
a
AM
W. T. SULAIMAN
468
for all , 0 . Proof. Since by Lemma 2.2,
b
f
b
h x dq x f h x g h x dq x
a x
x
a
a
g h x Dq g h t d q t Dq g , h Dq h t d q t ,
then, we have
b
f h x g h x dq x
a
a
x
h x Dq g
f
a
, h Dq h t d q t d q x
Dq g , h Dq h t f h x d q x d q t a t
b b Dq g , h Dq h t g h x d q x d q t a t
x
b
g h x dq x .
f
h x g
h x
f h x g h x . b
f a
b
x
a
a
a
b
1
f h x
b
a
b
f h x g h x dq x .
x
s
f h t dqt g h t dq t, then Copyright © 2011 SciRes.
a
1 b g h x dq x , a
b
g h x dq x
b
f h x dq x .
a
a
(2.12)
b
a
g h x d q x, a b
and the above implies, by (2.12), b
Theorem 2.6. Let f , g , h be nonnegative functions, h defined on a, b q and f , g are defined on R h , g , h are nondecreasing with g a h a 0 . If b
g h x dq x
a
a
b
b
Again, by the AG inequality, and via integration
a
a
1
b
a
f h x dq x
b
1 g h x d q x f h x g h x dxq
f h x g h x dq x f h x dq x
f h x g h x d q x.
and hence b
1 g h x f h x g 1 h x .
On integrating the above inequality with the using of (2.11) leads to
b
a
which implies
f h x g h x dq x f h x dq x
(2.11)
g h x dq x .
a
b
b
b
which implies via (2.9),
b , h Dq h t g h x d q x d q t t
g h x Dq g 1 , h Dq h t d q t d q x
b
a
1
a
h x dq x g h x dq x
f h x g h x d q x,
Dq g
Integrating the above inequality yields
, h Dq h t d q t d q x
Now, applying the AG inequality, we have for 1 ,
Using the arithmetic-geometric inequality yields
a
1
a
a
a
b
(2.9)
g h x d q x Dq g , h Dq h t d q t d q x a a b
x
b
b
f h x Dq g
b b Dq g 1 , h Dq h t f h x d q x d q t a t
a
b
for all 1, 0. Proof. For 1 , we have 1 f h x g h x dq x
b
b
(2.10)
a
t a, b q ,
a
f h x g h x dq x
b
f
h x dq x .
a
The other way round direction may be obtained from the following. Theorem 2.7. Let f, g, h be nonnegative functions, h defined on a, b q and f, g are defined on R h , g, h are nondecreasing with g a h a 0 . If AM
W. T. SULAIMAN b
b
x
s
f h t dqt g h t dq t,
t a, b q ,
a
f
b
h x d q x f h x g h x d q x
a
for all 0 . Proof. We have, via Lemma 2.2, f h x g h x dq x
a
a
f h x Dq g , h Dq h t d q t d q x
b Dq g , h Dq h t f h x d q x d q t a t (2.14) b b Dq g , h Dq h t g h x d q x d q t a t b x g h x d q x Dq g , h Dq h t d q t d q x a a b
b
g h x d q x .
Making use of the AG inequality,
f
b
f x g h x dq x
a
b
f x g h x dq x
g h x
h x g h x .
Integrating the above inequality, with making use of (2.14) yield
b
f x g h x dq x .
a
Y. Miao and F. Qi, “Several q-Integral Inequalities,” Journal of Mathematical Inequalities, Vol. 3, No. 1, 2009, pp. 115-121.
[2]
K. Brahim, N. Bettaibi and M. Sellemi, “On Some Feng Qi Type q-Intagral Inequlities,” Pure Applied Mathematics, Vol. 9, No. 2, 2008, Art. 43.
[3]
E. W. Weisstein, “q-Derivative,” Math World-A Wolfram Web Resource,” 2010. http://mathword. Wolfram .com/q-Derivative.html
[4]
E. W. Weisstein, “q-Integral,” Math World-A Wolfram Web Resource,” 2010. http://mathword. Wolfram .com/q-integral.html
[5]
F. H. Jackson, “On q-Definite Integrals,” Pure Applied Mathematics, Vol. 41, No. 15, 1910, pp. 193-203.
[6]
V. Kac and P. Cheung, “Quantum Calculus,” Universitext, Springer-Verlag, New York, 2003.
Copyright © 2011 SciRes.
f x g h x dq x
[1]
a
f h x
a
3. References
b
a
a
h x d q x g h x d q x
b
b
x
a
(2.13)
a
b
b
f
then b
469
AM
