Hindawi Publishing Corporation
Journal of Inequalities and Applications
Volume 2009, Article ID 135693, 8 pages
doi:10.1155/2009/135693
Research Article
An Interchangeable Theorem of q-Integral
Hongshun Ruan
Department of Applied Mathematics, Jiangsu Polytechnic University, Changzhou City 213164,
Jiangsu Province, China
Correspondence should be addressed to Hongshun Ruan, rhs@em.jpu.edu.cn
Received 25 August 2008; Accepted 3 January 2009
Recommended by Ondrej Dosly
We give a sufficient condition for the interchangeability of the order of sum and q-integral by
using inequality technique. As the application of the theorem, some interesting results on the
hypergeometric series are obtained.
Copyright q 2009 Hongshun Ruan. 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 and Some Lemmas
q-series, which are also called basic hypergeometric series, plays a very important role in
many fields, such as affine root systems, Lie algebras and groups, number theory, orthogonal
polynomials, and physics. Inequality technique is one of the useful tools in the study of
special functions. There are many papers about it see 1–6. First, we recall some definitions,
notations, and known results which will be used in this paper. Throughout this paper, it is
supposed that 0 < q < 1. The q-shifted factorials are defined as
a; q0 1,
a; qn a; q∞ n−1
1 − aqk ,
k0
∞
n 1, 2, . . . ,
1.1
1 − aqk .
k0
We also adopt the following compact notation for multiple q-shifted factorial:
a1 , a2 , . . . , am ; q
where n is an integer or ∞.
n
a1 ; q n a2 ; q n · · · am ; q n ,
1.2
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Journal of Inequalities and Applications
The q-binomial theorem 2 tells us that
∞ a; q zk
k
q; qk
k0
az; q∞
,
z; q∞
|z| < 1.
1.3
Replace a with 1/a, and z with az and then set a 0, we get
∞
−1k qkk−1/2 zk
k0
q; qk
z; q∞ .
1.4
Heine 2 introduced the basic hypergeometric series 2 φ1 , which is defined by
∞
a1 , a2 ; q k k
z .
2 φ1 a1 , a2 ; b1 ; q, z q, b1 ; q k
k0
1.5
Thomae 7 defined the q-integral on interval 0, 1 by
1
ftdq t 1 − q
0
∞
f qn qn ,
1.6
n0
provided that the series converges.
Fubini’s theorem. Suppose that fij is absolutely summary, that is
∞ ∞ fij < ∞,
1.7
i1 j1
then
∞
∞ i1 j1
fij ∞
∞ fij .
1.8
j1 i1
In order to prove the main result, we need to introduce two lemmas.
Lemma 1.1. Let b be a given real number, satisfying b < 1. Then, for 0 ≤ x ≤ 1, one has
1
≤ e|b|/1−bx .
1 − bx
1.9
fx 1 − bxeb/1−bx ,
1.10
Proof. Let
Journal of Inequalities and Applications
3
since
f x b2 1 − x b/1−bx
e
≥ 0,
1−b
0 ≤ x ≤ 1.
1.11
fx is monotonous increasing function with respect to 0 ≤ x ≤ 1. Hence,
1
≤ eb/1−bx ,
1 − bx
1.12
1.9 is proved.
Lemma 1.2. Let ai , bi be some real numbers, satisfying bi < 1 with i 1, 2, . . . , r. Then, for all
nonnegative integer n, one has
a1 , a2 , . . . , ar ; q r
n
≤ e1/1−q i1 |ai ||bi |/1−bi .
b1 , b2 , . . . , br ; q n
1.13
Proof. When n 0, it is obvious that 1.13 holds; when n ≥ 1, for 0 ≤ x ≤ 1 and 1 ≤ i ≤ r, we
have
1 − ai x ≤ 1 ai x ≤ e|ai |x ,
1.14
and by Lemma 1.1, we have
1 − a qk k
i
≤ e|ai ||bi |/1−bi q ,
1 − bi qk k 0, 1, 2, . . ..
1.15
Consequently,
ai ; q n−1
n
≤ e|ai ||bi |/1−bi 1q···q ≤ e1/1−q|ai ||bi |/1−bi bi ; q n
i 1, 2, . . . , r.
1.16
Thus, 1.13 follows. We complete the proof.
2. Main Result and Its Proof
We know that, whether the order of sum and q-integral is interchangeable is an important
problem in the study of q-series. We obtain following result on the interchangeability.
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Journal of Inequalities and Applications
Theorem 2.1. Let ai , bi be some real numbers, satisfying bi < 1 with i 1, 2, . . . , r. Suppose real
1
function fn t is q-integrable absolutely with n 0, 1, . . . and series ∞
n0 0 |fn t|dq t is convergent.
Then
1 ∞
∞ 1 a , a , . . . , a ; q
a1 , a2 , . . . , ar ; q n
1 2
r
fn tdq t n fn tdq t.
,
b
,
.
.
.
,
b
;
q
,
b
,
.
.
.
,
b
;
q
b
b
1 2
r
1 2
r
0 n0
n0 0
n
n
2.1
Proof. Using 1.13 and 1.6, we have
∞ ∞ a ,a ,...,a ;q
m m
r
1 2
n
1 − q
fn q q
b1 , b2 , . . . , br ; q
n0 m0
n
r
≤ e1/1−q
i1 |ai ||bi |/1−bi ∞ ∞
1 − q fn qm qm
n0
e
∞ 1
1/1−q ri1 |ai ||bi |/1−bi fn tdq t.
n0
Since, the series
2.2
m0
∞ 1
n0 0 |fn t|dq t
0
is convergent, the series
∞ ∞
a1 , a2 , . . . , ar ; q n m m
fn q q
n0 m0 b1 , b2 , . . . , br ; q n
2.3
is absolutely convergent. Hence, by the Fubini’s theorem, we have
∞
∞
∞ ∞ a1 , a2 , . . . , ar ; q n m m a1 , a2 , . . . , ar ; q n m m
fn q q fn q q .
n0 m0 b1 , b2 , . . . , br ; q n
m0 n0 b1 , b2 , . . . , br ; q n
2.4
From 2.4 and 1.6, 2.1 holds. The proof is completed.
3. Applications
As the application of Theorem 2.1, in this section, we obtain some results. First, we give
following lemma.
Lemma 3.1. Let a be a real number, satisfying a < 1. Then, for all nonnegative integer n, one has
1
qt; q∞ n
1 − q q; qn
t dq t .
1 − a aq; qn
0 at; q∞
3.1
Journal of Inequalities and Applications
5
Proof. By 1.3 and 1.6, we have
∞ a; q
aqn1 ; q ∞ k n1k
q
n1 q;
q
q ;q ∞
k
k0
k1 ∞
q ;q
a; q∞
1 − q
k ∞ qnk qk
1 − qq; q∞
k0 aq ; q ∞
a; q∞
1 − qq; q∞
1
qt; q∞ n
t dq t
at;
q∞
0
n1 1
aq ; q ∞ qt; q∞ n
a; qn1
t dq t.
1 − qq; qn qn1 ; q ∞ 0 at; q∞
3.2
From 3.2, 3.1 holds.
Theorem 3.2. Let a, b be two real numbers, satisfying a < 1, |b| < 1. Then
n1
∞
q , abqn1 ; q ∞ n
1
.
q n
n
1 − a1 − b
aq , bq ; q ∞
n0
3.3
n1
1
∞
q , abqn1 ; q ∞ n
qt, abqt; q∞
1
q dq t.
n
n
1
−
q
aq , bq ; q ∞
0 at, bt; q∞
n0
3.4
Proof. By 1.6, we have
By 1.3, we have
1
qt, abqt; q∞
dq t 0 at, bt; q∞
1
∞ aq; q
qt; q∞ m
btm dq t.
at;
q
q;
q
0
∞ m0
m
3.5
Using Theorem 2.1, we have
1
1
∞ aq; q
∞ aq; q
qt; q∞ qt; q∞ m
m
m m
btm dq t b
t dq t.
q; qm
0 at; q∞ m0 q; qm
0 at; q∞
m0
3.6
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Journal of Inequalities and Applications
By Lemma 3.1, we have
1
∞ aq; q
∞ aq; q
qt; q∞ m
m m
m m 1 − qq; qm
b
t dq t b
q; qm
q; qm
1 − aaq; qm
0 at; q∞
m0
m0
∞
bm
m0
1−q
1−a
3.7
1−q
.
1 − a1 − b
Combining 3.4–3.7, 3.3 holds.
In 3.5, replacing abq by c, we obtain the following result.
Corollary 3.3. Let a, b, c be some real numbers, satisfying a < 1, |b| < 1. Then
1
∞ c/b; q
qt, ct; q∞
1−q
m m
dq t b .
at,
bt;
q
1
−
a
aq;
q
0
∞
m
m0
3.8
Corollary 3.4. Let c be a real number. Then
∞
n0
c/q; qn qn ∞
−1n qnn−1/2
cn .
3.9
∞
c/q; qn qn .
3.10
q; qn1
n0
Proof. Taking a 0, b q in 3.8, we have
1
0
ct; q∞ dq t 1 − q
n0
On the other hand, by 1.4 and Theorem 2.1, we have
1
0
ct; q∞ dq t 1 ∞
0 n0
−1n qnn−1/2 ctn
dq t
q; qn
∞
−1n qnn−1/2 cn 1
n0
1 − q
q; qn
t n dq t
0
∞
−1n qnn−1/2
n0
which by combining with 3.10, implies 3.9.
Take c 1, 3.9 implies the following result.
q; qn1
cn ,
3.11
Journal of Inequalities and Applications
7
Corollary 3.5. The following equation holds:
∞
−1n qnn−1/2
q; qn1
n0
q.
3.12
Take c 1/q, 3.9 implies the following result.
Corollary 3.6. The following equation holds:
∞
−1n qnn−3/2
n0
q; qn1
q2 .
3.13
Remark 3.7. Taking c q−k , where k is positive integer, 3.9 readily yields many equations.
Corollary 3.8. Let a be a real number, satisfying |a| < 1. Then
∞
n0
∞
qn
an
.
a; qn1 n0 q; qn1
3.14
Proof. Taking b q, c 0 in 3.8, we have
1
∞
qn
1 − q
1
dq t .
1 − a n0 aq; qn
0 at; q∞
3.15
On the other hand, by Theorem 2.1 and set a 0 then replace z with at in 1.3, we have
1
1
dq t at;
q∞
0
1 ∞
0 n0
∞
1−q
atn
an
dq t .
q; qn
q; qn 1 − qn1
n0
3.16
Combining 3.15 and 3.16, 3.14 follows.
Theorem 3.9. Let a, b be two real numbers, satisfying |ab| < 1. Then
2 φ1
1−q
q/a, q/b; q2 ; q, ab 2 φ a, b; abq; q, q.
1 − ab 1
3.17
Proof. We recall the Heines transformation formula 7
2 φ1 a, b; c; q, z
abz/c; q∞
2 φ1 c/a, c/b; c; q, abz/c.
z; q∞
3.18
In 3.18, replacing c, z by q, qt, respectively, 3.18 yields
qt; q∞
2 φ a, b; q; q, qt 2 φ1 q/a, q/b; q; q, abt.
abt; q∞ 1
3.19
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Journal of Inequalities and Applications
Taking the q-integral on both sides of 3.19 with respect to variable t, we have
1
qt; q∞
2 φ a, b; q; q, qtdq t abt;
q∞ 1
0
1
0
2 φ1 q/a, q/b; q; q, abtdq t.
3.20
Applying 1.5 to 3.20 yields
1
1 ∞ q/a, q/b; q
∞ a, b; q
qt; q∞ n
n
qtn dq t abtn dq t.
abt;
q
q,
q;
q
q,
q;
q
0
0 n0
∞ n0
n
n
3.21
Applying Theorem 2.1 and Lemma 3.1 to 3.21, we have
∞ a, b; q
n
n0
q, q; qn
qn
∞ q/a, q/b; q
1 − q q; qn
1−q
n
abn
,
1 − ab abq; qn n0 q, q; qn
1 − qn1
3.22
hence,
∞
∞ q/a, q/b; q
a, b; qn n 1−q q 2 n abn .
1 − ab n0 q, abq; qn
q, q ; q n
n0
3.23
From 3.23 and 1.5, 3.17 follows.
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