# Note on aq-Contour Integral Formula of Gasper-Rahman

Authors: Jian-Ping Fang
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

We use theq-Chu-Vandermonde formula and transformation technique to derive a more general q-integral equation given by Gasper and Rahman, which involves the Cauchy polynomial. In addition, some applications of the general formula are presented in this paper.


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## Body

## 1. Introduction and Main Result

It is well known that theq-integral is an important branch of q-series theory. There are many techniques to achieve the ends; for instance, combinatorics method (cf. [1]), analysis methods (cf. [2–4]), and method of transformation (cf. [5–7]) are usually used. In 1989, Gasper and Rahman applied some analysis techniques to derive the following q-contour integral formula (cf. [8, Equation (3.17)]):
(1)12πi∫C‍(γ/z,bqz,qz/γ;q)∞(α/z,az,bz;q)∞(qz/γqm0;q)m0×∏i=1r‍(aiz;q)midzz=(γ/α,αq/γ,qb/a;q)∞(aα,q/aα,bα;q)∞(q/bγqm0;q)m0(bα)∑i=0rmi×∏i=1r‍(ai/b;q)mi.
Inspired by [7, 8], we employ the above equation and transformation technique to derive a more general q-contour integral equation. The main result of this paper is stated as follows.Theorem 1.
If
m
0
,
m
1
,
…
,
m
r
, and
h
are nonnegative integers and
q
=
a
γ
q
∑
i
=
0
r
‍
m
i
, then
(2)
1
2
π
i
∫
C
‍
(
γ
/
z
,
b
q
z
,
q
z
/
γ
;
q
)
∞
(
α
/
z
,
a
z
,
b
z
;
q
)
∞
(
q
z
/
γ
q
m
0
;
q
)
m
0
×
∏
i
=
1
r
(
a
i
z
;
q
)
m
i
∏
j
=
1
h
+
1
P
n
j
(
1
/
z
;
d
j
)
d
z
z
=
∏
l
=
0
h
(
α
d
l
;
q
)
n
l
α
∑
i
=
0
h
n
i
(
γ
/
α
,
q
α
/
γ
,
q
b
/
a
;
q
)
∞
(
a
α
,
q
/
a
α
,
b
α
;
q
)
∞
×
(
q
/
b
γ
q
m
0
;
q
)
m
0
(
b
α
)
∑
i
=
0
r
m
i
×
∏
j
=
1
r
(
a
j
/
b
;
q
)
m
j
×
∏
i
=
0
h
∑
k
i
=
0
n
i
(
q
-
n
i
,
q
A
i
b
α
,
q
A
i
+
n
i
+
1
d
i
+
1
α
,
…
,
q
A
i
+
n
h
d
h
α
;
q
)
k
i
(
q
,
q
A
i
d
i
α
,
q
A
i
d
i
+
1
α
,
…
,
q
A
i
d
h
α
;
q
)
k
i
×
q
k
i
(
1
-
∑
j
=
i
+
1
h
n
j
)
,
provided that |γ/α|<1 and C is a deformation of the unit circle so that the poles of 1/(az,bz;q)∞ lie outside the contour and the origin and the poles of 1/(α/z;q)∞ lie inside the contour. Where Pn(a;b) denotes the Cauchy polynomial defined as (7), one denotes that Ai=∑j=0i-1‍kj, and when i=0, one sets one  A0=∑j=0i-1‍kj=0.


## 2. Notations and Lemmas

We adopt the custom notations given in [9]. It is supposed that 0<|q|<1 in this paper. We use N to denote the set of all nonnegative integers.For any complex parametera, the q-shifted factorials are defined as
(3)(a;q)0=1,(a;q)n=∏k=0n-1(1-aqk),n=1,2,…,(a;q)∞=∏k=0∞(1-aqk).
For brevity, we also use the following notation:
(4)(a1,a2,…,am;q)n=(a1;q)n(a2;q)n⋯(am;q)n.
The q-binomial coefficient and the q-binomial theorem are given by
(5)[nk]=(q;q)n(q;q)k(q;q)n-k,∑n=0∞(a;q)nxn(q;q)n=(ax;q)∞(x;q)∞,|x|<1.
The basic hypergeometric seriessΦt  is given by
(6)sΦt=(a1,a2,…,asb1,b2,…,bt;q,x)=∑k=0∞(a1,a2…,as;q)k(q,b1,…,bt;q)k[(-1)kq(k2)]1+t-sxk.
In this paper, we denote that (n2)=n(n-1)/2 and k, m, n, s, t∈N.Leta,b be any complex variables; then, the Cauchy polynomial Pn(a;b) is defined as
(7)P0(a;b)=1,Pn(a;b)=(a-b)(a-bq)⋯(a-bqn-1),P0(a;b)=1,Pn(a;b)=fffffffff(a-b)(a-bq)⋯n≥1.Recall thatq-Chu-Vandermonde’s identity (cf. [9, page 14, Equation (1.5.3)]) is given as follows:
(8)2Φ1(q-n,af;q,q)==an(f/a;q)n(f;q)n.
As we know, it is one of the fundamental formulas in the basic hypergeometric series. Some applications of it were introduced in [5, 10, 11]. We will apply this identity to start our proof in the following. Since we assume that the integrals are the same established condition as the theorem, we omit the condition in the following.Lemma 2.
One has(9)12πi∫C(γ/z,bqz,qz/γ;q)∞(α/z,az,bz;q)∞(qz/γqm0;q)m0×∏i=1r(aiz;q)miPn(1/z;f)zdz=(fα;q)nαn(γ/α,αq/γ,qb/a;q)∞(aα,q/aα,bα;q)∞(q/bγqm0;q)m0×(bα)∑i=0rmi∏i=1r(ai/b;q)mi×∑k=0n(q-n,bα;q)k(aγq∑i=0rmi)k(q,fα;q)k.Proof.
We rewrite (8) as follows:
(10)∑k=0n(q-n;q)k(q,f;q)kqk1(aqk;q)∞=an(f/a;q)n(f;q)n(a;q)∞.
Replacing (a,c) by (α/z,fα), respectively, we have
(11)∑k=0n(q-n;q)k(q,fα;q)kqk1(qkα/z;q)∞=αn(fα;q)nPn(1/z;f)(α/z;q)∞.
Both sides of (11) multiply by
(12)(γ/z,bqz,qz/γ;q)∞(az,bz;q)∞(qz/γqm0;q)m0∏i=1r(aiz;q)mi1z.
Then, we have
(13)∑k=0n(q-n;q)kqk(q,fα;q)k(γ/z,bqz,qz/γ;q)∞(qkα/z,az,bz;q)∞(qz/γqm0;q)m0×∏i=1r(aiz;q)mi1z=αn(fα;q)n(γ/z,bqz,qz/γ;q)∞(α/z,az,bz;q)∞(qz/γqm0;q)m0×∏i=1r(aiz;q)miPn(1/z;f)z.
Taking the q-integral on both sides of (13) with respect to variable z, we get
(14)∑k=0n(q-n;q)kqk(q,fα;q)k∫C(γ/z,bqz,qz/γ;q)∞(qkα/z,az,bz;q)∞×(qz/γqm0;q)m0∏i=1r(aiz;q)mi1zdz=αn(fα;q)n∫C‍(γ/z,bqz,qz/γ;q)∞(α/z,az,bz;q)∞×(qz/γqm0;q)m0∏i=1r(aiz;q)miPn(1/z;f)zdz.
Employing (1) to the left side of (14), we have the desired result after some simplification.On the other hand, if we multiply (13) by Pn1(1/z;g), we have
(15)∑k=0n(q-n;q)kqk(q,fα;q)k(γ/z,bqz,qz/γ;q)∞(qkα/z,az,bz;q)∞(qz/γqm0;q)m0×∏i=1r(aiz;q)miPn1(1/z;g)z=αn(fα;q)n(γ/z,bqz,qz/γ;q)∞(α/z,az,bz;q)∞(qz/γqm0;q)m0×∏i=1r(aiz;q)miPn1(1/z;g)Pn(1/z;f)z.Taking theq-integral on both sides of (15) with respect to variable z, we use (9) in the resulting equation. After simple rearrangements, noting that q=aγq∑i=0r‍mi, we get the following.Lemma 3.
One has(16)12πi∫C(γ/z,bqz,qz/γ;q)∞(α/z,az,bz;q)∞(qz/γqm0;q)m0×∏i=1r(aiz;q)miPn1(1/z;g)Pn(1/z;f)zdz=(gα;q)n1(fα;q)nαn1αn(γ/α,qα/γ,qb/a;q)∞(aα,q/aα,bα;q)∞×(q/bγqm0;q)m0(bα)∑i=0rmi∏i=1r(ai/b;q)mi×∑k=0n(q-n,bα,qn1gα;q)kqk(1-n1)(q,fα,gα;q)k×∑k1=0n1(q-n1,bαqk;q)k1qk1(q,gαqk;q)k1.Both sides of (11) multiply by
(17)(γ/z,bqz,qz/γ;q)∞(az,bz;q)∞(qz/γqm0;q)m0∏i=1r(aiz;q)mi×∏j=1hPnj(1/z;dj)1z.
Then, taking the q-integral on both sides of the result equation with respect to variable z, we find the following.Lemma 4.
On has(18)∫C(γ/z,bqz,qz/γ;q)∞(α/z,az,bz;q)∞(qz/γqm0;q)m0∏i=1r(aiz;q)mi×∏j=1h+1Pnj(1/z;dj)dzz=(αdh+1;q)nh+1αnh+1×∑k=0nh+1(q-nh+1;q)kqk(q,αdh+1;q)k×∫C(γ/z,bqz,qz/γ;q)∞(qkα/z,az,bz;q)∞(qz/γqm0;q)m0×∏i=1r(aiz;q)mi∏j=1hPnj(1/z;dj)dzz,
where (nh+1,dh+1) denote (n,f), respectively.


## 3. Proof and Some Applications

Now, we return to the proof of Theorem1.The following result can be easily derived from (16) and (18):
(19)∫C(γ/z,bqz,qz/γ;q)∞(α/z,az,bz;q)∞(qz/γqm0;q)m0×∏i=1r(aiz;q)mi∏j=13Pnj(1/z;dj)dzz=∏i=13(αdi;q)niα∑i=13ni(γ/α,qα/γ,qb/a;q)∞(aα,q/aα,bα;q)∞×(q/bγqm0;q)m0(bα)∑i=0rmi∏j=1r(aj/b;q)mj×∑k=0n(q-n,bα,qn1d1α,qn2d2α;q)k(q,fα,d1α,d2α;q)kqk(1-n1-n2)×∑k1=0n1(q-n1,qkbα,qn2+kd2α;q)k1(q,qkd1α,qkd2α;q)k1qk1(1-n2)×∑k2=0n2(q-n2,qk+k1bα;q)k2(q,qk+k1d2α;q)k2qk2.Lettingn=n0, k=k0, and f=d0 and combining (19) with (18), by induction, similar proof can be performed to get the desired result.Takingn1=n2=⋯=nh+1=0 in (2), the theorem goes back to formula (1). Putting n1=⋯=nh=0 in (2), we have the  following.Corollary 5.
One has(20)12πi∫C(γ/z,bqz,qz/γ;q)∞(α/z,az,bz;q)∞(qz/γqm0;q)m0×∏i=1r(aiz;q)miPn(1/z;d0)zdz=(γ/α,αq/γ,qb/a;q)∞(aα,q/aα,bα;q)∞(q/bγqm0;q)m0(bα)∑i=0rmi×∏i=1r(ai/b;q)mi(d0/b;q)nbn.Lettingn2=⋯=nh=0 in (2), we get the following.Corollary 6.
One has(21)12πi∫C(γ/z,bqz,qz/γ;q)∞(α/z,az,bz;q)∞(qz/γqm0;q)m0×∏i=1r(aiz;q)miPn1(1/z;d1)Pn(1/z;d0)zdz=(γ/α,αq/γ,qb/a;q)∞(aα,q/aα,bα;q)∞(q/bγqm0;q)m0(bα)∑i=0rmi×∏i=1r(ai/b;q)mi(d0/b;q)n(d1/b;q)n1bn+n1.Combining (21) with (18), by induction and applying (2), we can conclude the following.Theorem 7.
One has(22)12πi∫C(γ/z,bqz,qz/γ;q)∞(α/z,az,bz;q)∞(qz/γqm0;q)m0×∏i=1r(aiz;q)mi∏j=0hPnj(1/z;dj)dzz=(γ/α,qα/γ,qb/a;q)∞(aα,q/aα,bα;q)∞(q/bγqm0;q)m0(bα)∑i=0rmib∑i=0hni×∏i=1r(ai/b;q)mi∏j=0h(dj/b;q)nj.Comparing (2) and (22), we have the following interesting identity.Corollary 8.
If
m
0
,
m
1
,
…
,
m
r
, and
h
are nonnegative integers, then
(23)
∑
k
0
=
0
n
0
(
q
-
n
0
,
b
α
,
q
n
1
d
1
α
,
…
,
q
n
h
d
h
α
;
q
)
k
0
(
q
,
d
0
α
,
d
1
α
,
…
,
d
h
α
;
q
)
k
0
q
k
0
(
1
-
∑
j
=
1
h
n
j
)
×
∏
i
=
1
h
‍
∑
k
i
=
0
n
i
(
q
-
n
i
,
q
A
i
b
α
,
q
A
i
+
n
i
+
1
d
i
+
1
α
,
…
,
q
A
i
+
n
h
d
h
α
;
q
)
k
i
(
q
,
q
A
i
d
i
α
,
q
A
i
d
i
+
1
α
,
…
,
q
A
i
d
h
α
;
q
)
k
i
×
q
k
i
(
1
-
∑
j
=
i
+
1
h
n
j
)
=
∏
i
=
0
h
(
d
i
/
b
;
q
)
n
i
(
d
i
α
;
q
)
n
i
(
b
α
)
n
0
+
n
1
+
⋯
+
n
h
.Takingh=1 and d0=d1=qb in (23), we have
(24)∑k0=0n0[n0k0](bα,qn1+1bα;q)k0(qbα,qbα;q)k0(-1)k0q(k0+12)-k0(n0+n1)×∑k1=0n1[n1k1](qk0bα;q)k1(qk0+1bα;q)k1(-1)k1q(k1+12)-k1n1=(q;q)n0(q;q)n1(qbα;q)n0(qbα;q)n1(bα)n0+n1.
Setting bα=q, then letting q→1 in the above identity, we have  the following.Corollary 9.
Ifn0, n1∈N, then
(25)∑k0=0n0(n0k0)(n1+2)k0(2)k0(-1)k0∑k1=0n1(n1k1)1k0+k1+1(-1)k1=1(n0+1)(n1+1),where(a)0=1and(a)n=a(a+1)⋯(a+n-1),n≥1,n∈N.Takingh=2 and d0=d1=d2=qb in (23), we have
(26)∑k=0n0[n0k0](bα,qn1+1bα,qn2+1bα;q)k0(qbα,qbα,qbα;q)k0(-1)k0×q(k0+12)-k0(n0+n1+n2)×∑k1=0n1[n1k1](qk0bα,qn2+k0+1bα;q)k1(qk0+1bα,qk0+1bα;q)k1(-1)k1×q(k1+12)-k1(n1+n2)×∑k2=0n2[n2k2](qk0+k1bα;q)k2(qk0+k1+1bα;q)k2(-1)k2×q(k2+12)-k2n2=(q;q)n0(q;q)n1(q;q)n2(qbα;q)n0(qbα;q)n1(qbα;q)n2(bα)n0+n1+n2.
Setting bα=q, then letting q→1 in the above identity, we have the following.Corollary 10.
Ifn0, n1, n2∈N, then
(27)∑k0=0n0(n0k0)(n1+2)k0(n2+2)k0(2)k0(2)k0(-1)k0×∑k1=0n1(n1k1)(n2+k0+2)k1(k0+2)k1(-1)k1×∑k2=0n2(n2k2)1k0+k1+k2+1(-1)k2=1(n0+1)(n1+1)(n2+1),where(a)0=1and(a)n=a(a+1)⋯(a+n-1),n≥1,n∈N.More general, we have the following identity.Corollary 11.
Ifh, n0, n1, …, nh∈N, then
(28)∑k0,…,kh‍∏i=0h-1(niki)(Ai+ni+1+2)ki⋯(Ai+nh+2)ki(Ai+2)ki⋯(Ai+2)ki×(-1)k0+⋯+khAh+kh+2=∏i=0h1(ni+1),
where 0≤ki≤ni, i=0,…,h.Both sides of (20) multiply by 1/(q;q)n; then, summing n from 0 to ∞ and using the q-binomial theorem, we find the following.Corollary 12.
Ifmax{|1/z|,|b|}<1, then
(29)12πi∫C‍(γ/z,bqz,qz/γ;q)∞(α/z,1/z,az,bz;q)∞(qz/γqm0;q)m0×∏i=1r(aiz;q)midzz=(γ/α,αq/γ,qb/a;q)∞(aα,q/aα,bα,b;q)∞(q/bγqm0;q)m0(bα)∑i=0rmi×∏i=1r(ai/b;q)mi.Remark 13.
Ifn1=n2=⋯=nh=0, identity (23) becomes the q-Chu-Vandermonde formula.



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*Source: 840704-2013-09-10.xml*