On a Reduction Formula for a Kind of Double q-Integrals

Abstract

Using the q-integral representation of Sears’ nonterminating extension of the q-Saalschütz summation, we derive a reduction formula for a kind of double q-integrals. This reduction formula is used to derive a curious double q-integral formula, and also allows us to prove a general q-beta integral formula including the Askey–Wilson integral formula as a special case. Using this double q-integral formula and the theory of q-partial differential equations, we derive a general q-beta integral formula, which includes the Nassrallah–Rahman integral as a special case. Our evaluation does not require the orthogonality relation for the q-Hermite polynomials and the Askey–Wilson integral formula.

1. A Double q-Integral Formula

Throughout the paper, we assume that $0<q<1$. For $a\in \mathbb{C}$, we define the q-shifted factorials as follows:

$${(a;q)}_0=1,{(a;q)}_n=\prod _{k=0}^{n-1}(1-a{q}^k),n=0,1,2,\cdots ,$$

and

$${(a;q)}_{\infty }=\underset{n\to \infty }{lim}{(a;q)}_n=\prod _{k=0}^{\infty }(1-a{q}^k).$$

If n is an integer or ∞, the multiple q-shifted factorials are defined as:

$${(a_1,a_2,...,a_m;q)}_n={(a_1;q)}_n{(a_2;q)}_n\dots {(a_m;q)}_n.$$

The q-binomial coefficients are the q-analogs of the binomial coefficients, which are defined by

$${\displaystyle {\left[\genfrac{}{}{0pt}{}{n}{k}\right]}_q={\displaystyle \frac{{(q;q)}_n}{{(q;q)}_k{(q;q)}_{n-k}}}.}$$

Definition 1. 

For$x=cos\theta$, we define $h(x;a)$ and $h(x;a_1,a_2,\dots ,a_m)$ as follows:

$$\begin{array}{cc}& h(x;a)={(a{e}^{i\theta },a{e}^{-i\theta };q)}_{\infty }=\prod _{k=0}^{\infty }(1-2q^{k}ax+q^{2k}{a}^2),\hfill \\ & h(x;a_1,a_2,\dots ,a_m)=h(x;a_1)h(x;a_2)\cdots h(x;a_m).\hfill \end{array}$$

Definition 2. 

For simplicity, we use $\Delta (u,v)$ to denote the theta function:

$$(1-q)v{(q,u/v,qv/u;q)}_{\infty }.$$

As usual, the basic hypergeometric series or q-hypergeometric series ${}_r{\varphi }_s$ is defined by:

$${}_r{\varphi }_s\left(\genfrac{}{}{0pt}{}{a_1,a_2,...,a_r}{b_1,b_2,...,b_r};q,z\right)=\sum _{n=0}^{\infty }{\displaystyle \frac{{(a_1,a_2,...,a_r;q)}_n}{{(q,b_1,b_2,...,b_s;q)}_n}}{\left({(-1)}^n{q}^{n(n-1)/2}\right)}^{1+s-r}{z}^n.$$

Now, we introduce the definition of the Thomae–Jackson q-integral in q-calculus, which was introduced by Thomae [1] and Jackson [2].

Definition 3. 

Given a function $f(x)$, the Thomae–Jackson q-integral is defined by:

$${\int }_a^{b}f(x)d_{q}x=(1-q)\sum _{n=0}^{\infty }[bf(b{q}^n)-af(a{q}^n)]q^n.$$

Using the q-integral notation, one can write some q-formulas in more compact forms.

On making use of the q-exponential operator method, we [3] proved the following proposition (some misprints have been corrected here):

Proposition 1. 

For $max\{|au|,|bu|,|cu|,|av|,|bv|,|cv|,|d/b|\}<1$, we have:

$$\begin{array}{cc}\hfill {\int }_u^v{\displaystyle \frac{{(qx/u,qx/v,dx;q)}_{\infty }}{{(ax,bx,cx;q)}_{\infty }}}{d}_{q}x& ={\displaystyle \frac{\Delta (u,v){(abuv,bcuv,d/b;q)}_{\infty }}{{(au,av,bu,bv,cu,cv;q)}_{\infty }}}\hfill \\ & \times {}_3{\varphi }_2\left(\genfrac{}{}{0pt}{}{bu,bv,abcuv/d}{abuv,bcuv};q,{\displaystyle \frac{d}{b}}\right).\hfill \end{array}$$

When $d=abcuv$ in Proposition 1, upon noting that ${(1;q)}_k={\delta }_{0,k}$, the ${}_3{\varphi }_2$ series in the proposition reduces to 1 and thus the proposition becomes the Al-Salam–Verma formula, which is the q-integral representation of Sears’ nonterminating extension of the q-Saalschütz summation [4], see also [5] (p. 52).

Proposition 2. 

If there are no zero factors in the denominator of the integral and $max\{|au|,|av|,|bu|,|bv|,|cu|,|cv|\}<1$, then we have:

$${\int }_u^v{\displaystyle \frac{{(qx/u,qx/v,abcuvx;q)}_{\infty }}{{(ax,bx,cx;q)}_{\infty }}}{d}_{q}x={\displaystyle \frac{\Delta (u,v){(abuv,acuv,bcuv;q)}_{\infty }}{{(au,bu,cu,av,bv,cv;q)}_{\infty }}}.$$

When $c=0,$ this q-integral formula reduces to the following q-integral formula due to Andrews and Askey [6], which can be derived from Ramanujan ${}_1{\psi }_1$ summation.

Proposition 3. 

If there are no zero factors in the denominator of the integral and $max\{|au|,|av|,|bu|,|bv|\}<1$, then we have:

$${\int }_u^v{\displaystyle \frac{{(qx/u,qx/v;q)}_{\infty }}{{(ax,bx;q)}_{\infty }}}{d}_{q}x={\displaystyle \frac{\Delta (u,v){(abuv;q)}_{\infty }}{{(au,bu,av,bv;q)}_{\infty }}}.$$

The main purpose of this paper is to study double q-integrals. There are not many studies on this subject, and it is difficult to find the definition of the double q-integrals in the literature. We now give the definition of the double q-integral of a two-variable function $f(x,y)$ over the rectangular region.

Definition 4. 

If $f(x,y)$ is a two variable function, the double q-integral of f over $[a,b]\times [c,d]$ is formally defined as:

$$\begin{array}{cc}& {\int \int }_{[a,b]\times [c,d]}f(x,y)d_{q}x{d}_{q}y\hfill \\ & ={(1-q)}^2\sum _{m,n=0}^{\infty }\left(bdf(b{q}^m,d{q}^n)+acf(a{q}^m,c{q}^n)\right)q^{m+n}\hfill \\ & -{(1-q)}^2\sum _{m,n=0}^{\infty }\left(bcf(b{q}^m,c{q}^n)+adf(a{q}^m,d{q}^n)\right)q^{m+n}.\hfill \end{array}$$

Proposition 4. 

If for any $x\in [a,b]$ and $y\in [c,d]$, the double series,

$$\sum _{m,n=0}^{\infty }f(x{q}^m,y{q}^n)q^{m+n}$$

is absolutely convergent, then it is equal to each of the two iterated q-integrals, namely,

$${\int \int }_{[a,b]\times [c,d]}f(x,y)d_{q}x{d}_{q}y={\int }_a^b{d}_{q}x{\int }_c^{d}f(x,y)d_{q}y={\int }_c^d{d}_{q}y{\int }_a^{b}f(x,y)d_{q}x.$$

Proof. 

If the conditions of the proposition are satisfied, then the q-double integral of $f(x,y)$ represents an absolutely convergent double series, and we can interchange the order of summation to complete the proof of Proposition 4. ☐

In order to determine the absolute convergence of double series, one can use the ratio test for double series (see, for example, [7] (Corollary 7.35)):

Proposition 5. 

Let $(a_{k,l})$ be a double sequence of nonzero real numbers such that either $|a_{k,l+1}|/|a_{k,l}|\to a$ or $|a_{k+1,l}|/|a_{k,l}|\to \tilde{a}$ as $(k,l)\to (\infty ,\infty )$, where $a,\tilde{a}\in R\cup \infty .$ If each row-series as well as each column-series corresponding to ${\sum }_{k,l}{a}_{k,l}$ is absolutely convergent and $a<1$ or $\tilde{a}<1$, then ${\sum }_{k,l}{a}_{k,l}$ is absolutely convergent.

The principal result of this paper is the following general iterated q-integral formula:

Theorem 1. 

Suppose that $f(y)$ is a function of y which satisfies,

$$\underset{n\to \infty }{lim}|f(y{q}^n)/f(y{q}^{n-1})|\le 1.$$

Then, we have the iterated q-integral identity,

$$\begin{array}{cc}& {\int }_u^v{\displaystyle \frac{{(qx/u,qx/v;q)}_{\infty }}{{(ax,bx;q)}_{\infty }}}{d}_{q}x{\int }_c^d{\displaystyle \frac{f(y){(qy/c,qy/d,abuvxy;q)}_{\infty }}{{(xy;q)}_{\infty }}}{d}_{q}y\hfill \\ & ={\displaystyle \frac{\Delta (u,v){(abuv;q)}_{\infty }}{{(au,av,bu,bv;q)}_{\infty }}}{\int }_c^d{\displaystyle \frac{f(y){(qy/c,qy/d,auvy,buvy;q)}_{\infty }}{{(uy,vy;q)}_{\infty }}}{d}_{q}y.\hfill \end{array}$$

Proof. 

For simplicity, we use $I(u,v,c,d)$ to denote the following q-double integral:

$${\int \int }_{[u,v]\times [c,d]}{\displaystyle \frac{f(y){(qx/u,qx/v,qy/c,qy/d,abuvxy;q)}_{\infty }}{{(ax,bx,xy;q)}_{\infty }}}{d}_{q}x{d}_{q}y.$$

Using the ratio test for double series in Proposition 5, we can prove that $I(u,v,c,d)$ is absolutely convergent.

Interchanging the order of the iterated integral on the left-hand side of the equation in Theorem 1, we find that:

$${\int }_c^d{(qy/c,qy/d;q)}_{\infty }f(y)d_{q}y{\int }_u^v{\displaystyle \frac{{(qx/u,qx/v,abuvxy;q)}_{\infty }}{{(ax,bx,xy;q)}_{\infty }}}{d}_{q}x.$$

On replacing c by y in the Al-Salam–Verma integral in Proposition 2, we immediately find that

$${\int }_u^v{\displaystyle \frac{{(qx/u,qx/v,abuvxy;q)}_{\infty }}{{(ax,bx,xy;q)}_{\infty }}}{d}_{q}x={\displaystyle \frac{\Delta (u,v){(abuv,auvy,buvy;q)}_{\infty }}{{(au,av,bu,bv,uy,vy;q)}_{\infty }}}.$$

Combining these two equations, we complete the proof of Theorem 1. ☐

Theorem 1 can be used to derive some double q-integral evaluation formulas in Theorems 2–7, some of whose proofs are given in the later sections.

Theorem 2. 

For ${max}_{s\in \{a,b,c,d\}}\{|su|,|sv|,|cw|,|dw|,|r/w|\}<1,$ we have:

$$\begin{array}{cc}& {\int }_u^v{\displaystyle \frac{{(qx/u,qx/v;q)}_{\infty }}{{(ax,bx;q)}_{\infty }}}{d}_{q}x{\int }_c^d{\displaystyle \frac{{(qy/c,qy/d,abuvxy,ry;q)}_{\infty }}{{(auvy,buvy,xy,wy;q)}_{\infty }}}{d}_{q}y\hfill \\ & ={\displaystyle \frac{\Delta (u,v)\Delta (c,d){(abuv,cduw,cdvw,r/w;q)}_{\infty }}{{(au,av,bu,bv,cu,cv,du,dv,cw,dw;q)}_{\infty }}}\hfill \\ & \times {}_3{\varphi }_2\left(\genfrac{}{}{0pt}{}{cw,dw,cduvw/r}{cduw,cdvw};q,{\displaystyle \frac{r}{w}}\right).\hfill \end{array}$$

When $r=cduvw$, the ${}_3{\varphi }_2$ series in the above equation reduces to 1, and hence we have the following curious double q-integral formula.

Proposition 6. 

For $max\{|a|,|b|,|c|,|d|,|u|,|v|,|w|\}<1,$ we have:

$$\begin{array}{cc}& {\int }_u^v{\displaystyle \frac{{(qx/u,qx/v;q)}_{\infty }}{{(ax,bx;q)}_{\infty }}}{d}_{q}x{\int }_c^d{\displaystyle \frac{{(qy/c,qy/d,abuvxy,cduvwy;q)}_{\infty }}{{(auvy,buvy,xy,wy;q)}_{\infty }}}{d}_{q}y\hfill \\ & ={\displaystyle \frac{\Delta (u,v)\Delta (c,d){(abuv,cduv,cduw,cdvw;q)}_{\infty }}{{(au,av,bu,bv,cu,cv,du,dv,cw,dw;q)}_{\infty }}}.\hfill \end{array}$$

Using Theorem 1, we can prove the following general q-beta integral formula that includes the Askey–Wilson integral as a special case.

Theorem 3. 

Suppose that $f(y)$ is a function of y that satisfies:

$$\underset{n\to \infty }{lim}|f(y{q}^n)/f(y{q}^{n-1})|\le 1.$$

Then, we have the following general q-beta integral formula:

$$\begin{array}{cc}& {\int }_0^{\pi }{\displaystyle \frac{h(cos2\theta ;1)}{h(cos\theta ;a,b)}}\left({\int }_c^d{\displaystyle \frac{f(y){(qy/c,qy/d,wy;q)}_{\infty }}{h(cos\theta ;y){(ry;q)}_{\infty }}}{d}_{q}y\right)d\theta \hfill \\ & ={\displaystyle \frac{2\pi }{{(q,ab;q)}_{\infty }}}{\int }_c^d{\displaystyle \frac{f(y){(qy/c,qy/d,wy;q)}_{\infty }}{{(ay,by,ry;q)}_{\infty }}}{d}_{q}y.\hfill \end{array}$$

Using this theorem and the theory of q-partial differential equations developed recently by us, we can prove the following general q-beta integral formula, which includes the Nassrallah–Rahman integral as a special case. Our evaluation does not require the orthogonality relation for the q-Hermite polynomials and the Askey–Wilson integral formula.

Theorem 4. 

For $\{|a|,|b|,|c|,|d|,|u|,|abs/t|\}<1,$ we have the q-formula:

$$\begin{array}{cc}& {\int }_0^{\pi }{\displaystyle \frac{h(cos2\theta ;1)h(cos\theta ;w)}{h(cos\theta ;a,b,c,d,u)}}{}_3{\varphi }_2\left(\genfrac{}{}{0pt}{}{t{e}^{i\theta },t{e}^{-i\theta },t/s}{at,bt};q,{\displaystyle \frac{abs}{t}}\right)d\theta \hfill \\ & ={\displaystyle \frac{2\pi {(w/c,w/d;q)}_{\infty }}{\Delta (c,d){(q,cd,au,bu,abs/t;q)}_{\infty }}}\hfill \\ & \times {\int }_c^d{\displaystyle \frac{{(qy/c,qy/d,wy,absuy/t;q)}_{\infty }}{{(ay,by,uy,wy/cd;q)}_{\infty }}}{}_3{\varphi }_2\left(\genfrac{}{}{0pt}{}{t/s,t/u,t/y}{at,bt};q,{\displaystyle \frac{absuy}{t}}\right)d_{q}y.\hfill \end{array}$$

Upon taking $s=t$ in this theorem and upon noting that ${(1;q)}_k={\delta }_{0,k}$, the two ${}_3{\varphi }_2$ series in the above Theorem both have the value 1; thus, we obtain the following integral formula due to the Nassrallah and Rahman [5] (pp. 157–158).

Theorem 5. 

For $max\{|a|,|b|,|c|,|d|,|u|\}<1,$ we have:

$$\begin{array}{cc}& {\int }_0^{\pi }{\displaystyle \frac{h(cos2\theta ;1)h(cos\theta ;w)}{h(cos\theta ;a,b,c,d,u)}}d\theta \hfill \\ & ={\displaystyle \frac{2\pi {(w/c,w/d;q)}_{\infty }}{\Delta (c,d){(q,ab,au,bu,cd;q)}_{\infty }}}{\int }_c^d{\displaystyle \frac{{(qy/c,qy/d,wy,abuy;q)}_{\infty }}{{(ay,by,uy,wy/cd;q)}_{\infty }}}{d}_{q}y.\hfill \end{array}$$

The proof of this formula given in [5] (pp. 157–158) needs to know the Askey–Wilson integral formula in advance.

Theorem 6. 

(Nassrallah–Rahman Integral). For $max\{|a|,|b|,|c|,|d|,|u|\}<1,$ we have:

$$\begin{array}{cc}& {\int }_0^{\pi }{\displaystyle \frac{h(cos2\theta ;1)h(cos\theta ;abcdu)d\theta }{h(cos\theta ;a,b,c,d,u)}}\hfill \\ & ={\displaystyle \frac{2\pi {(abcu,abcd,abdu,acdu,bcdu;q)}_{\infty }}{{(q,ab,ac,ad,au,bc,bd,bu,cd,cu,du;q)}_{\infty }}}.\hfill \end{array}$$

Proof. 

Setting $w=abcdu$ in Theorem 5, we immediately deduce that:

$$\begin{array}{cc}& {\int }_0^{\pi }{\displaystyle \frac{h(cos2\theta ;1)h(cos\theta ;abcdu)}{h(cos\theta ;a,b,c,d,u)}}d\theta \hfill \\ & ={\displaystyle \frac{2\pi {(abcu,abdu;q)}_{\infty }}{\Delta (c,d){(q,ab,au,bu,cd;q)}_{\infty }}}{\int }_c^d{\displaystyle \frac{{(qy/c,qy/d,abcduy;q)}_{\infty }}{{(ay,by,uy;q)}_{\infty }}}{d}_{q}y.\hfill \end{array}$$

Using the Al-Salam–Verma integral in Proposition 2, we immediately have:

$${\int }_c^d{\displaystyle \frac{{(qy/c,qy/d,abcduy;q)}_{\infty }}{{(ay,by,uy;q)}_{\infty }}}{d}_{q}y={\displaystyle \frac{\Delta (c,d)(abcd,acdu,bcdu;q)}{{(ac,ad,bc,bd,cu,du;q)}_{\infty }}}.$$

Combining these two equations, we complete the proof of the theorem. ☐

Putting $t=u,$ then the ${}_3{\varphi }_2$ series on the right-hand side of the equation in Theorem 4 reduces to $1.$ Hence, we arrive at the following proposition.

Theorem 7. 

For $\{|a|,|b|,|c|,|d|,|u|,|abs/u|\}<1,$ we have the q-formula:

$$\begin{array}{cc}& {\int }_0^{\pi }{\displaystyle \frac{h(cos2\theta ;1)h(cos\theta ;w)}{h(cos\theta ;a,b,c,d,u)}}{}_3{\varphi }_2\left(\genfrac{}{}{0pt}{}{t{e}^{i\theta },t{e}^{-i\theta },u/s}{at,bt};q,{\displaystyle \frac{abs}{u}}\right)d\theta \hfill \\ & ={\displaystyle \frac{2\pi {(w/c,w/d;q)}_{\infty }}{\Delta (c,d){(q,cd,au,bu,abs/u;q)}_{\infty }}}\hfill \\ & \times {\int }_c^d{\displaystyle \frac{{(qy/c,qy/d,wy,absy;q)}_{\infty }}{{(ay,by,uy,wy/cd;q)}_{\infty }}}{d}_{q}y.\hfill \end{array}$$

The remainder of the paper is organized as follows: Section 2 is devoted to the proof of Theorem 2, and Theorem 3 is proved in Section 3. Theorem 4 is established in Section 4. In Section 5, we give a general double q-integral formula.

2. The Proof of Theorem 2

Proof. 

If $f(y)={(ry;q)}_{\infty }/{(auvy,buvy,wy;q)}_{\infty }$, then, it is easily seen that:

$${\displaystyle \frac{f(q^{n}y)}{f(q^{n-1}y)}}={\displaystyle \frac{(1-auvy{q}^{n-1})(1-buvy{q}^{n-1})(1-wy{q}^{n-1})}{(1-ry{q}^{n-1})}}.$$

It follows that ${lim}_{n\to \infty }|f(y{q}^n)/f(y{q}^{n-1})|=1.$ Thus, we can choose:

$$f(y)={(ry;q)}_{\infty }/{(auvy,buvy,wy;q)}_{\infty }$$

in Theorem 1 to obtain:

$$\begin{array}{cc}& {\int }_u^v{\displaystyle \frac{{(qx/u,qx/v;q)}_{\infty }}{{(ax,bx;q)}_{\infty }}}{d}_{q}x{\int }_c^d{\displaystyle \frac{{(qy/c,qy/d,abuvxy,ry;q)}_{\infty }}{{(auvy,buvy,wy,xy;q)}_{\infty }}}{d}_{q}y\hfill \\ & ={\displaystyle \frac{\Delta (u,v){(abuv;q)}_{\infty }}{{(au,av,bu,bv;q)}_{\infty }}}{\int }_c^d{\displaystyle \frac{{(qy/c,qy/d,ry;q)}_{\infty }}{{(uy,wy,vy;q)}_{\infty }}}{d}_{q}y.\hfill \end{array}$$

If we use I to denote the q-integral in the right-hand side of the above equation, then, appealing to Proposition 1, we find that:

$$\begin{array}{c}\hfill I={\displaystyle \frac{\Delta (c,d){(cduw,cdvw,r/w;q)}_{\infty }}{{(cu,du,cv,dv,cw,dw;q)}_{\infty }}}{}_3{\varphi }_2\left(\genfrac{}{}{0pt}{}{cw,dw,cduvw/r}{cduw,cdvw};q,{\displaystyle \frac{r}{w}}\right).\end{array}$$

Combining these two equations, we complete the proof of Theorem 2. ☐

3. The Proof of Theorem 3 and the Askey–Wilson Integral

Proof. 

If $f(y)$ is replaced by $f(y){(wy;q)}_{\infty }/{(auvy,buvy,ry;q)}_{\infty }$ in Theorem 1, then, we deduce that:

$$\begin{array}{c}{\int }_u^v{\displaystyle \frac{{(qx/u,qx/v;q)}_{\infty }}{{(ax,bx;q)}_{\infty }}}{d}_{q}x{\int }_c^d{\displaystyle \frac{f(y){(qy/c,qy/d,abuvxy,wy;q)}_{\infty }}{{(auvy,buvy,xy,ry;q)}_{\infty }}}{d}_{q}y\hfill \\ ={\displaystyle \frac{\Delta (u,v){(abuv;q)}_{\infty }}{{(au,av,bu,bv;q)}_{\infty }}}{\int }_c^d{\displaystyle \frac{f(y){(qy/c,qy/d,wy;q)}_{\infty }}{{(uy,vy,ry;q)}_{\infty }}}{d}_{q}y.\hfill \end{array}$$

(1)

Noting the definition of $\Delta (u,v)$ in Definition 2 and by a direct computation, we deduce that:

$$(e^{i\theta }-e^{-i\theta })\Delta (e^{i\theta },e^{-i\theta })=(1-q){(q;q)}_{\infty }h(cos2\theta ;1).$$

Keeping this in mind and replacing $(u,v)$ by $(e^{i\theta },e^{-i\theta })$ in (1), we conclude that:

$$\begin{array}{c}(e^{i\theta }-e^{-i\theta }){\int }_{e^{i\theta }}^{e^{-i\theta }}{(qx{e}^{i\theta },qx{e}^{-i\theta };q)}_{\infty }I(x)d_{q}x\hfill \\ ={\displaystyle \frac{(1-q){(q,ab;q)}_{\infty }h(cos2\theta ;1)}{h(cos\theta ;a,b)}}{\int }_c^d{\displaystyle \frac{f(y){(qy/c,qy/d,wy;q)}_{\infty }}{h(cos\theta ;y){(ry;q)}_{\infty }}}{d}_{q}y,\hfill \end{array}$$

(2)

where

$$I(x):={\displaystyle \frac{1}{{(ax,bx;q)}_{\infty }}}{\int }_c^d{\displaystyle \frac{f(y){(qy/c,qy/d,abxy,wy;q)}_{\infty }}{{(ay,by,xy,ry;q)}_{\infty }}}{d}_{q}y.$$

It is easily seen that $I(x)$ is analytic near $x=0$. Thus, there exists a sequence ${\{a_k\}}_{k=0}^{\infty }$ independent of x such that:

$$I(x)=a_0+\sum _{k=1}^{\infty }{a}_k{x}^k.$$

By setting $x=0$ in the above equation, we immediately conclude that:

$$a_0={\int }_c^d{\displaystyle \frac{f(y){(qy/c,qy/d,wy;q)}_{\infty }}{{(ay,by,ry;q)}_{\infty }}}{d}_{q}y.$$

(3)

Using the definition of the q-integral, we find that the left-hand side of (2) equals:

$$\begin{array}{cc}& (1-q)(1-e^{-2i\theta })\sum _{n=0}^{\infty }{(q^{n+1};q)}_{\infty }{(q^{n+1}{e}^{-2i\theta };q)}_{\infty }I(q^n{e}^{-i\theta })q^n\hfill \\ & +(1-q)(1-e^{2i\theta })\sum _{n=0}^{\infty }{(q^{n+1};q)}_{\infty }{(q^{n+1}{e}^{2i\theta };q)}_{\infty }I(q^n{e}^{i\theta })q^n.\hfill \end{array}$$

Inspecting the first series in the above equation, we see that this series can be expanded in terms of the negative powers of ${\{e^{-ki\theta }\}}_{k=0}^{\infty }$, and the constant term of the Fourier expansion of this series is $(1-q)a_0$, since

$$\begin{array}{c}\hfill \sum _{n=0}^{\infty }{(q^{n+1};q)}_{\infty }{q}^n=\sum _{n=0}^{\infty }{\displaystyle \frac{{(q;q)}_{\infty }}{{(q;q)}_n}}{q}^n={\displaystyle \frac{{(q;q)}_{\infty }}{{(q;q)}_{\infty }}}=1\end{array}$$

by the binomial theorem (see, for example [5] (1.3.15)). Thus, there exists a sequence ${\{{\alpha }_k\}}_{k=1}^{\infty }$ independent of θ such that the first series equals

$$(1-q)a_0+\sum _{k=1}^{\infty }{\alpha }_k{e}^{-ik\theta }.$$

On replacing θ by $-\theta$, we immediately find that the second series is equal to

$$(1-q)a_0+\sum _{k=1}^{\infty }{\alpha }_k{e}^{ik\theta }.$$

It follows that:

$$\begin{array}{cc}& (e^{i\theta }-e^{-i\theta }){\int }_{e^{i\theta }}^{e^{-i\theta }}{(qx{e}^{i\theta },qx{e}^{-i\theta };q)}_{\infty }I(x)d_{q}x\hfill \\ & =2(1-q)a_0+2\sum _{k=1}^{\infty }{\alpha }_{k}cosk\theta .\hfill \end{array}$$

Comparing this equation with (2), we are led to the Fourier series expansion

$$\begin{array}{cc}& 2(1-q)a_0+2\sum _{k=1}^{\infty }{\alpha }_{k}cosk\theta \hfill \\ & ={\displaystyle \frac{(1-q){(q,ab;q)}_{\infty }h(cos2\theta ;1)}{h(cos\theta ;a,b)}}{\int }_c^d{\displaystyle \frac{f(y){(qy/c,qy/d,wy;q)}_{\infty }}{h(cos\theta ;y){(ry;q)}_{\infty }}}{d}_{q}y.\hfill \end{array}$$

If we integrate this equation with respect to θ over $[-\pi ,\pi ],$ using the well known fact

$${\int }_{-\pi }^{\pi }(cosk\theta )d\theta =2\pi {\delta }_{k,0},$$

and noting that the integrand is an even function of θ, we immediately deduce that

$${\displaystyle \frac{2\pi a_0}{{(q,ab;q)}_{\infty }}}={\int }_0^{\pi }{\displaystyle \frac{h(cos2\theta ;1)}{h(cos\theta ;a,b)}}{\int }_c^d\left({\displaystyle \frac{f(y){(qy/c,qy/d,wy;q)}_{\infty }}{h(cos\theta ;y){(ry;q)}_{\infty }}}{d}_{q}y\right)d\theta$$

Substituting (3) into the left-hand side of this equation, we complete the proof of Theorem 3. ☐

Theorem 3 can be used to give a very simple derivation of the Askey–Wilson integral [8].

Theorem 8. 

If max$\{|a|,|b|,|c|,|d|\}<1$, then we have:

$${\int }_0^{\pi }{\displaystyle \frac{h(cos2\theta ;1)}{h(cos\theta ;a,b,c,d)}}d\theta ={\displaystyle \frac{2\pi {(abcd;q)}_{\infty }}{{(q,ab,ac,ad,bc,bd,cd;q)}_{\infty }}}.$$

Proof. 

Choosing $f(y)={(ry;q)}_{\infty }/{(wy;q)}_{\infty }$ in Theorem 3, we conclude that

$$\begin{array}{cc}& {\int }_0^{\pi }{\displaystyle \frac{h(cos2\theta ;1)}{h(cos\theta ;a,b)}}\left({\int }_c^d{\displaystyle \frac{{(qy/c,qy/d;q)}_{\infty }}{h(cos\theta ;y)}}{d}_{q}y\right)d\theta \hfill \\ & ={\displaystyle \frac{2\pi }{{(q,ab;q)}_{\infty }}}{\int }_c^d{\displaystyle \frac{{(qy/c,qy/d;q)}_{\infty }}{{(ay,by;q)}_{\infty }}}{d}_{q}y.\hfill \end{array}$$

Appealing to the Andrews–Askey integral in Proposition 3, we arrive at

$${\int }_c^d{\displaystyle \frac{{(qy/c,qy/d;q)}_{\infty }}{h(cos\theta ;y)}}{d}_{q}y={\displaystyle \frac{\Delta (c,d){(cd;q)}_{\infty }}{h(cos\theta ;c,d)}},$$

$${\int }_c^d{\displaystyle \frac{{(qy/c,qy/d;q)}_{\infty }}{{(ay,by;q)}_{\infty }}}{d}_{q}y={\displaystyle \frac{\Delta (c,d){(abcd;q)}_{\infty }}{{(ac,ad,bc,bd;q)}_{\infty }}}.$$

Combining these three equations, we complete the proof of Theorem 8. ☐

For other proofs of this integral formula, see [9,10,11,12,13,14,15,16,17,18].

Theorem 9. 

For $max\{|a|,|b|,|c|,|d|\}<1$, we have the q-beta integral formula:

$$\begin{array}{cc}& {\int }_0^{\pi }{\displaystyle \frac{h(cos2\theta ;1)h(cos\theta ;w)}{h(cos\theta ;a,b,c,d)}}d\theta \hfill \\ & ={\displaystyle \frac{2\pi {(w/c,w/d;q)}_{\infty }}{\Delta (c,d){(q,ab,cd;q)}_{\infty }}}{\int }_c^d{\displaystyle \frac{{(qy/c,qy/d,wy;q)}_{\infty }}{{(ay,by,wy/cd;q)}_{\infty }}}{d}_{q}y.\hfill \end{array}$$

Proof. 

Choosing $f(y)={(ry;q)}_{\infty }/{(wy/cd;q)}_{\infty }$ in Theorem 3, we arrive at

$$\begin{array}{cc}& {\int }_0^{\pi }{\displaystyle \frac{h(cos2\theta ;1)}{h(cos\theta ;a,b)}}\left({\int }_c^d{\displaystyle \frac{{(qy/c,qy/d,wy;q)}_{\infty }}{h(cos\theta ;y){(wy/cd;q)}_{\infty }}}{d}_{q}y\right)d\theta \hfill \\ & ={\displaystyle \frac{2\pi }{{(q,ab;q)}_{\infty }}}{\int }_c^d{\displaystyle \frac{{(qy/c,qy/d,wy;q)}_{\infty }}{{(ay,by,wy/cd;q)}_{\infty }}}{d}_{q}y.\hfill \end{array}$$

On using the Al-Salam–Verma integral in Proposition 2, we conclude that

$${\int }_c^d{\displaystyle \frac{{(qy/c,qy/d,wy;q)}_{\infty }}{h(cos\theta ;y){(wy/cd;q)}_{\infty }}}{d}_{q}y={\displaystyle \frac{\Delta (c,d)h(cos\theta ;w){(cd;q)}_{\infty }}{h(cos\theta ;c,d){(w/c,w/d;q)}_{\infty }}}.$$

Combining the above two equations, we complete the proof of Theorem 9. ☐

4. The Rogers–Szego Polynomials and the Proof Theorem 4

4.1. The Rogers–Szego Polynomials and q-Hermite Polynomials

The Rogers–Szego polynomials play an important role in the theory of orthogonal polynomials, which are defined by (see, for example, [19] (Definition 1.2)):

$$h_n(a,b|q)=\sum _{k=0}^n{\displaystyle {\left[\genfrac{}{}{0pt}{}{n}{k}\right]}_q}{a}^k{b}^{n-k}.$$

By multiplying two copies of the q-binomial theorem (see, for example, [5] (p. 8, Equation (1.3.2))), we readily find that

$$\sum _{n=0}^{\infty }{h}_n(a,b|q){\displaystyle \frac{t^n}{{(q;q)}_n}}={\displaystyle \frac{1}{{(at,bt;q)}_{\infty }}},\left|at\right|<1,\left|bt\right|<1.$$

(4)

The continuous q-Hermite polynomials $H_n(cos\theta |q)$ is defined by:

$$H_n(cos\theta |q):=h_n(e^{-i\theta },e^{i\theta }|q)=\sum _{k=0}^n{\displaystyle {\left[\genfrac{}{}{0pt}{}{n}{k}\right]}_q}{e}^{i(n-2k)\theta }.$$

Using (4), one can easily find the following generating function for the q-Hermite polynomials [16] (Equation (5.3)):

$$\sum _{n=0}^{\infty }{H}_n(cos\theta |q){\displaystyle \frac{t^n}{{(q;q)}_n}}={\displaystyle \frac{1}{{(t{e}^{i\theta },t{e}^{-i\theta };q)}_{\infty }}}={\displaystyle \frac{1}{h(cos\theta ;t)}},\text{for}\left|t\right|<1.$$

(5)

Proposition 7. 

For $\left|t\right|<1$, the following series converges uniformly on $0\le \theta \le \pi$:

$$\sum _{n=0}^{\infty }{H}_n(cos\theta |q){\displaystyle \frac{t^n}{{(q;q)}_n}},$$

and we also have

$${\displaystyle \frac{1}{|{(t{e}^{i\theta };t{e}^{-i\theta };q)}_{\infty }|}}\le {\displaystyle \frac{1}{{(|t|;q)}_{\infty }^2}}.$$

Proof. 

It is easily seen that $|H_n(x|q)|\le H_n(1|q)$ for $-1\le x\le 1$, and the series in (5) converges at $x=1.$ Thus, the series in (5) converges for $\left|t\right|<1$ uniformly on $\theta \in [-\pi ,\pi ]$.

Using $|H_n(x|q)|\le H_n(1|q)$ for $-1\le x\le 1,$ we easily find that for any $\left|t\right|<1,$

$$\left|{\displaystyle \frac{1}{{(t{e}^{i\theta };t{e}^{-i\theta };q)}_{\infty }}}\right|\le \sum _{n=0}^{\infty }{H}_n(1|q){\displaystyle \frac{{\left|t\right|}^n}{{(q;q)}_n}}={\displaystyle \frac{1}{{(|t|;q)}_{\infty }^2}}.$$

 ☐

Proposition 8. 

If k is an non-negative integer, then, for $max\{|a|,|b|,|c|,|d|,|u|\}<1,$ $A_k(\theta )$ is bounded on $[-\pi ,\pi ],$ where $A_k(\theta )$ is given by:

$$A_k(\theta )={\displaystyle \frac{h(cos2\theta ;1)h(cos\theta ;w){(t{e}^{i\theta };t{e}^{-i\theta };q)}_k}{h(cos\theta ;a,b,c,d,u)}}.$$

Proof. 

On applying the inequality in Proposition 7, we immediately deduce that

$$\left|{\displaystyle \frac{1}{h(cos\theta ;a,b,c,d,u)}}\right|\le {\displaystyle \frac{1}{{(|a|,|b|,|c|,|d|,|u|;q)}_{\infty }^2}}.$$

Using the triangle inequality, we easily find that $\left|h(cos2\theta ;1)\right|\le 4{(-q;q)}_{\infty }^2$, and

$$\left|h(cos\theta ;w)\right|\le {(-|w|;q)}_{\infty }^2,|{(t{e}^{i\theta };t{e}^{-i\theta };q)}_k{|\le (-\left|t\right|;q)}_{\infty }^2.$$

(6)

It follows that $A_k(\theta )$ is bounded on $[-\pi ,\pi ].$ This completes the proof of the proposition. ☐

Proposition 9. 

If k is an non-negative integer, then, for $max\{|a|,|b|,|c|,|d|,|u|\}<1,$ ${\int }_0^{\pi }{A}_k(\theta )d\theta$ is an analytic function of $a,b,c,d,u$.

Proof. 

From Proposition 7, we know that, for $\left|a\right|<1$, the following series is convergent uniformly on $0\le \theta \le \pi$:

$$\sum _{n=0}^{\infty }{H}_n(cos\theta |q){\displaystyle \frac{a^n}{{(q;q)}_n}}={\displaystyle \frac{1}{{(a{e}^{i\theta },a{e}^{-i\theta };q)}_{\infty }}}.$$

Thus, for $\left|a\right|<1$, the integrand of the integral in Proposition 9 can be expanded into the following series, which is convergent uniformly on $0\le \theta \le \pi$:

$$\sum _{n=0}^{\infty }{\displaystyle \frac{a^n{H}_n(cos\theta |q)h(cos2\theta ;1)h(cos\theta ;w){(t{e}^{i\theta };t{e}^{-i\theta };q)}_k}{h(cos\theta ;b,c,d,u){(q;q)}_n}}.$$

On replacing the integrand of the integral in Proposition 9 by this series and then integrating term by term, one can see that the resulting series is of the form

$$\sum _{n=0}^{\infty }{\displaystyle \frac{a^n}{{(q;q)}_n}}{\int }_0^{\pi }{\displaystyle \frac{H_n(cos\theta |q)h(cos2\theta ;1)h(cos\theta ;w){(t{e}^{i\theta };t{e}^{-i\theta };q)}_k}{h(cos\theta ;b,c,d,u)}}d\theta .$$

(7)

Using the triangle inequality and inequalities in Proposition 7 and (6), we find that

$$\begin{array}{cc}& \left|\sum _{n=0}^{\infty }{\displaystyle \frac{a^n}{{(q;q)}_n}}{\int }_0^{\pi }{\displaystyle \frac{H_n(cos\theta |q)h(cos2\theta ;1)h(cos\theta ;w){(t{e}^{i\theta };t{e}^{-i\theta };q)}_k}{h(cos\theta ;b,c,d,u)}}d\theta \right|\hfill \\ & \le {\displaystyle \frac{4\pi {(-q,-|w|,-|t|;q)}_{\infty }^2}{{(|a|,|b|,|c|,|d|,|u|;q)}_{\infty }^2}}.\hfill \end{array}$$

It follows that the infinite series in (7) is uniformly and absolutely convergent. Thus, the integral in Proposition 9 is an analytic function of a for $\left|a\right|<1.$ By symmetry, we know that the integral is also analytic for $b,c,d,u.$ ☐

Proposition 10. 

For $max\{|a|,|b|,|c|,|d|,|u|,|abs/t|\}<1,$ the following integral is an analytic function of a and b:

$${\int }_0^{\pi }{\displaystyle \frac{h(cos2\theta ;1)h(cos\theta ;w)}{h(cos\theta ;a,b,c,d,u)}}{}_3{\varphi }_2\left(\genfrac{}{}{0pt}{}{t{e}^{i\theta },t{e}^{-i\theta },t/s}{at,bt};q,{\displaystyle \frac{abs}{t}}\right)d\theta .$$

Proof. 

Using the definition of $A_k(\theta )$ in Proposition 8, we find that the integrand of the integral in Proposition 10 can be written as:

$$\sum _{k=0}^{\infty }{\displaystyle \frac{{(t/s;q)}_k}{{(q,at,bt;q)}_k}}{\left({\displaystyle \frac{abs}{t}}\right)}^k{A}_k(\theta ).$$

(8)

By the ratio test, we easily find that the following series is absolutely convergent for $|abs/t|<1:$

$$\sum _{k=0}^{\infty }{\displaystyle \frac{{(t/s;q)}_k}{{(q,at,bt;q)}_k}}{\left({\displaystyle \frac{abs}{t}}\right)}^k.$$

Thus, the series in (8) is absolutely and uniformly convergent on $[-\pi ,\pi ].$ Substituting the series in (8) into the integral in Proposition 10 and then integrating term by term, we obtain

$$\sum _{k=0}^{\infty }{\displaystyle \frac{{(t/s;q)}_k}{{(q,at,bt;q)}_k}}{\left({\displaystyle \frac{abs}{t}}\right)}^k{\int }_0^{\pi }{A}_k(\theta )d\theta .$$

The above series is uniformly convergent and every term is an analytic function of a and b, so the series converges to an analytic function of a and b. This completes the proof of the proposition. ☐

Using the same argument used in Proposition 10, we can prove the following proposition.

Proposition 11. 

The following q-integral is an analytic function of a and b at $(0,0)\in {\mathbb{C}}^2$:

$${\int }_c^d{\displaystyle \frac{{(qy/c,qy/d,wy,absuy/t;q)}_{\infty }}{{(ay,by,uy,wy/cd;q)}_{\infty }}}{}_3{\varphi }_2\left(\genfrac{}{}{0pt}{}{t/s,t/u,t/y}{at,bt};q,{\displaystyle \frac{absuy}{t}}\right)d_{q}y.$$

4.2. q-Partial Differential Equations

For any function $f(x)$ of one variable, the q-derivative of $f(x)$ with respect to $x,$ is defined as:

$${\mathcal{D}}_{q,x}\{f(x)\}={\displaystyle \frac{f(x)-f(qx)}{x}},$$

and we further define ${\mathcal{D}}_{q,x}^0\{f\}=f,$ and for $n\ge 1$, ${\mathcal{D}}_{q,x}^n\{f\}={\mathcal{D}}_{q,x}\{{\mathcal{D}}_{q,x}^{n-1}\{f\}\}.$

Now, we give the definitions of the q-partial derivative and the q-partial differential equations.

Definition 5. 

A q-partial derivative of a function of several variables is its q-derivative with respect to one of those variables, regarding other variables as constants. The q-partial derivative of a function f with respect to the variable x is denoted by ${\partial }_{q,x}\{f\}$.

Definition 6. 

A q-partial differential equation is an equation that contains unknown multivariable functions and their q-partial derivatives.

It turns out that the q-partial differential equation methods are useful for deriving q-formulas [16,19,20,21]. The following useful expansion theorem for q-series can be found in [19] (Proposition 1.6).

Theorem 10. 

If $f(x,y)$ is a two-variable analytic function at $(0,0)\in {\mathbb{C}}^2,$ then f can be expanded in terms of $h_n(x,y|q)$ if and only if f satisfies the q-partial differential equation ${\partial }_{q,x}\{f\}={\partial }_{q,y}\{f\}.$

Proposition 12. 

If we use $L(a,b,u,v,s,t)$ to denote

$${\displaystyle \frac{{(av,bv,abstu/v;q)}_{\infty }}{{(as,at,au,bs,bt,bu;q)}_{\infty }}}{}_3{\varphi }_2\left(\genfrac{}{}{0pt}{}{v/s,v/t,v/u}{av,bv};q,{\displaystyle \frac{abstu}{v}}\right),$$

then $L(a,b,u,v,s,t)$ satisfies the q-partial differential equation ${\partial }_{q,a}\{L\}={\partial }_{q,b}\{L\}.$

Proof. 

If we set $v=cst$, then the q-integral formula in [19] (Proposition 13.8) becomes

$$\begin{array}{cc}\hfill {\int }_s^t{\displaystyle \frac{{(qx/s,qx/t,abux;q)}_{\infty }}{{(ax,bx,vx/st;q)}_{\infty }}}{d}_{q}x& ={\displaystyle \frac{\Delta (s,t){(av,bv,abstu/v;q)}_{\infty }}{{(as,bs,at,bt,v/s,v/t;q)}_{\infty }}}\hfill \\ & \times {}_3{\varphi }_2\left(\genfrac{}{}{0pt}{}{v/s,v/t,v/u}{av,bv};q,{\displaystyle \frac{abstu}{v}}\right).\hfill \end{array}$$

It follows that

$$L(a,b,u,v,s,t)={\displaystyle \frac{{(v/s,v/t;q)}_{\infty }}{\Delta (s,t){(au,bu;q)}_{\infty }}}{\int }_s^t{\displaystyle \frac{{(qx/s,qx/t,abux;q)}_{\infty }}{{(ax,bx,vx/st;q)}_{\infty }}}{d}_{q}x.$$

By a direct computation, we find that

$$\begin{array}{c}{\partial }_{q,a}\{L\}={\partial }_{q,b}\{L\}\hfill \\ ={\displaystyle \frac{{(v/s,v/t;q)}_{\infty }}{\Delta (s,t){(au,bu;q)}_{\infty }}}{\int }_s^t{\displaystyle \frac{(x+u-(a+b)ux){(qx/s,qx/t,abuxq;q)}_{\infty }}{{(ax,bx,vx/st;q)}_{\infty }}}{d}_{q}x.\hfill \end{array}$$

This completes the proof of Proposition 12. ☐

4.3. The Proof of Theorem 4

Proof. 

From Propositions 10 and 11, we know that both sides of the equation in Theorem 4 are all analytic functions of a and b at $(0,0)\in {\mathbb{C}}^2$.

Using the definition of $L(a,b,u,v,s,t)$ in Proposition 12 and a simple calculation, we find that

$$\begin{array}{cc}& L(a,b,s,t,e^{i\theta },e^{-i\theta })\hfill \\ & ={\displaystyle \frac{{(at,bt,abs/t;q)}_{\infty }}{{(as,bs;q)}_{\infty }h(cos\theta ;a,b)}}{}_3{\varphi }_2\left(\genfrac{}{}{0pt}{}{t{e}^{i\theta },t{e}^{-i\theta },t/s}{at,bt};q,{\displaystyle \frac{abs}{t}}\right)\hfill \end{array}$$

$$\begin{array}{cc}& L(a,b,u,t,s,y)\hfill \\ & ={\displaystyle \frac{{(at,bt,absyu/t;q)}_{\infty }}{{(as,ay,au,bs,by,bu;q)}_{\infty }}}{}_3{\varphi }_2\left(\genfrac{}{}{0pt}{}{t/s,t/y,t/u}{at,bt};q,{\displaystyle \frac{absuy}{t}}\right).\hfill \end{array}$$

Using these two equations, the q-integral formula in Theorem 4 can be rewritten as:

$$\begin{array}{cc}& {\int }_0^{\pi }{\displaystyle \frac{L(a,b,s,t,e^{i\theta },e^{-i\theta })h(cos2\theta ;1)h(cos\theta ;w)}{h(cos\theta ;c,d,u)}}d\theta \hfill \\ & ={\displaystyle \frac{2\pi {(w/c,w/d;q)}_{\infty }}{\Delta (c,d){(q,cd;q)}_{\infty }}}{\int }_c^{d}L(a,b,u,t,s,y){\displaystyle \frac{{(qy/c,qy/d,wy;q)}_{\infty }}{{(uy,wy/cd;q)}_{\infty }}}{d}_{q}y.\hfill \end{array}$$

(9)

If we use $f(a,b)$ to denote the left-hand side of (9), then it satisfies the partial q-differential equation ${\partial }_{q,a}\{f\}={\partial }_{q,b}\{f\}$ by Proposition 12. Thus, there exists a sequence $\{{\alpha }_n\}$ independent of a and b such that

$$f(a,b)=\sum _{n=0}^{\infty }{\alpha }_n{h}_n(a,b|q).$$

Setting $b=0$ in this equation, using the obvious fact $h_n(a,0|q)=a^n$ and the identity

$$L(a,0,s,t,e^{i\theta },e^{-i\theta })={\displaystyle \frac{{(at;q)}_{\infty }}{{(as;q)}_{\infty }h(cos\theta ;a)}}$$

in the resulting equation, we immediately conclude that

$$f(a,0)={\displaystyle \frac{{(at;q)}_{\infty }}{{(as;q)}_{\infty }}}{\int }_0^{\pi }{\displaystyle \frac{h(cos2\theta ;1)h(cos\theta ;w)}{h(cos\theta ;a,c,d,u)}}d\theta =\sum _{n=0}^{\infty }{\alpha }_n{a}^n.$$

(10)

If we use $g(a,b)$ to denote the right-hand side of (9), then it satisfies the partial q-differential equation ${\partial }_{q,a}\{g\}={\partial }_{q,b}\{g\}$ by Proposition 12. Thus, there exists a sequence $\{{\beta }_n\}$ independent of a and b such that

$$g(a,b)=\sum _{n=0}^{\infty }{\beta }_n{h}_n(a,b|q).$$

Putting $b=0$ in this equation, substituting $h_n(a,0|q)=a^n$ and the identity

$$L(a,0,u,t,s,y)={\displaystyle \frac{{(at;q)}_{\infty }}{{(as,ay,au;q)}_{\infty }}}$$

in the resulting equation, we are led to the expansion formula

$${\displaystyle \frac{2\pi {(w/c,w/d,at;q)}_{\infty }}{\Delta (c,d){(q,au,cd,as;q)}_{\infty }}}{\int }_c^d{\displaystyle \frac{{(qy/c,qy/d,wy;q)}_{\infty }}{{(ay,uy,wy/cd;q)}_{\infty }}}{d}_{q}y=\sum _{n=0}^{\infty }{\beta }_n{a}^n.$$

On replacing b by u in the equation in Theorem 9, we conclude that

$$\begin{array}{cc}& {\int }_0^{\pi }{\displaystyle \frac{h(cos2\theta ;1)h(cos\theta ;w)d\theta }{h(cos\theta ;a,u,c,d)}}\hfill \\ & ={\displaystyle \frac{2\pi {(w/c,w/d;q)}_{\infty }}{\Delta (c,d){(q,au,cd;q)}_{\infty }}}{\int }_c^d{\displaystyle \frac{{(qy/c,qy/d,wy;q)}_{\infty }}{{(ay,uy,wy/cd;q)}_{\infty }}}{d}_{q}y.\hfill \end{array}$$

Combining these two equations, we deduce that

$${\displaystyle \frac{{(at;q)}_{\infty }}{{(as;q)}_{\infty }}}{\int }_0^{\pi }{\displaystyle \frac{h(cos2\theta ;1)h(cos\theta ;w)d\theta }{h(cos\theta ;a,u,c,d)}}=\sum _{n=0}^{\infty }{\beta }_n{a}^n.$$

Combining this equation with (10), we obtain the power series identity

$$\sum _{n=0}^{\infty }{\alpha }_n{a}^n=\sum _{n=0}^{\infty }{\beta }_n{a}^n.$$

It follows that ${\alpha }_n={\beta }_n$ for $n=0,1,\dots$, which completes the proof of Theorem 4. ☐

5. Conclusions

Using the same method used in the proof of Theorem 4, we can also prove the following double q-integral formula.

Theorem 11. 

If we use $\Phi (x,y)$ to denote the following function of x and y:

$${\displaystyle \frac{{(abuvxy,rbuvxy,cduvwxy;q)}_{\infty }}{{(auvy,buvy,xy,wy,ruvy;q)}_{\infty }}}{}_3{\varphi }_2\left(\genfrac{}{}{0pt}{}{buvy,bx,xy}{abuvxy,rbuvxy};q,aruv\right),$$

then, we have the double q-integral formula

$$\begin{array}{cc}& {\int }_u^v{\displaystyle \frac{{(qx/u,qx/v;q)}_{\infty }}{{(ax,bx,rx;q)}_{\infty }}}{d}_{q}x{\int }_c^d{(qy/c,qy/d;q)}_{\infty }\Phi (x,y)d_{q}y\hfill \\ & ={\displaystyle \frac{\Delta (c,d)\Delta (u,v){(abuv,cduv,cduw,cdvw,rbuv;q)}_{\infty }}{{(au,av,bu,bv,cu,cv,du,dv,ru,rv,cw,dw;q)}_{\infty }}}.\hfill \end{array}$$

Next, we will use Proposition 12 and Theorem 10 to give a proof of the theorem.

Proof. 

Using the definition of $L(a,b,u,v,s,t)$ in Proposition 12, we easily find that

$$L(a,r,x,buvxy,uvy,buv)$$

(11)

$$={\displaystyle \frac{{(abuvxy,rbuvxy,aruv;q)}_{\infty }}{{(ax,rx,abuv,rbuv,auvy,ruvy;q)}_{\infty }}}{}_3{\varphi }_2\left(\genfrac{}{}{0pt}{}{buvy,bx,xy}{abuvxy,rbuvxy};q,aruv\right).$$

For the sake of brevity, we temporarily use $I(a,r,x,y)$ to denote the expression

$${\displaystyle \frac{{(cduvwxy;q)}_{\infty }}{{(buvy,xy,wy;q)}_{\infty }}}L(a,r,x,buvxy,uvy,buv).$$

Using this expression and (11), the q-integral formula in Theorem 11 can be rewritten as:

$$\begin{array}{cc}& {\int }_u^v{\displaystyle \frac{{(qx/u,qx/v;q)}_{\infty }}{{(bx;q)}_{\infty }}}{d}_{q}x{\int }_c^d{(qy/c,qy/d;q)}_{\infty }I(a,r,x,y)d_{q}y\hfill \\ & ={\displaystyle \frac{\Delta (c,d)\Delta (u,v){(cduv,cduw,cdvw,aruv;q)}_{\infty }}{{(au,av,bu,bv,cu,cv,du,dv,ru,rv,cw,dw;q)}_{\infty }}}.\hfill \end{array}$$

(12)

If we use $f(a,r)$ to denote the right-hand of the above equation, then, using the same method as that used in the proof of Proposition 10, we can prove that $f(a,r)$ is analytic at $(0,0)\in {\mathbb{C}}^2$. By Proposition 12, one can show that $f(a,r)$ satisfies the partial q-differential equation ${\partial }_{q,a}\{f\}={\partial }_{q,r}\{f\}$. Hence, by Theorem 10, there exists a sequence $\{{\lambda }_n\}$ independent of a and r such that

$$f(a,r)=\sum _{n=0}^{\infty }{\lambda }_n{h}_n(a,r|q).$$

Setting $r=0$ in this equation, substituting the equations $h_n(a,0|q)=a^n$ and

$$L(a,0,x,buvxy,uvy,buv)={\displaystyle \frac{{(abuvxy;q)}_{\infty }}{{(ax,auvy,abuv;q)}_{\infty }}}$$

in the resulting equation, we conclude that

$$\begin{array}{cc}& {\int }_u^v{\displaystyle \frac{{(qx/u,qx/v;q)}_{\infty }}{{(ax,bx;q)}_{\infty }}}{d}_{q}x{\int }_c^d{\displaystyle \frac{{(qy/c,qy/d,abuvxy,cduvwxy;q)}_{\infty }}{{(auvy,buvy,xy,wy;q)}_{\infty }}}{d}_{q}y\hfill \\ & ={(abuv;q)}_{\infty }\sum _{n=0}^{\infty }{\lambda }_n{a}^n.\hfill \end{array}$$

Using Proposition 6, we know that the left-hand side of the above equation equals

$${\displaystyle \frac{\Delta (u,v)\Delta (c,d){(abuv,cduv,cduw,cdvw;q)}_{\infty }}{{(au,av,bu,bv,cu,cv,du,dv,cw,dw;q)}_{\infty }}}.$$

It follows that

$$\sum _{n=0}^{\infty }{\lambda }_n{a}^n={\displaystyle \frac{\Delta (u,v)\Delta (c,d){(cduv,cduw,cdvw;q)}_{\infty }}{{(au,av,bu,bv,cu,cv,du,dv,cw,dw;q)}_{\infty }}}.$$

(13)

If we use $g(a,r)$ to denote the right-hand side of (12), then, by a direct computation, we find that

$${\partial }_{q,a}\{g(a,r)\}={\partial }_{q,r}\{g(a,r)\}={\displaystyle \frac{u+v-(a+r)uv}{1-aruv}}g(a,r).$$

Thus, by Theorem 10, there exists a sequence $\{{\delta }_n\}$ independent of a and r such that

$$g(a,r)=\sum _{n=0}^{\infty }{\delta }_n{h}_n(a,r|q).$$

Setting $r=0$ in this equation and using the fact $h_n(a,0|q)=a^n,$ we arrive at

$$g(a,0)=\sum _{n=0}^{\infty }{\delta }_n{a}^n={\displaystyle \frac{\Delta (u,v)\Delta (c,d){(cduv,cduw,cdvw;q)}_{\infty }}{{(au,av,bu,bv,cu,cv,du,dv,cw,dw;q)}_{\infty }}}.$$

(14)

Comparing Equations (13) and (14), we have the power series identity

$$\sum _{n=0}^{\infty }{\lambda }_n{a}^n=\sum _{n=0}^{\infty }{\delta }_n{a}^n.$$

Hence, we find for $n=0,1,\dots ,$ that ${\lambda }_n={\delta }_n.$ This completes the proof of Theorem 11. ☐

Remark 1. 

Proceedings through the same steps used to derive Theorem 3 from Theorem 1, setting $u=e^{i\theta }$ and $v=e^{-i\theta }$ in the equation in Theorem 11, and then integrating both sides of the resulting equation with respect to θ over $[-\pi ,\pi ],$ we can give a new proof of Theorem 5.