A Further Extension for Ramanujan’s Beta Integral and Applications

Abstract

In 1915, Ramanujan stated the following formula ${\int }_0^{\infty }{t}^{x-1}\frac{{(-at;q)}_{\infty }}{{(-t;q)}_{\infty }}dt=\frac{\pi }{sin\pi x}\frac{{(q^{1-x},a;q)}_{\infty }}{{(q,a{q}^{-x};q)}_{\infty }},$ where $0<q<1$, $x>0$, and $0<a<q^x$. The above formula is called Ramanujan’s beta integral. In this paper, by using q-exponential operator, we further extend Ramanujan’s beta integral. As some applications, we obtain some new integral formulas of Ramanujan and also show some new representation with gamma functions and q-gamma functions.

1. Introduction, Preliminaries and Main Results

The gamma function is the most natural extension of the factorial

$$n!=1·2\cdots n.$$

Euler’s original definition is

$$\Gamma (x+1)=\prod _{k=1}^{\infty }\frac{k}{k+x}{\left(\frac{k+1}{k}\right)}^x.$$

(1)

The integral representation of the gamma function is the following form

$$\Gamma \left(x\right)={\int }_0^{\infty }{t}^{x-1}{e}^{-t}dt,\Re \left(x\right)>0.$$

(2)

The q-shifted factorials are defined by

$$\begin{array}{c}{(a;q)}_0=1,{(a;q)}_n=\prod _{k=0}^{n-1}(1-a{q}^k),\hfill \end{array}$$

(3)

$$\begin{array}{c}{(a;q)}_{\infty }=\underset{n\to \infty }{lim}\prod _{k=0}^{n-1}(1-a{q}^k)=\prod _{k=0}^{\infty }(1-a{q}^k),n\ge 1.\hfill \end{array}$$

(4)

Clearly,

$$\begin{array}{c}{(a;q)}_n={\displaystyle \frac{{(a;q)}_{\infty }}{{(a{q}^n;q)}_{\infty }}}.\end{array}$$

(5)

Analogously with $\Gamma \left(x\right)$, F. H. Jackson [1] defined ${\Gamma }_q\left(x\right)$ by

$${\Gamma }_q\left(x\right)=\frac{{(q;q)}_{\infty }}{{(q^x;q)}_{\infty }}{(1-q)}^{1-x},0<q<1.$$

(6)

${\Gamma }_q\left(x\right)$ is called the q-gamma function.

The functional equation for $\Gamma \left(x\right)$,

$$\Gamma (x+1)=x\Gamma \left(x\right),$$

(7)

becomes

$${\Gamma }_q(x+1)=\frac{1-q^x}{1-q}{\Gamma }_q\left(x\right)$$

(8)

for the q-gamma function. In the future, we will always take $0<q<1$.

We also adopt the following compact notations for the multiple q-shifted factorials:

$$\begin{array}{c}{(a_1,a_2,\dots ,a_m;q)}_n={(a_1;q)}_n{(a_2;q)}_n\cdots {(a_m;q)}_n,\hfill \\ {(a_1,a_2,\dots ,a_m;q)}_{\infty }={(a_1;q)}_{\infty }{(a_2;q)}_{\infty }\cdots {(a_m;q)}_{\infty }.\hfill \end{array}$$

The basic hypergeometric series, or q-series ${}_r{\varphi }_s$ is usually defined by

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

(9)

with $\left(\genfrac{}{}{0pt}{}{n}{2}\right)=n(n-1)/2$, where $q\ne 0$, when $r>s+1$. Clearly, we have

$${}_{r+1}{\varphi }_r\left(\genfrac{}{}{0pt}{}{a_1,a_2,\dots ,a_{r+1}}{b_1,b_2,\dots ,,b_r};q,z\right)=\sum _{n=0}^{\infty }\frac{{(a_1,a_2,\dots ,a_{r+1};q)}_n}{{(q,b_1,b_2,\dots ,b_r;q)}_n}{z}^n.$$

(10)

The usual q-differential operator, or q-derivative operator $D_q$ is defined by (see ([2], p. 177, (2.1)) or [1,3,4,5]).

$$\begin{array}{c}{D}_q\left\{f\left(a\right)\right\}=\frac{f\left(a\right)-f\left(aq\right)}{a},\hfill \end{array}$$

(11)

$$\begin{array}{c}{D}_q^n\left\{f\left(a\right)\right\}=D_q\left\{D_q^{(n-1)}\left\{f\left(a\right)\right\}\right\}.\hfill \end{array}$$

(12)

The q-shift operator $\eta$ is (see ([6], p. 112)):

$$\begin{array}{c}\eta \left\{f\right(a\left)\right\}=f\left(aq\right),\hfill \end{array}$$

(13)

$$\begin{array}{c}{\eta }^{-1}\left\{f\left(a\right)\right\}=f\left(a{q}^{-1}\right).\hfill \end{array}$$

(14)

The operator $\theta$ is (see [7]):

$$\theta ={\eta }^{-1}{D}_q.$$

(15)

The q-exponential operator $E\left(b\theta \right)$ is defined by (see ([6], p. 112))

$$\begin{array}{c}E\left(b\theta \right)=\sum _{n=0}^{\infty }\frac{{\left(b\theta \right)}^n{q}^{\left(\genfrac{}{}{0pt}{}{n}{2}\right)}}{{(q;q)}_n}.\end{array}$$

(16)

Recently, Fang further generalized the q-exponential operator $E\left(b\theta \right)$ in the following form (see [8], or ([9], p. 1394, Equation (5))):

$${}_1{\varphi }_0\left(\genfrac{}{}{0pt}{}{b}{-};q,-c\theta \right)=\sum _{n=0}^{\infty }\frac{{(b;q)}_n{(-c\theta )}^n}{{(q;q)}_n}$$

(17)

and obtained two q-operator identities as follows:

$$\begin{array}{c}{}_1{\varphi }_0\left(\genfrac{}{}{0pt}{}{b}{-};q,-c\theta \right)\left\{{(as;q)}_{\infty }\right\}=\frac{{(as,bcs;q)}_{\infty }}{{(cs;q)}_{\infty }},\hfill \end{array}$$

(18)

$$\begin{array}{c}{}_1{\varphi }_0\left(\genfrac{}{}{0pt}{}{b}{-};q,-c\theta \right)\left\{\frac{{(as;q)}_{\infty }}{{(a\omega ;q)}_{\infty }}\right\}=\frac{{(as;q)}_{\infty }}{{(a\omega ;q)}_{\infty }}{}_2{\varphi }_1\left(\genfrac{}{}{0pt}{}{b,s/\omega }{q/a\omega };q,qc/a\right).\hfill \end{array}$$

(19)

In 1915, Ramanujan stated the following formula in [10,11]:

$$\begin{array}{c}{\int }_0^{\infty }{t}^{x-1}\frac{{(-at;q)}_{\infty }}{{(-t;q)}_{\infty }}dt=\frac{\pi }{sin\pi x}\frac{{(q^{1-x},a;q)}_{\infty }}{{(q,a{q}^{-x};q)}_{\infty }},\end{array}$$

(20)

where $0<q<1$, $x>0$, and $0<a<q^x$. The right-hand side must be interpreted using a limit when x is an integer. The above formula is called Ramanujan’s beta integral.

Hardy gave the first proof of (20) in [12]. He closed this paper with the evaluation of “another curious integral”, which is another important integral. Hardy gave a nice treatment of Ramanujan’s method of evaluating integrals of this type in his book on Ramanujan [13]. Rahman and Suslov gave a simple proof of (20) in ([14], pp. 109–110) by Ramanujan’s sum formula ${}_1{\psi }_1$. Askey ([15], p. 349) gave an elementary proof of (20) and obtained the following formula when $a=q^{x+y}$ in (20):

$$\begin{array}{c}{\int }_0^{\infty }{t}^{x-1}\frac{{(-t{q}^{x+y};q)}_{\infty }}{{(-t;q)}_{\infty }}dt=\frac{{\Gamma }_q\left(y\right)\Gamma \left(x\right)\Gamma (1-x)}{{\Gamma }_q(x+y){\Gamma }_q(1-x)}\end{array}$$

(21)

in terms of the q-gamma function and the ordinary gamma function. When $q\to 1$, this reduces to

$$\begin{array}{c}{\int }_0^{\infty }\frac{t^{x-1}}{{(1+t)}^{x+y}}dt=\frac{\Gamma \left(x\right)\Gamma \left(y\right)}{\Gamma (x+y)}=B(x,y),\end{array}$$

(22)

where $B(x,y)$ denotes the beta function defined by

$$B(x,y)={\int }_0^1{t}^{x-1}{(1-t)}^{y-1}dt.$$

(23)

Recently, Chen and Liu ([6], p. 123. Equation (7.3)) gave an extension of (20) by the method of the operator as follows:

$$\begin{array}{c}{\int }_0^{\infty }{t}^{x-1}\frac{{(-at,-bt;q)}_{\infty }}{{(-t,-ab{q}^{-x-1}t;q)}_{\infty }}dt=\frac{\pi }{sin\pi x}\frac{{(q^{1-x},a,b;q)}_{\infty }}{{(q,a{q}^{-x},b{q}^{-x};q)}_{\infty }}.\end{array}$$

(24)

The aim of the present paper is to further generalize Ramanujan’s beta integral by the operator ${}_1{\varphi }_0\left(\genfrac{}{}{0pt}{}{b}{-};q,-c\theta \right)$ and to give some new formulas of Ramanujan’s beta integral. We also show the connections with gamma functions and q-gamma functions.

We now state our result as follows:

Theorem 1.

If $0<q<1$, $x>0$, $0<a,a_1,a_2,\dots ,a_{2r+2}<q^x$, $|q{a}_2/a|<1$ and $|q{a}_{2j+2}/a_{2j-1}|<1(j=1,2,\dots ,r)$; $a_1,a_2\ne a,a_{l+2},a_{l+3}\ne a_l(l=1,3,5,\dots ,2r-1)$, then we have

$$\begin{array}{ll}{\int }_0^{\infty }{t}^{x-1}& \frac{{(-at,-a_{1}t,-a_{3}t,\dots ,-a_{2r+1}t;q)}_{\infty }}{{(-t,-a_{2}t,-a_{4}t,\dots ,-a_{2r+2}t;q)}_{\infty }}dt\hfill \\ & =\frac{\pi }{sin\pi x}\frac{{(q^{1-x},a;q)}_{\infty }}{{(q,a{q}^{-x};q)}_{\infty }}\sum _{k=0}^{\infty }\frac{{(q^x,a_1/a_2;q)}_k{(q{a}_2/a)}^k}{{(q,q^{1+x}/a;q)}_k}\hfill \\ & \times \sum _{\begin{array}{c}{k}_1+k_2+\cdots +k_r=k\\ 0\le k_r\le k_{r-1}\le \cdots \le k_2\le k_1\le k\end{array}}\prod _{j=1}^r\frac{{(q^{-k_{j-1}},a_{2j+1}/a_{2j+2};q)}_{k_j}{(q{a}_{2j+2}/a_{2j-1})}^{k_j}}{{(q,q^{1-k_{j-1}}{a}_{2j}/a_{2j-1};q)}_{k_j}}.\hfill \end{array}$$

(25)

2. Proof of Theorem 1

Proof of Theorem 1.

Firstly, we write Ramanujan’s formula as follows:

$$\begin{array}{c}{\int }_0^{\infty }{t}^{x-1}\frac{{(-at;q)}_{\infty }}{{(-t;q)}_{\infty }}dt=\frac{\pi }{sin\pi x}\frac{{(q^{1-x},a;q)}_{\infty }}{{(q,a{q}^{-x};q)}_{\infty }}.\end{array}$$

(26)

Applying the operator

$${}_1{\varphi }_0\left(\genfrac{}{}{0pt}{}{a_1}{-};q,-a_2\theta \right)$$

on both sides of (26) with respect to variable a, we obtain

$$\begin{array}{l}{\int }_0^{\infty }{t}^{x-1}\frac{1}{{(-t;q)}_{\infty }}{}_1{\varphi }_0\left(\genfrac{}{}{0pt}{}{a_1}{-};q,-a_2\theta \right)\left\{{(-at;q)}_{\infty }\right\}dt\\ =\frac{\pi }{sin\pi x}\frac{{(q^{1-x};q)}_{\infty }}{{(q;q)}_{\infty }}{}_1{\varphi }_0\left(\genfrac{}{}{0pt}{}{a_1}{-};q,-a_2\theta \right)\left\{\frac{{(a;q)}_{\infty }}{{(a{q}^{-x};q)}_{\infty }}\right\}.\end{array}$$

(27)

By (18) and (19), we get that

$$\begin{array}{c}{\int }_0^{\infty }{t}^{x-1}\frac{{(-at,-a_1{a}_{2}t;q)}_{\infty }}{{(-t,-a_{2}t;q)}_{\infty }}dt=\frac{\pi }{sin\pi x}\frac{{(q^{1-x},a;q)}_{\infty }}{{(q,a{q}^{-x};q)}_{\infty }}{}_2{\varphi }_1\left(\genfrac{}{}{0pt}{}{a_1,q^x}{q^{1+x}/a};q,q{a}_2/a\right).\end{array}$$

(28)

By (5), we rewrite the formula (28) in the following form:

$$\begin{array}{c}{\int }_0^{\infty }{t}^{x-1}\frac{{(-at;q)}_{\infty }}{{(-t,-a_{2}t;q)}_{\infty }}{(-a_1{a}_{2}t;q)}_{\infty }dt=\frac{\pi }{sin\pi x}\frac{{(q^{1-x},a;q)}_{\infty }}{{(q,a{q}^{-x};q)}_{\infty }}\sum _{k=0}^{\infty }\frac{{(q^x;q)}_k{(q{a}_2/a)}^k}{{(q,q^{1+x}/a;q)}_k}\frac{{(a_1;q)}_{\infty }}{{(a_1{q}^k;q)}_{\infty }}.\end{array}$$

(29)

Next, by applying the operator

$${}_1{\varphi }_0\left(\genfrac{}{}{0pt}{}{a_3}{-};q,-a_4\theta \right)$$

on both sides of the Equation (29) with respect to variable $a_1$, we arrive at

$$\begin{array}{ll}{\int }_0^{\infty }{t}^{x-1}& \frac{{(-at,-a_1{a}_{2}t,-a_2{a}_3{a}_{4}t;q)}_{\infty }}{{(-t,-a_{2}t,-a_2{a}_{4}t;q)}_{\infty }}dt\hfill \\ & =\frac{\pi }{sin\pi x}\frac{{(q^{1-x},a;q)}_{\infty }}{{(q,a{q}^{-x};q)}_{\infty }}\sum _{k=0}^{\infty }\frac{{(q^x,a_1;q)}_k{(q{a}_2/a)}^k}{{(q,q^{1+x}/a;q)}_k}\sum _{k_1=0}^k\frac{{(q^{-k},a_3;q)}_{k_1}{(q{a}_4/a_1)}^{k_1}}{{(q,q^{1-k}/a_1;q)}_{k_1}}.\hfill \end{array}$$

(30)

We rewrite (30) in the following form:

$$\begin{array}{ll}{\int }_0^{\infty }{t}^{x-1}& \frac{{(-at,-a_1{a}_{2}t;q)}_{\infty }}{{(-t,-a_{2}t,-a_2{a}_{4}t;q)}_{\infty }}{(-a_2{a}_3{a}_{4}t;q)}_{\infty }dt\hfill \\ & =\frac{\pi }{sin\pi x}\frac{{(q^{1-x},a;q)}_{\infty }}{{(q,a{q}^{-x};q)}_{\infty }}\sum _{k=0}^{\infty }\frac{{(q^x,a_1;q)}_k{(q{a}_2/a)}^k}{{(q,q^{1+x}/a;q)}_k}\sum _{k_1=0}^k\frac{{(q^{-k};q)}_{k_1}{(q{a}_4/a_1)}^{k_1}}{{(q,q^{1-k}/a_1;q)}_{k_1}}\frac{{(a_3;q)}_{\infty }}{{(a_3{q}^{k_1};q)}_{\infty }}.\hfill \end{array}$$

(31)

Applying the operator

$${}_1{\varphi }_0\left(\genfrac{}{}{0pt}{}{a_5}{-};q,-a_6\theta \right)$$

on both sides of the Equation (31) with respect to variable $a_3$, we have

$$\begin{array}{ll}\hfill {\int }_0^{\infty }{t}^{x-1}& \frac{{(-at,-a_1{a}_{2}t,-a_2{a}_3{a}_{4}t,-a_2{a}_4{a}_5{a}_{6}t;q)}_{\infty }}{{(-t,-a_{2}t,-a_2{a}_{4}t,-a_2{a}_4{a}_{6}t;q)}_{\infty }}dt\hfill \\ \hfill =& \frac{\pi }{sin\pi x}\frac{{(q^{1-x},a;q)}_{\infty }}{{(q,a{q}^{-x};q)}_{\infty }}\hfill \\ & \times \sum _{k=0}^{\infty }\frac{{(q^x,a_1;q)}_k{(q{a}_2/a)}^k}{{(q,q^{1+x}/a;q)}_k}\sum _{k_1=0}^k\frac{{(q^{-k},a_3;q)}_{k_1}{(q{a}_4/a_1)}^{k_1}}{{(q,q^{1-k}/a_1;q)}_{k_1}}\sum _{k_2=0}^{k_1}\frac{{(q^{-k_1},a_5;q)}_{k_2}{(q{a}_6/a_3)}^{k_2}}{{(q,q^{1-k_1}/a_3;q)}_{k_2}}.\hfill \end{array}$$

(32)

By the mathematical induction, iterating $r+1$ times, and applying the operator

$${}_1{\varphi }_0\left(\genfrac{}{}{0pt}{}{a_{2r+1}}{-};q,-a_{2r+2}\theta \right)$$

and noting that (18) and (19), we obtain

$$\begin{array}{ll}\hfill {\int }_0^{\infty }& t^{x-1}\frac{{(-at,-a_1{a}_{2}t,-a_2{a}_3{a}_{4}t,-a_2{a}_4{a}_5{a}_{6}t,\dots ,-a_2{a}_4{a}_6{a}_8\cdots a_{2r-2}{a}_{2r}{a}_{2r+1}{a}_{2r+2}t;q)}_{\infty }}{{(-t,-a_{2}t,-a_2{a}_{4}t,-a_2{a}_4{a}_{6}t,\dots ,-a_2{a}_4{a}_6{a}_8\cdots a_{2r-2}{a}_{2r}{a}_{2r+2}t;q)}_{\infty }}dt\hfill \\ \hfill =& \frac{\pi }{sin\pi x}\frac{{(q^{1-x},a;q)}_{\infty }}{{(q,a{q}^{-x};q)}_{\infty }}\sum _{k=0}^{\infty }\frac{{(q^x,a_1;q)}_k{(q{a}_2/a)}^k}{{(q,q^{1+x}/a;q)}_k}\sum _{k_1=0}^k\frac{{(q^{-k},a_3;q)}_{k_1}{(q{a}_4/a_1)}^{k_1}}{{(q,q^{1-k}/a_1;q)}_{k_1}}\hfill \\ & \times \sum _{k_2=0}^{k_1}\frac{{(q^{-k_1},a_5;q)}_{k_2}{(q{a}_6/a_3)}^{k_2}}{{(q,q^{1-k_1}/a_3;q)}_{k_2}}\cdots \sum _{k_r=0}^{k_{r-1}}\frac{{(q^{-k_{r-1}},a_{2r+1};q)}_{k_r}{(q{a}_{2r+2}/a_{2r-1})}^{k_r}}{{(q,q^{1-k_{r-1}}/a_{2r-1};q)}_{k_r}}.\hfill \end{array}$$

(33)

Letting $a_{2j-1}⟼a_{2j-1}/a_{2j}$ and $a_{2j+2}⟼a_{2j+2}/a_{2j}(j=1,2,\cdots r$) in (33), we show that

$$\begin{array}{ll}{\int }_0^{\infty }{t}^{x-1}& \frac{{(-at,-a_{1}t,-a_{3}t,\dots ,-a_{2r+1}t;q)}_{\infty }}{{(-t,-a_{2}t,-a_{4}t,\dots ,-a_{2r+2}t;q)}_{\infty }}dt\hfill \\ & =\frac{\pi }{sin\pi x}\frac{{(q^{1-x},a;q)}_{\infty }}{{(q,a{q}^{-x};q)}_{\infty }}\sum _{k=0}^{\infty }\frac{{(q^x,a_1/a_2;q)}_k{(q{a}_2/a)}^k}{{(q,q^{1+x}/a;q)}_k}\hfill \\ & \times \sum _{\begin{array}{c}{k}_1+k_2+\cdots +k_r=k\\ 0\le k_r\le k_{r-1}\le \cdots \le k_2\le k_1\le k\end{array}}\prod _{j=1}^r\frac{{(q^{-k_{j-1}},a_{2j+1}/a_{2j+2};q)}_{k_j}{(q{a}_{2j+2}/a_{2j-1})}^{k_j}}{{(q,q^{1-k_{j-1}}{a}_{2j}/a_{2j-1};q)}_{k_j}}.\hfill \end{array}$$

(34)

The proof of Theorem 1 is complete.

3. Some Applications

In this section, we will obtain the corresponding new integral formulas from (25).

Taking $r=0$ in (25) and defining the empty sum equal to 1, we obtain the following integral formula:

Corollary 1.

For $0<q<1$, $x>0$, $0<a,a_1,a_2<q^x$ and $|q{a}_2/a|<1$; $a_1,a_2\ne a$, we have

$$\begin{array}{c}{\int }_0^{\infty }{t}^{x-1}\frac{{(-at,-a_{1}t;q)}_{\infty }}{{(-t,-a_{2}t;q)}_{\infty }}dt=\frac{\pi }{sin\pi x}\frac{{(q^{1-x},a;q)}_{\infty }}{{(q,a{q}^{-x};q)}_{\infty }}{}_2{\varphi }_1\left(\genfrac{}{}{0pt}{}{q^x,a_1/a_2}{q^{1+x}/a};q,q{a}_2/a\right).\hfill \end{array}$$

(35)

Remark 1.

If setting $a_2=a{a}_1{q}^{-x-1}$ in (35), we get

$$\begin{array}{c}{\int }_0^{\infty }{t}^{x-1}\frac{{(-at,-a_{1}t;q)}_{\infty }}{{(-t,-a{a}_1{q}^{-x-1}t;q)}_{\infty }}dt=\frac{\pi }{sin\pi x}\frac{{(q^{1-x},a;q)}_{\infty }}{{(q,a{q}^{-x};q)}_{\infty }}{}_1{\varphi }_0\left(\genfrac{}{}{0pt}{}{q^x}{-};q,a_1{q}^{-x}\right).\end{array}$$

(36)

Applying the q-binomial theorem (see ([16], p. 8. (1.3.2)))

$$\begin{array}{c}{}_1{\varphi }_0\left(\genfrac{}{}{0pt}{}{a}{-};q,z\right)=\frac{{(az;q)}_{\infty }}{{(z;q)}_{\infty }}\end{array}$$

(37)

in (36), we obtain

$$\begin{array}{c}{\int }_0^{\infty }{t}^{x-1}\frac{{(-at,-a_{1}t;q)}_{\infty }}{{(-t,-a{a}_1{q}^{-x-1}t;q)}_{\infty }}dt=\frac{\pi }{sin\pi x}\frac{{(q^{1-x},a,a_1;q)}_{\infty }}{{(q,a{q}^{-x},a_1{q}^{-x};q)}_{\infty }}.\end{array}$$

(38)

Setting $a_1=b$ in (38), we obtain (24) immediately. Hence, we say that the formula (35) is an extension of result (24) of Chen and Liu.

Corollary 2.

For $0<q<1$, $x>0$, $0<a,b<q^x$; $a\ne b$, we have

$$\begin{array}{c}{\int }_0^{\infty }{t}^{x-1}\frac{{(-at;q)}_{\infty }}{{(-bt;q)}_{\infty }}dt=\frac{\pi }{sin\pi x}\frac{{(q^{1-x},a,q/a,b{q}^{1+x}/a;q)}_{\infty }}{{(q,a{q}^{-x},qb/a,q^{1+x}/a;q)}_{\infty }}.\end{array}$$

(39)

Proof. 

Setting $a_1\to 1,a_2\to b$ in (35) and applying q-Gauss sum formula ([16], p. 14, Equation (1.5.1)):

$$\begin{array}{c}{}_2{\varphi }_1\left(\genfrac{}{}{0pt}{}{a,b}{c};q,c/ab\right)=\frac{{(c/a,c/b;q)}_{\infty }}{{(c,c/ab;q)}_{\infty }},|c/ab|<1,\end{array}$$

we obtain (39).

Remark 2.

If letting $b\to 1$ in (39), we obtain (20) immediately. Hence, the formula (39) is also an extension of (20):

Taking $r=1$ in (25), we have

Corollary 3.

For $0<q<1$, $x>0$, $0<a,a_1,a_2,a_3,a_4<q^x$, $|q{a}_2/a|<1$ and $|q{a}_4/a_1|<1$; $a_1,a_2\ne a,a_3,a_4\ne a_1$, we have

$$\begin{array}{ll}{\int }_0^{\infty }{t}^{x-1}& \frac{{(-at,-a_{1}t,-a_{3}t;q)}_{\infty }}{{(-t,-a_{2}t,-a_{4}t;q)}_{\infty }}dt\hfill \\ & =\frac{\pi }{sin\pi x}\frac{{(q^{1-x},a;q)}_{\infty }}{{(q,a{q}^{-x};q)}_{\infty }}\sum _{k=0}^{\infty }\frac{{(q^x,a_1/a_2;q)}_k{(q{a}_2/a)}^k}{{(q,q^{1+x}/a;q)}_k}{}_2{\varphi }_1\left(\genfrac{}{}{0pt}{}{q^{-k},a_3/a_4}{a_2{q}^{1-k}/a_1};q,q{a}_4/a_1\right).\hfill \end{array}$$

(40)

Theorem 2.

If $0<q<1$, $0<a,a_1,a_2,\dots ,a_{2r+2}<q^n(n=1,2,\cdots )$, $|q{a}_2/a|<1$ and $|q{a}_{2j+2}/a_{2j-1}|<1$, $(j=1,2,\dots ,r)$; $a_1,a_2\ne a,a_{l+2},a_{l+3}\ne a_l(l=1,3,5,\dots ,2r-1)$, then we have

$$\begin{array}{ll}{\int }_0^{\infty }{t}^{n-1}& \frac{{(-at,-a_{1}t,-a_{3}t,\dots ,-a_{2r+1}t;q)}_{\infty }}{{(-t,-a_{2}t,-a_{4}t,\dots ,-a_{2r+2}t;q)}_{\infty }}dt\hfill \\ & =\frac{{(-1)}^{n-1}{(q;q)}_n{q}^{n}logq}{{(a^{-1}q;q)}_n{a}^n(1-q^n)}\sum _{k=0}^{\infty }\frac{{(q^n,a_1/a_2;q)}_k{(q{a}_2/a)}^k}{{(q,q^{1+n}/a;q)}_k}\hfill \\ & \times \sum _{\begin{array}{c}{k}_1+k_2+\cdots +k_r=k\\ 0\le k_r\le k_{r-1}\le \cdots \le k_2\le k_1\le k\end{array}}\prod _{j=1}^r\frac{{(q^{-k_{j-1}},a_{2j+1}/a_{2j+2};q)}_{k_j}{(q{a}_{2j+2}/a_{2j-1})}^{k_j}}{{(q,q^{1-k_{j-1}}{a}_{2j}/a_{2j-1};q)}_{k_j}}.\hfill \end{array}$$

(41)

Proof. 

By (4), we easily obtain

$$\begin{array}{c}{(q^{1-x};q)}_{\infty }=(1-q^{1-x})(1-q^{2-x})\cdots (1-q^{n-1-x})(1-q^{n-x})(1-q^{n+1-x})(1-q^{n+2-x})\cdots ,\hfill \end{array}$$

(42)

$$\begin{array}{c}{(a{q}^{-x};q)}_{\infty }=(1-a{q}^{-x})(1-a{q}^{1-x})(1-a{q}^{2-x})\cdots (1-a{q}^{n-1-x})(1-a{q}^{n-x})(1-a{q}^{n+1-x})(1-a{q}^{n+2-x})\cdots .\hfill \end{array}$$

(43)

Noting (42), (43) and using the L’Hospital rule, via some simple calculation, we get

$$\begin{array}{cc}\underset{x\to n}{lim}& \frac{\pi }{sin\pi x}\frac{{(q^{1-x},a;q)}_{\infty }}{{(q,a{q}^{-x};q)}_{\infty }}\hfill \\ =& \frac{{(a;q)}_{\infty }}{{(q;q)}_{\infty }}\underset{x\to n}{lim}\frac{\pi }{sin\pi x}\frac{{(q^{1-x};q)}_{\infty }}{{(a{q}^{-x};q)}_{\infty }}\hfill \\ =& \frac{{(a;q)}_{\infty }}{{(q;q)}_{\infty }}\frac{{(-1)}^{n-1}{q}^{-1-2-\cdots -(n-1)}(1-q)(1-q^2)\cdots (1-q^{n-1})(1-q)(1-q^2)\cdots }{{(-1)}^n{q}^{-1-2-\cdots -n}(a-q)(a-q^2)\cdots (a-q^n)(1-a)(1-aq)(1-a{q}^2)\cdots }\underset{x\to n}{lim}\frac{\pi (1-q^{n-x})}{sin\pi x}\hfill \\ =& \frac{{(a;q)}_{\infty }}{{(q;q)}_{\infty }}\frac{-(1-q)(1-q^2)\cdots (1-q^{n-1}){(q;q)}_{\infty }}{q^{-n}(a-q)(a-q^2)\cdots (a-q^n){(a;q)}_{\infty }}\frac{logq}{cosn\pi }\hfill \\ =& \frac{{(-1)}^{n-1}{(q;q)}_n{q}^{n}logq}{{(a^{-1}q;q)}_n{a}^n(1-q^n)}.\hfill \end{array}$$

Letting $x⟶n$ on both sides of (25) and using the above limit, we obtain (41) immediately.

Taking $r=0$ in (41) and defining the empty sum equal to 1, we obtain the following integral formula:

Corollary 4.

For $0<q<1$, $0<a,a_1,a_2<q^n(n=1,2,\cdots )$ and $|q{a}_2/a|<1$; $a_1,a_2\ne a$, we have

$$\begin{array}{c}{\int }_0^{\infty }{t}^{n-1}\frac{{(-at,-a_{1}t;q)}_{\infty }}{{(-t,-a_{2}t;q)}_{\infty }}dt=\frac{{(-1)}^{n-1}{(q;q)}_n{q}^{n}logq}{{(a^{-1}q;q)}_n{a}^n(1-q^n)}\sum _{k=0}^{\infty }\frac{{(q^n,a_1/a_2;q)}_k{(q{a}_2/a)}^k}{{(q,q^{1+n}/a;q)}_k}.\end{array}$$

(44)

Remark 3.

Taking $a_1=a_2$ in (44), we deduce

$$\begin{array}{c}{\int }_0^{\infty }{t}^{n-1}\frac{{(-at;q)}_{\infty }}{{(-t;q)}_{\infty }}dt=\frac{{(-1)}^{n-1}{(q;q)}_n{q}^{n}logq}{{(a^{-1}q;q)}_n{a}^n(1-q^n)}=\frac{-{(q;q)}_{n-1}{q}^{n}logq}{(q-a)(q^2-a)\cdots (q^n-a)},\end{array}$$

(45)

which is exactly Askey’s result in ([15], p. 349, (2.9)).

Taking $r=1$ in (41), we get

Corollary 5.

For $0<q<1$, $0<a,a_1,a_2,a_3,a_4<q^n(n=1,2,\cdots )$, $|q{a}_2/a|<1$ and $|q{a}_4/a_1|<1$; $a_1,a_2\ne a,a_3,a_4\ne a_1$, we have

$$\begin{array}{ll}\hfill {\int }_0^{\infty }{t}^{n-1}& \frac{{(-at,-a_{1}t,-a_{3}t;q)}_{\infty }}{{(-t,-a_{2}t,-a_{4}t;q)}_{\infty }}dt\hfill \\ \hfill =& \frac{{(q;q)}_n{q}^{n}logq}{(a-q)(a-q^2)\cdots (a-q^n)(1-q^n)}\sum _{k=0}^{\infty }\frac{{(q^n,a_1/a_2;q)}_k{(q{a}_2/a)}^k}{{(q,q^{1+n}/a;q)}_k}{}_2{\varphi }_1\left(\genfrac{}{}{0pt}{}{q^{-k},a_3/a_4}{a_2{q}^{1-k}/a_1};q,q{a}_4/a_1\right).\hfill \end{array}$$

(46)

Theorem 3.

If $0<q<1$, $0<a,a_1,a_2,\dots ,a_{2r+2}<q$, $|q{a}_2/a|<1$ and $|q{a}_{2j+2}/a_{2j-1}|<1$ $(j=1,2,\dots ,r)$; $a_1,a_2\ne a,a_{l+2},a_{l+3}\ne a_l$ $(l=1,3,5,\dots ,2r-1)$, then we have

$$\begin{array}{ll}{\int }_0^{\infty }& \frac{{(-at,-a_{1}t,-a_{3}t,\dots ,-a_{2r+1}t;q)}_{\infty }}{{(-t,-a_{2}t,-a_{4}t,\dots ,-a_{2r+2}t;q)}_{\infty }}dt\hfill \\ & =\frac{qlogq}{a-q}\sum _{k=0}^{\infty }\frac{{(a_1/a_2;q)}_k{(q{a}_2/a)}^k}{{(q^2/a;q)}_k}\hfill \\ & \times \sum _{\begin{array}{c}{k}_1+k_2+\cdots +k_r=k\\ 0\le k_r\le k_{r-1}\le \cdots \le k_2\le k_1\le k\end{array}}\prod _{j=1}^r\frac{{(q^{-k_{j-1}},a_{2j+1}/a_{2j+2};q)}_{k_j}{(q{a}_{2j+2}/a_{2j-1})}^{k_j}}{{(q,q^{1-k_{j-1}}{a}_{2j}/a_{2j-1};q)}_{k_j}}.\hfill \end{array}$$

(47)

Proof. 

Taking $n=1$ in (41), we obtain (47) immediately.

Taking $r=0$ in (47) and defining the empty sum equal to 1, we obtain the following integral formula:

Corollary 6.

For $0<q<1$, $0<a,a_1,a_2<q$ and $|q{a}_2/a|<1$; $a_1,a_2\ne a$, we have

$$\begin{array}{c}{\int }_0^{\infty }\frac{{(-at,-a_{1}t;q)}_{\infty }}{{(-t,-a_{2}t;q)}_{\infty }}dt=\frac{qlogq}{a-q}{}_2{\varphi }_1\left(\genfrac{}{}{0pt}{}{q,a_1/a_2}{q^2/a};q,q{a}_2/a\right).\end{array}$$

(48)

Taking $r=1$ in (47), we deduce

Corollary 7.

For $0<q<1$, $0<a,a_1,a_2,a_3,a_4<q$, $|q{a}_2/a|<1$ and $|q{a}_4/a_1|<1$; $a_1,a_2\ne a,a_3,a_4\ne a_1$, we have

$$\begin{array}{c}{\int }_0^{\infty }\frac{{(-at,-a_{1}t,-a_{3}t;q)}_{\infty }}{{(-t,-a_{2}t,-a_{4}t;q)}_{\infty }}dt=\frac{qlogq}{a-q}\sum _{k=0}^{\infty }\frac{{(a_1/a_2;q)}_k{(q{a}_2/a)}^k}{{(q^2/a;q)}_k}{}_2{\varphi }_1\left(\genfrac{}{}{0pt}{}{q^{-k},a_3/a_4}{a_2{q}^{1-k}/a_1};q,q{a}_4/a_1\right).\end{array}$$

(49)

4. Connections with the q-Gamma Function

In this section, we give the corresponding formulas with the q-gamma function from (25).

Theorem 4.

If $0<q<1$, $x>0$, $0<a_1,a_2,\dots ,a_{2r+2}<q^x$, $|a_2/q^{x+y-1}|<1$ and $|q{a}_{2j+2}/a_{2j-1}|<1$, $(j=1,2,\dots ,r)$; $a_1,a_2\ne q^{x+y},a_{l+2},a_{l+3}\ne a_l$ $(l=1,3,5,\dots ,2r-1)$, then we have

$$\begin{array}{ll}{\int }_0^{\infty }{t}^{x-1}& \frac{{(-t{q}^{x+y},-a_{1}t,-a_{3}t,\dots ,-a_{2r+1}t;q)}_{\infty }}{{(-t,-a_{2}t,-a_{4}t,\dots ,-a_{2r+2}t;q)}_{\infty }}dt\hfill \\ & =\frac{{\Gamma }_q\left(y\right){\Gamma }_q(1-y)\Gamma \left(x\right)\Gamma (1-x)}{{\Gamma }_q(x+y){\Gamma }_q\left(x\right){\Gamma }_q(1-x)}\sum _{k=0}^{\infty }\frac{{\Gamma }_q(k+x){(a_1/a_2;q)}_k}{{\Gamma }_q(k+1-y){(q;q)}_k}{(a_2/q^{x+y-1})}^k\hfill \\ & \times \sum _{\begin{array}{c}{k}_1+k_2+\cdots +k_r=k\\ 0\le k_r\le k_{r-1}\le \cdots \le k_2\le k_1\le k\end{array}}\prod _{j=1}^r\frac{{(q^{-k_{j-1}},a_{2j+1}/a_{2j+2};q)}_{k_j}{(q{a}_{2j+2}/a_{2j-1})}^{k_j}}{{(q,q^{1-k_{j-1}}{a}_{2j}/a_{2j-1};q)}_{k_j}}.\hfill \end{array}$$

(50)

Proof. 

Taking $a=q^{x+y}$ in (25) and noting that

$${\Gamma }_q\left(x\right)=\frac{{(q;q)}_{\infty }}{{(q^x;q)}_{\infty }}{(1-q)}^{1-x}and\Gamma \left(x\right)\Gamma (1-x)=\frac{\pi }{sin\pi x},$$

we easily obtain (50).

Taking $r=0$ in (50) and defining the empty sum equal to 1, we obtain the following integral formula.

Corollary 8.

For $0<q<1$, $x>0$, $0<a_1,a_2<q^x$ and $|a_2/q^{x+y-1}|<1$; $a_1,a_2\ne q^{x+y}$, we have

$$\begin{array}{c}{\int }_0^{\infty }{t}^{x-1}\frac{{(-t{q}^{x+y},-a_{1}t;q)}_{\infty }}{{(-t,-a_{2}t;q)}_{\infty }}dt=\frac{{\Gamma }_q\left(y\right){\Gamma }_q(1-y)\Gamma \left(x\right)\Gamma (1-x)}{{\Gamma }_q(x+y){\Gamma }_q\left(x\right){\Gamma }_q(1-x)}\sum _{k=0}^{\infty }\frac{{\Gamma }_q(k+x){(a_1/a_2;q)}_k}{{\Gamma }_q(k+1-y){(q;q)}_k}{(a_2/q^{x+y-1})}^k.\hfill \end{array}$$

(51)

Taking $a_1=a_2$ in (51), we deduce the result of Askey as follows:

Corollary 9

([15], p. 350, Equation (2.10)). For $0<q<1$, $x>0$, we have

$$\begin{array}{c}{\int }_0^{\infty }{t}^{x-1}\frac{{(-t{q}^{x+y};q)}_{\infty }}{{(-t;q)}_{\infty }}dt=\frac{{\Gamma }_q\left(y\right)\Gamma \left(x\right)\Gamma (1-x)}{{\Gamma }_q(x+y){\Gamma }_q(1-x)}.\hfill \end{array}$$

(52)

Remark 4.

Letting $a\to 0$ and $b\to 1$ in (39), we have

$$\begin{array}{c}{\int }_0^{\infty }\frac{t^{x-1}}{{(-t;q)}_{\infty }}dt=\frac{\pi }{sin\pi x}\frac{{(q^{1-x};q)}_{\infty }}{{(q;q)}_{\infty }}.\end{array}$$

(53)

Applying

$${\Gamma }_q\left(x\right)=\frac{{(q;q)}_{\infty }}{{(q^x;q)}_{\infty }}{(1-q)}^{1-x}and\Gamma \left(x\right)\Gamma (1-x)=\frac{\pi }{sin\pi x},$$

we obtain

$$\begin{array}{c}{\int }_0^{\infty }\frac{t^{x-1}}{{(-t;q)}_{\infty }}dt=\frac{\Gamma \left(x\right)\Gamma (1-x){(1-q)}^x}{{\Gamma }_q(1-x)},\end{array}$$

(54)

which is exactly the result of Askey (see ([15], p. 353, Equation (4.2))).

5. Conclusions

In this paper, by applying q-exponential operator

$${}_1{\varphi }_0\left(\genfrac{}{}{0pt}{}{b}{-};q,-c\theta \right)=\sum _{n=0}^{\infty }\frac{{(b;q)}_n{(-c\theta )}^n}{{(q;q)}_n},$$

we further extend the following Ramanujan’s beta integral [10]

$$\begin{array}{c}{\int }_0^{\infty }{t}^{x-1}\frac{{(-at;q)}_{\infty }}{{(-t;q)}_{\infty }}dt=\frac{\pi }{sin\pi x}\frac{{(q^{1-x},a;q)}_{\infty }}{{(q,a{q}^{-x};q)}_{\infty }}.\end{array}$$

Especially, we obtain two new integral formulas

$$\begin{array}{c}{\int }_0^{\infty }{t}^{x-1}\frac{{(-at,-a_{1}t;q)}_{\infty }}{{(-t,-a_{2}t;q)}_{\infty }}dt=\frac{\pi }{sin\pi x}\frac{{(q^{1-x},a;q)}_{\infty }}{{(q,a{q}^{-x};q)}_{\infty }}{}_2{\varphi }_1\left(\genfrac{}{}{0pt}{}{q^x,a_1/a_2}{q^{1+x}/a};q,q{a}_2/a\right)\hfill \end{array}$$

and

$$\begin{array}{c}{\int }_0^{\infty }{t}^{x-1}\frac{{(-at;q)}_{\infty }}{{(-bt;q)}_{\infty }}dt=\frac{\pi }{sin\pi x}\frac{{(q^{1-x},a,q/a,b{q}^{1+x}/a;q)}_{\infty }}{{(q,a{q}^{-x},qb/a,q^{1+x}/a;q)}_{\infty }}.\end{array}$$

We also show that Ramanujan’s beta integral can be represented with q-gamma functions [15].