Rahmanの超幾何積分

$ABCq=(abcst)^2$として, non-terminating Jacksonの${}_8\phi_7$和公式 より,
\begin{align} &\Q87{abcst^2/q,q\sqrt{abcst^2/q},-q\sqrt{abcst^2/q},A,B,C,te^{i\theta},te^{-i\theta}}{\sqrt{abcst^2/q},-\sqrt{abcst^2/q},abcst^2/A,abcst^2/B,abcst^2/C,abcste^{i\theta},abcste^{-i\theta}}q\\ &\qquad+\frac{(Aq/B,Aq/C,Aqe^{i\theta}/t,Aqe^{-i\theta}/t,abcst^2,B,C,te^{i\theta},te^{-i\theta},Aq/abcst^2;q)_{\infty}}{(A^2q^2/abcst^2,ABq/abcst^2,ACq/abcst^2,Aqe^{i\theta}/abcst,Aqe^{-i\theta}/abcst;q)_{\infty}}\\ &\qquad\cdot\frac 1{(abcst^2/B,abcst^2/C,abcste^{i\theta},abcste^{-i\theta},abcst^2/Aq;q)_{\infty}}\\ &\qquad\cdot\Q87{A^2q/abcst^2,q\sqrt{A^2q/abcst^2},-q\sqrt{A^2q/abcst^2},A,ABq/abcst^2,ACq/abcst^2,Aqte^{i\theta}/abcst,Aqe^{-i\theta}/abcst}{\sqrt{A^2q/abcst^2},-\sqrt{A^2q/abcst^2},A/abcst^2,Aq/B,Aq/C,Aqe^{i\theta}/t,Aqe^{-i\theta}/t}q\\ &=\frac{(abcst^2,Aq/abcst^2,abcst^2/BC,abcste^{i\theta}/B,abcste^{-i\theta}/B,abcste^{i\theta}/C,abcste^{-i\theta}/C,abcs;q)_{\infty}}{(abcst^2/B,abcst^2/C,abcste^{i\theta},abcste^{-i\theta},ABq/abcst^2,ACq/abcst^2,Aqe^{i\theta}/abcst,Aqe^{-i\theta}/abcst;q)_{\infty}} \end{align}
ここで, これに
\begin{align} \frac 1{2\pi}\left|\frac{(e^{2i\theta},abcste^{i\theta};q)_{\infty}}{(ae^{i\theta},be^{i\theta},ce^{i\theta},se^{i\theta},te^{i\theta};q)_{\infty}}\right|^2\frac{1}{\sqrt{1-x^2}} \end{align}
を掛けて積分すると,
\begin{align} &\frac 1{2\pi}\int_{-1}^1\left|\frac{(e^{2i\theta},abcste^{i\theta};q)_{\infty}}{(ae^{i\theta},be^{i\theta},ce^{i\theta},se^{i\theta},te^{i\theta};q)_{\infty}}\right|^2\,\frac{(te^{i\theta},te^{-i\theta};q)_r}{(abcste^{i\theta},abcste^{-i\theta};q)_r}\frac{dx}{\sqrt{1-x^2}}\\ &=\frac{(abcs,abct,abst,acst,bcst;q)_{\infty}}{(q,ab,ac,bc,as,bs,cs,at,bt,ct,st;q)_{\infty}}\frac{(at,bt,ct,st;q)_r}{(abct,abst,acst,bcst;q)_r}\\ &\frac 1{2\pi}\int_{-1}^1\left|\frac{(e^{2i\theta},abcste^{i\theta};q)_{\infty}}{(ae^{i\theta},be^{i\theta},ce^{i\theta},se^{i\theta},te^{i\theta};q)_{\infty}}\right|^2\,\frac{(Aqe^{i\theta}/t,Aqe^{-i\theta}/t,te^{i\theta},te^{-i\theta};q)_{\infty}}{(Aqe^{i\theta}/abcst,Aqe^{-i\theta}/abcst,abcste^{i\theta},abcste^{-i\theta};q)_{\infty}}\\ &\qquad\cdot\frac{(Aqe^{i\theta}/abcst^2,Aqe^{-i\theta}/abcst^2;q)_r}{(Aqe^{i\theta}/t,Aqe^{-i\theta}/t;q)_r}\frac{dx}{\sqrt{1-x^2}}\\ &=\frac 1{2\pi}\int_{-1}^1\left|\frac{(e^{2i\theta},Aq^{r+1}e^{i\theta}/t;q)_{\infty}}{(ae^{i\theta},be^{i\theta},ce^{i\theta},se^{i\theta},Aq^{r+1}e^{i\theta}/abcst;q)_{\infty}}\right|^2\,\frac{dx}{\sqrt{1-x^2}}\\ &=\frac{(abcs,Aq/at,Aq/bt,Aq/ct,Aq/st;q)_{\infty}}{(q,ab,ac,bc,as,bs,cs,Aq/abst,Aq/acst,Aq/bcst,Aq/abct;q)_{\infty}}\\ &\qquad\cdot\frac{(Aq/abst,Aq/acst,Aq/bcst,Aq/abct;q)_r}{(Aq/at,Aq/bt,Aq/ct,Aq/st;q)_r} \end{align}
が成り立つから,

\begin{align} &\frac 1{2\pi}\frac{(abcst^2,Aq/abcst^2,abcst^2/BC,abcs;q)_{\infty}}{(abcst^2/B,abcst^2/C,ABq/abcst^2,ACq/abcst^2;q)_{\infty}}\int_{-1}^1\left|\frac{(e^{2i\theta},abcste^{i\theta}/B,abcste^{i\theta}/C;q)_{\infty}}{(ae^{i\theta},be^{i\theta},ce^{i\theta},se^{i\theta},te^{i\theta},Aqe^{i\theta}/abcst;q)_{\infty}}\right|^2\,\frac{dx}{\sqrt{1-x^2}}\\ &=\frac{(abcs,abct,abst,acst,bcst;q)_{\infty}}{(q,ab,ac,bc,as,bs,cs,at,bt,ct,st;q)_{\infty}}\\ &\qquad\cdot\Q{10}{9}{abcst^2/q,q\sqrt{abcst^2/q},-q\sqrt{abcst^2/q},A,B,C,at,bt,ct,st}{\sqrt{abcst^2/q},-\sqrt{abcst^2/q},abcst^2/A,abcst^2/B,abcst^2/C,abct,abst,acst,bcst}q\\ &\qquad+\frac{(Aq/B,Aq/C,abcst^2,B,C,Aq/abcst^2;q)_{\infty}}{(A^2q^2/abcst^2,ABq/abcst^2,ACq/abcst^2,abcst^2/B,abcst^2/C,abcst^2/Aq;q)_{\infty}}\\ &\qquad\cdot\frac{(abcs,Aq/at,Aq/bt,Aq/ct,Aq/st;q)_{\infty}}{(q,ab,ac,bc,as,bs,cs,Aq/abst,Aq/acst,Aq/bcst,,Aq/abct;q)_{\infty}}\\ &\qquad\cdot\Q{10}9{A^2q/abcst^2,q\sqrt{A^2q/abcst^2},-q\sqrt{A^2q/abcst^2},A,ABq/abcst^2,ACq/abcst^2,Aq/abst,Aq/acst,Aq/bcst,Aq/abct}{\sqrt{A^2q/abcst^2},-\sqrt{A^2q/abcst^2},A/abcst^2,Aq/B,Aq/C,Aq/at,Aq/bt,Aq/ct,Aq/st}q \end{align}
より,
\begin{align} &\frac 1{2\pi}\int_{-1}^1\left|\frac{(e^{2i\theta},abcste^{i\theta}/B,abcste^{i\theta}/C;q)_{\infty}}{(ae^{i\theta},be^{i\theta},ce^{i\theta},se^{i\theta},te^{i\theta},Aqe^{i\theta}/abcst;q)_{\infty}}\right|^2\,\frac{dx}{\sqrt{1-x^2}}\\ &=\frac{(abcst^2/B,abcst^2/C,ABq/abcst^2,ACq/abcst^2;q)_{\infty}}{(abcst^2,Aq/abcst^2,abcst^2/BC,abcs;q)_{\infty}}\frac{(abcs,abct,abst,acst,bcst;q)_{\infty}}{(q,ab,ac,bc,as,bs,cs,at,bt,ct,st;q)_{\infty}}\\ &\qquad\cdot\Q{10}{9}{abcst^2/q,q\sqrt{abcst^2/q},-q\sqrt{abcst^2/q},A,B,C,at,bt,ct,st}{\sqrt{abcst^2/q},-\sqrt{abcst^2/q},abcst^2/A,abcst^2/B,abcst^2/C,abct,abst,acst,bcst}q\\ &\qquad+\frac{(abcst^2/B,abcst^2/C,ABq/abcst^2,ACq/abcst^2;q)_{\infty}}{(abcst^2,Aq/abcst^2,abcst^2/BC,abcs;q)_{\infty}}\\ &\qquad\cdot\frac{(Aq/B,Aq/C,abcst^2,B,C,Aq/abcst^2;q)_{\infty}}{(A^2q^2/abcst^2,ABq/abcst^2,ACq/abcst^2,abcst^2/B,abcst^2/C,abcst^2/Aq;q)_{\infty}}\\ &\qquad\cdot\frac{(abcs,Aq/at,Aq/bt,Aq/ct,Aq/st;q)_{\infty}}{(q,ab,ac,bc,as,bs,cs,Aq/abst,Aq/acst,Aq/bcst,,Aq/abct;q)_{\infty}}\\ &\qquad\cdot\Q{10}9{A^2q/abcst^2,q\sqrt{A^2q/abcst^2},-q\sqrt{A^2q/abcst^2},A,ABq/abcst^2,ACq/abcst^2,Aq/abst,Aq/acst,Aq/bcst,Aq/abct}{\sqrt{A^2q/abcst^2},-\sqrt{A^2q/abcst^2},A/abcst^2,Aq/B,Aq/C,Aq/at,Aq/bt,Aq/ct,Aq/st}q\\ &=\frac{(abcst^2/B,abcst^2/C,ABq/abcst^2,ACq/abcst^2,abct,abst,acst,bcst;q)_{\infty}}{(abcst^2,Aq/abcst^2,abcst^2/BC,q,ab,ac,bc,as,bs,cs,at,bt,ct,st;q)_{\infty}}\\ &\qquad\cdot\Q{10}{9}{abcst^2/q,q\sqrt{abcst^2/q},-q\sqrt{abcst^2/q},A,B,C,at,bt,ct,st}{\sqrt{abcst^2/q},-\sqrt{abcst^2/q},abcst^2/A,abcst^2/B,abcst^2/C,abct,abst,acst,bcst}q\\ &\qquad\cdot\frac{(Aq/B,Aq/C,B,C,Aq/at,Aq/bt,Aq/ct,Aq/st;q)_{\infty}}{(abcst^2/BC,A^2q^2/abcst^2,abcst^2/Aq,q,ab,ac,bc,as,bs,cs,Aq/abst,Aq/acst,Aq/bcst,Aq/abct;q)_{\infty}}\\ &\qquad\cdot\Q{10}9{A^2q/abcst^2,q\sqrt{A^2q/abcst^2},-q\sqrt{A^2q/abcst^2},A,ABq/abcst^2,ACq/abcst^2,Aq/abst,Aq/acst,Aq/bcst,Aq/abct}{\sqrt{A^2q/abcst^2},-\sqrt{A^2q/abcst^2},A/abcst^2,Aq/B,Aq/C,Aq/at,Aq/bt,Aq/ct,Aq/st}q \end{align}
となって定理が示された.