Formula:KLS:14.10:02

$\displaystyle {\displaystyle \frac{1}{2\cpi}\int_{-1}^1\frac{w(x)}{\sqrt{1-x^2}} \ctsqJacobi{\alpha}{\beta}{m}@{x}{q}\ctsqJacobi{\alpha}{\beta}{n}@{x}{q}\,dx {}=\frac{\qPochhammer{q^{\frac{1}{2}(\alpha+\beta+2)},q^{\frac{1}{2}(\alpha+\beta+3)}}{q}{\infty}}{\qPochhammer{q,q^{\alpha+1},q^{\beta+1},-q^{\frac{1}{2}(\alpha+\beta+1)} -q^{\frac{1}{2}(\alpha+\beta+2)}}{q}{\infty}}\,\frac{1-q^{\alpha+\beta+1}}{1-q^{2n+\alpha+\beta+1}} {}\frac{\qPochhammer{q^{\alpha+1},q^{\beta+1},-q^{\frac{1}{2}(\alpha+\beta+3)}}{q}{n}} {\qPochhammer{q,q^{\alpha+\beta+1},-q^{\frac{1}{2}(\alpha+\beta+1)}}{q}{n}}q^{(\alpha+\frac{1}{2})n}\,\Kronecker{m}{n} }$

Substitution(s)

$\displaystyle {\displaystyle w(x):=w(x;q^{\alpha},q^{\beta}|q) =\left|\frac{\qPochhammer{\expe^{2\iunit\theta}}{q}{\infty}} {\qPochhammer{q^{\frac{1}{2}\alpha+\frac{1}{4}}\expe^{\iunit\theta},q^{\frac{1}{2}\alpha+\frac{3}{4}}\expe^{\iunit\theta} -q^{\frac{1}{2}\beta+\frac{1}{4}}\expe^{\iunit\theta},-q^{\frac{1}{2}\beta+\frac{3}{4}}\expe^{\iunit\theta}}{q}{\infty}}\right|^2 =\left|\frac{\qPochhammer{\expe^{\iunit\theta},-\expe^{\iunit\theta}}{q^{\frac{1}{2}}}{\infty}}{\qPochhammer{q^{\frac{1}{2}\alpha+\frac{1}{4}}\expe^{\iunit\theta} -q^{\frac{1}{2}\beta+\frac{1}{4}}\expe^{\iunit\theta}}{q^{\frac{1}{2}}}{\infty}}\right|^2 =\frac{h(x,1)h(x,-1)h(x,q^{\frac{1}{2}})h(x,-q^{\frac{1}{2}})} {h(x,q^{\frac{1}{2}\alpha+\frac{1}{4}})h(x,q^{\frac{1}{2}\alpha+\frac{3}{4}}) h(x,-q^{\frac{1}{2}\beta+\frac{1}{4}})h(x,-q^{\frac{1}{2}\beta+\frac{3}{4}})}}$ &
${\displaystyle{\displaystyle{\displaystyle x=\cos\theta}}}$

Proof

We ask users to provide proof(s), reference(s) to proof(s), or further clarification on the proof(s) in this space.