# Formula:KLS:14.10:89

${\displaystyle{\displaystyle{\displaystyle\frac{1}{2\pi}\int_{-1}^{1}\frac{w(x% ;1|q)}{\sqrt{1-x^{2}}}P_{m}\!\left(x|q\right)P_{n}\!\left(x|q\right)\,dx{}=% \frac{\left(q^{\frac{1}{2}};q\right)_{\infty}}{\left(q,q,-q^{\frac{1}{2}},-q;q% \right)_{\infty}}\frac{q^{\frac{1}{2}n}}{1-q^{n+\frac{1}{2}}}\,\delta_{m,n}}}}$

## Substitution(s)

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

## Proof

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