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# Definite integral of Hypergeometric function 2F1

2018-03-12 20:24:28

I am looking for a way to evaluate

$\mathcal{I}(z,x)=\int_0^1 \int_0^1 \frac{s^{ix}(1-s)^{-ix}}{\sqrt{s(1-s)t(1-t)}} \,_2F_1\left( \frac{1}{2},1 ; \frac{3}{2} ; z^2 \,s^{1+2ix}(1-s)^{1-2ix} t(1-t) \right)$

where $x\geq0$ in closed form.

For $x=0$, I know that this integral evaluates to

$\mathcal{I}(z,0)=\int_0^1 \int_0^1 \frac{1}{\sqrt{s(1-s)t(1-t)}} \,_2F_1\left( \frac{1}{2},1 ; \frac{3}{2} ; z^2 \,s(1-s) t(1-t) \right) = \frac{1}{z^2}\sum_{j\geq1} \frac{z^2j}{2j-1}Beta\left(j-\frac{1}{2},j-\frac{1}{2}\right)^2 = \pi^2 \,_3F_2\left( \frac{1}{2},\frac{1}{2},\frac{1}{2} ;1, \frac{3}{2} ; \left(\frac{z}{4}\right)^2 \right)$

(This is found by using the integral representation of Beta functions and the series-expression for the hypergeometric function, and then mathematica evaluates the sum straight away), but this approach doesn't seem to work when $x\neq 0$. Therefore, I am looking for a way to see this straight from the integral which might be easier to generali