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# Definite integral of Hypergeometric function 2F1

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