# 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 generalise..?

So, can anyone help me with a good way to see why the above holds without using the infinite sum, (or, in the best possible scenario, help me with the general $\mathcal{I}(z,x)$)?

The approaches I have tried so far include looking for cleaver changes of variables in the integral, and/or using the integral representation of

$\,_3F_2\left( \frac{1}{2},\frac{1}{2},\frac{1}{2} ;1, \frac{3}{2} ; \left(\frac{z}{4}\right)^2 \right) \propto \int_{0}^1 u^{1/2-1}(1-u)^{1-1/2-1} \,_2F_1\left( \frac{1}{2},\frac{1}{2} ; \frac{3}{2} ; \left(\frac{z}{4}\right)^2 u \right) du$

but then I'm stuck. I know the hypergeometric functions have many nice properties, but I don't see how I can map this single integral to the double integral above, with the change of the parameter as well?

• Have you checked whether using the Mellin–Barnes representation of $_2F_1$ and integrating out $s$ and $t$ gives a Fox H-function? – K B Dave Mar 12 '18 at 17:33
• No, I haven't, thanks for the tip! If you could give me some more hints/details, that'd be very much appreciated! – Lou Mar 13 '18 at 13:09
• Thank you again for your hint, this allowed me to rewrite the integral, via an infinite sum of gamma functions, as a Fox-Wright function, which also works for non-vanishing x. Thanks! – Lou Mar 14 '18 at 19:19
• @Lou What about the variable $z$? Is it a real or complex number? Does it satisfy any inequalities or conditions? Such info could help avoid a lot of annoying casework... – David H Mar 15 '18 at 14:25
• @DavidH Yes, z>0, but what I'm really interested in is really the asymptotics at exponentially large z (the analytic continuation). I'm now trying to get it from the asymptotics of the Fox-Wright functions, but if you know a better way it would be highly appreciated! – Lou Mar 16 '18 at 16:17