# Integral of Hypergeometric series

Is it possible to calculate this integral ? $$I:=\int_0^{\gamma^2} \int_0^{\gamma^2} \left((1-x)(1-y)\right)^{s-2} {}_3F_2\left(s,s,s;1,1;xyt\right) dx dy$$ where $$\gamma,\; t,\; s\geq 0$$ and $${}_3F_2$$ is the hypergeometric series.

Although the integral looks neat and fairly simple in form. I tried evaluating it using integration by parts but without success. I also couldn't find the solution in the book, "Table of Integrals, Series, and Products" by Gradshteyn and Ryzhik and Prudnikov, Brychkov, - Integrals and Series 1-3.

Can anyone help me in solving this?

Thank you.

• If $s>1$ then for the special case of $\gamma=\pm 1$ your integral evaluates to $\frac{1}{(s-1)^2}{_5F_4}\left({s,s,s,s,s\atop 1,1,1,1};t\right)$. Feb 6 at 11:14
• thank you Aaron, but I can't see how you get this result ? Feb 6 at 13:58

I do not know about a solution for arbitary $$\gamma$$, but for the special case of $$\gamma=\pm 1$$ the integral in question reduces to $$I=\int_0^1\int_0^1((1-x)(1-y))^{s-2} {}_3F_2\left({s,s,s\atop 1,1};xyt\right)\,\mathrm dx\mathrm dy.$$ Through the use of DLMF 16.5.2 we may identify the integral w.r.t. $$x$$ using $$a_0=1$$ and $$b_0=s$$ to obtain $$I=\frac{1}{s-1}\int_0^1(1-y)^{s-2} {_4F_3}\left({s,s,s,s\atop 1,1,1};yt\right)\,\mathrm dy.$$ Note that the use of the DLMF integral representation requires $$\Re s>1$$. Application of DLMF 16.5.2, again for $$a_0=1$$ and $$b_0=s$$, then gives the final solution $$I=\frac{1}{(s-1)^2}{_5F_4}\left({s,s,s,s,s\atop 1,1,1,1};t\right).$$
• Now, i see, thank's Aaron. I have found a general formula in Anatolij P. Prudnikov, Yu. A. Brychkov, O. I. Marichev (transl. - Integrals and Series ) vol 3, p.316, Eq.20 for the Gaussian ${}_2F_1$ but unfortunately not for ${}_3F_2$. Feb 7 at 17:34
• @othmane the problem is that if you write 3F2 as a series and integrate term-wise you end up with integrals of the form $\int_0^{\gamma^2}x^k(1-x)^{s-2}\,\mathrm dx$ which are incomplete beta functions (2F1 functions in disguise). So the series does not evlauate to something pretty. On the other hand, when the upper bound of integration is one you get a complete beta function which sums nicely. Feb 7 at 18:17