# Challenging definite integral of hypergeometric functions

Given $$\ell\in\mathbb{N}\land\left(m,n\right)\in\mathbb{Z}_{\ge0}^{2}\land\left(a,b,c\right)\in\mathbb{R}^{3}\land0, define the function $$\mathcal{J}_{\ell,m,n}{\left(a,b,c\right)}$$ via the definite integral

$$\small{\mathcal{J}_{\ell,m,n}{\left(a,b,c\right)}:=\int_{0}^{1}\mathrm{d}t\,\frac{t^{m-\frac12}\left(1-t\right)^{\ell+n-\frac12}}{\left(1-at\right)^{\ell}}\,{_2F_1}{\left(\frac12,m+\frac12;m+\frac32;bt\right)}\,{_2F_1}{\left(\frac12,n+\frac12;n+\frac32;c\left(1-t\right)\right)}}.$$

The Gauss hypergeometric function $${_2F_1}$$ may be defined via the infinite series

$${_2F_1}{\left(\alpha,\beta;\gamma;z\right)}:=\sum_{n=0}^{\infty}\frac{\left(\alpha\right)_{n}\,\left(\beta\right)_{n}}{\left(\gamma\right)_{n}}\cdot\frac{z^{n}}{n!};~~~\small{\left|z\right|<1}.$$

Question: Does the integral above possess a closed form representation in terms of hypergeometric (or simpler) functions?

Possible starting point: Lacking any obvious way to evaluate this integral, one strategy that came to mind was to expand the integral as a multiple infinite series, which could then be converted into some hypergeometric representation (assuming such exists). As you can see below, things get messy, and it's hard to tell if any headway is being made.

Let $$\ell\in\mathbb{N}\land\left(m,n\right)\in\mathbb{Z}_{\ge0}^{2}\land\left(a,b,c\right)\in\mathbb{R}^{3}\land0. We find

\begin{align} \mathcal{J}_{\ell,m,n}{\left(a,b,c\right)} &=\int_{0}^{1}\mathrm{d}t\,\frac{t^{m-\frac12}\left(1-t\right)^{\ell+n-\frac12}}{\left(1-at\right)^{\ell}}\,{_2F_1}{\left(\frac12,m+\frac12;m+\frac32;bt\right)}\\ &~~~~~\times\,{_2F_1}{\left(\frac12,n+\frac12;n+\frac32;c\left(1-t\right)\right)}\\ &=\int_{0}^{1}\mathrm{d}t\,\frac{t^{m-\frac12}\left(1-t\right)^{\ell+n-\frac12}}{\left(1-at\right)^{\ell}}\,{_2F_1}{\left(\frac12,m+\frac12;m+\frac32;bt\right)}\\ &~~~~~\times\sum_{k=0}^{\infty}\frac{\left(\frac12\right)_{k}\,\left(n+\frac12\right)_{k}\,c^{k}}{\left(n+\frac32\right)_{k}\,k!}\left(1-t\right)^{k}\\ &=\small{\sum_{k=0}^{\infty}\frac{\left(\frac12\right)_{k}\,\left(n+\frac12\right)_{k}\,c^{k}}{\left(n+\frac32\right)_{k}\,k!}\int_{0}^{1}\mathrm{d}t\,\frac{t^{m-\frac12}\left(1-t\right)^{\ell+n+k-\frac12}}{\left(1-at\right)^{\ell}}\,{_2F_1}{\left(\frac12,m+\frac12;m+\frac32;bt\right)}}\\ &=\small{\sum_{k=0}^{\infty}\frac{\left(\frac12\right)_{k}\,\left(n+\frac12\right)_{k}\,c^{k}}{\left(n+\frac32\right)_{k}\,k!}\int_{0}^{1}\mathrm{d}t\,\sum_{j=0}^{\infty}\binom{j+\ell-1}{j}a^{j}t^{m+j-\frac12}\left(1-t\right)^{\ell+n+k-\frac12}}\\ &~~~~~\times\,{_2F_1}{\left(\frac12,m+\frac12;m+\frac32;bt\right)}\\ &=\sum_{k=0}^{\infty}\frac{\left(\frac12\right)_{k}\,\left(n+\frac12\right)_{k}\,c^{k}}{\left(n+\frac32\right)_{k}\,k!}\sum_{j=0}^{\infty}\binom{j+\ell-1}{j}a^{j}\\ &~~~~~\times\int_{0}^{1}\mathrm{d}t\,t^{m+j-\frac12}\left(1-t\right)^{\ell+n+k-\frac12}\,{_2F_1}{\left(\frac12,m+\frac12;m+\frac32;bt\right)}\\ &=\sum_{j=0}^{\infty}\sum_{k=0}^{\infty}\frac{\binom{j+\ell-1}{j}\left(\frac12\right)_{k}\,\left(n+\frac12\right)_{k}\,a^{j}\,c^{k}}{\left(n+\frac32\right)_{k}\,k!}\,\operatorname{B}{\left(\frac12+m+j,\frac12+\ell+n+k\right)}\\ &~~~~~\times\,{_3F_2}{\left(\frac12,\frac12+m,\frac12+m+j;\frac32+m,1+\ell+m+n+j+k;b\right)}\\ &=\sum_{j=0}^{\infty}\sum_{k=0}^{\infty}\frac{\binom{j+\ell-1}{j}\left(\frac12\right)_{k}\,\left(n+\frac12\right)_{k}\,a^{j}\,c^{k}}{\left(n+\frac32\right)_{k}\,k!}\,\operatorname{B}{\left(\frac12+m+j,\frac12+\ell+n+k\right)}\\ &~~~~~\times\sum_{p=0}^{\infty}\frac{\left(\frac12\right)_{p}\,\left(\frac12+m\right)_{p}\,\left(\frac12+m+j\right)_{p}\,b^{p}}{\left(\frac32+m\right)_{p}\,\left(1+\ell+m+n+j+k\right)_{p}\,p!}.\\ \end{align}

Does anybody have any ideas for finishing the derivation, or another approach entirely? This is a difficult and time-consuming problem, but also one that I'm quite eager to crack. Any help would be greatly appreciated!

For $m,n\in\mathbb Z_{\ge0}$ the integrand can be rewritten in terms of elementary functions using that $$z^{m-\frac12}{}_2F_1\left(\frac12,m+\frac12;m+\frac32;z\right)=\frac{\left(\frac12\right)_{m+1}}{m!}\left[\frac{2\arcsin\sqrt z}{z}-\sqrt{1-z}\cdot\sum_{k=1}^m\frac{(k-1)!\,z^{k-\frac32}}{\left(\frac12\right)_{k}}\right].$$ The initial integral therefore breaks down into finite number of pieces of 3 different types - however, neither of them looks computable.