How to solve this multiple integral of Hypergeometric function? Sorry for the typeset in the previous post, could you please help me with this integral?
kind regards
Ara
$$ \int _0^{2\pi }d\phi \int _0^{1}du\left(u\cos^2(\phi )-1\right) \,_2F_1\left(1,\Delta +3,5/2,\cos^2(\alpha )\left(1-u\cos^2(\phi )\right)\right) $$
 A: Let $t = \tan\phi$, $v = u\cos^2\phi$ and abbreviate
$\,_2F_1\left(1,\Delta +3;5/2;\cos^2(\alpha)x\right)$ as $F(x)$, we can rewrite the
integral $\mathscr{I}$ as
$$\begin{align}
\mathscr{I} 
= &-\int_0^{2\pi}\frac{d\phi}{\cos^2\phi}\int_0^{\cos^2\phi} dv\,(1-v)F(1-v)\\
= & -4\int_0^{\pi/2}\frac{d\phi}{\cos^2\phi}\int_0^{\cos^2\phi} dv\,(1-v)F(1-v)\\
= & -4\int_0^{\infty} dt \int_0^{\frac{1}{1+t^2}} dv\,(1-v)F(1-v)\\
= & -4\int_0^1 dv \int_0^{\sqrt{\frac{1}{v}-1}} dt\,(1-v)F(1-v)\\
= & -4\int_0^1 (1-v)^{3/2} v^{-\frac12} F(1-v) dv
\end{align}$$
Let $(x)_k$ be the rising Pochhammer symbol  $x(x+1)\cdots(x+k-1)$. We can expand 
$$F(x) \quad\text{ as }\quad 
\sum_{k=0}^{\infty} \alpha_k x^k \quad\text{ where }\quad
\alpha_k = \frac{(1)_k(\Delta+3)_k}{(\frac52)_k k!} \cos^{2k}(\alpha)$$
If we integrate this expansion of $F(x)$ term by term, we get
$$\begin{align}
\mathscr{I} 
= & -4 \sum_{k=0}^{\infty} \alpha_k \int_0^1 (1-v)^{k+\frac32}v^{-\frac12} dv\\
= & -4 \sum_{k=0}^{\infty} \alpha_k \frac{\Gamma(k+\frac52)\Gamma(\frac12)}{\Gamma(k+3)}\\
= & -\frac{3\pi}{2} \sum_{k=0}^{\infty} \alpha_k \frac{(\frac52)_k}{(3)_k}\\
= & -\frac{3\pi}{2} \sum_{k=0}^{\infty} \frac{(1)_k(\Delta+3)_k}{(3)_k k!} 
\cos^{2k}(\alpha)\\
= & -\frac{3\pi}{2} \!\,_2F_1(1,\Delta+3;\,3;\,\cos^2(\alpha))\\
= & -\frac{3\pi}{(\Delta+1)(\Delta+2)\cos^4(\alpha)}
\left[
    \frac{1}{\sin^{2(\Delta+1)}(\alpha)} - 1 - (\Delta+1)\cos^2(\alpha)
\right]
\end{align}$$
