Solution to this beta function integral-within-an-integral (Continued) Continuing on from my earlier post, here, I have re-evaluated all the work-up to the integral and managed to re-formulate everything to be slightly different. Now, the aim is to solve
$$
I = \int_0^\infty \frac{t^{\frac{3}{\alpha}-1}}{(1+t)^8}\left(\int_0^t
\frac{t'^{\frac{2}{\alpha}-1}}{(1+t')^8}dt'\right)d t.$$
Once again, we have the condition that $2 \leq \alpha \leq 4$, where $\alpha$ is real. I'm not sure if the same problem stands that the inside integral diverges if $\alpha < 1/2$, as in my previous question or if this has managed to help somehow as the exponent is different.
Thanks!
Edit:
This once again can be written out as the integral of a hypergeometric function,
$$
I = \frac{\alpha}{2}\int_0^\infty \frac{t^{\frac{5}{\alpha}-1}}{(1+t)^8}\space_2F_1(8,\frac{2}{\alpha};\frac{\alpha+2}{\alpha};-t) \space dt.
$$

If $\operatorname{Re}(a) > 0$ and $- 1 < \operatorname{Re}(b) < \min (\operatorname{Re}(c)+7,\operatorname{Re}(a) + \operatorname{Re}(c) - 1)$, then
\begin{align*}
& \int_0^{ + \infty } {\frac{{t^b }}{{(1 + t)^c }}{}_2F_1 \left( {8,a;a + 1; - t} \right)dt}  = \int_0^{ + \infty } {\frac{{t^b }}{{(1 + t)^{c + 8} }}{}_2F_1 \!\left( {8,1;a + 1;\frac{t}{{t + 1}}} \right)dt} 
\\ & = \frac{{\Gamma (a + 1)}}{{7!}}\sum\limits_{n = 0}^\infty  {\frac{{\Gamma (8 + n)}}{{\Gamma (a + 1 + n)}}\int_0^{ + \infty } {\frac{{t^{b + n} }}{{(1 + t)^{c + 8 + n} }}dt} } 
\\ & = \frac{{\Gamma (a + 1)\Gamma (c - b + 7)}}{{7!}}\sum\limits_{n = 0}^\infty  {\frac{{\Gamma (8 + n)\Gamma (b + 1 + n)}}{{\Gamma (a + 1 + n)\Gamma (c + 8 + n)}}} 
\\ & = \frac{{\Gamma (b + 1)\Gamma (c - b + 7)}}{{\Gamma (8 + c)}}{}_3F_2 (1,8,b + 1;a + 1,c + 8;1).
\end{align*}
In your case $a=\frac{2}{\alpha}$, $b=\frac{5}{\alpha}-1$, $c=8$ and $2\leq\alpha\leq 4$ fulfill the conditions.
