On page $46$ of this paper, the authors say that
By writing the Gauss Hypergeometric function in series, integrating and summing again we find the following expression
$$\int_0^\infty \frac1{(1+z)^{2+\gamma}} \ _2F_1\left(1+\frac\gamma2,\frac32;3;\frac{4z}{(1+z)^2}\right)\,dz=\frac{\Gamma\left(-\frac\gamma2\right)\Gamma\left(\frac{\gamma-1}2\right)}{\Gamma\left(\frac{1-\gamma}2\right)\Gamma\left(\frac\gamma2\right)}.$$
I can do the first two steps: writing the Gauss Hypergeometric function in series and integrating, but I don't know how to "sum again" to get the neat result. Let me first explain the notations. Here $\gamma\in(0,1)$ is a given parameter and $\Gamma$ is the Gamma function. The Gauss Hypergeometric function is defined for $|z|<1$ by the series
$$_2F_1(a,b;c;z)=\sum_{n=0}^\infty\frac{(a)_n(b)_n}{(c)_n}\frac{z^n}{n!},$$
where the (rising) Pochhammer symbol $(x)_n$ is defined by
$$(x)_n=\begin{cases} 1, & n=0,\\ x(x+1)\cdots(x+n-1), & n\geq1.\end{cases}$$
It follows from definition that $(x)_n=\Gamma(x+n)/\Gamma(x)$ for $x>0$.
My calculation. By definition,
\begin{align*}
_2F_1\left(1+\frac\gamma2,\frac32;3;\frac{4z}{(1+z)^2}\right)&=\sum_{n=0}^\infty \frac{\left(1+\frac\gamma2\right)_n\left(\frac32\right)_n}{(3)_n}\frac1{n!}\left(\frac{4z}{(1+z)^2}\right)^n\\
&=\sum_{n=0}^\infty \frac{\Gamma\left(1+\frac\gamma2+n\right)}{\Gamma\left(1+\frac\gamma2\right)}\frac{\Gamma\left(\frac32+n\right)}{\Gamma\left(\frac32\right)}\frac{\Gamma\left(3\right)}{\Gamma\left(3+n\right)}\frac{4^n}{n!}\frac{z^n}{(1+z)^{2n}}.
\end{align*}
Now, we do the integration. Change the variable $z=\tan^2\theta$ gives that $\frac{dz}{d\theta}=2\tan\theta\frac1{\cos^2\theta}$ and thus
\begin{align*}
\int_0^\infty \frac{z^n}{(1+z)^{2+\gamma+2n}}\,dz&=2\int_0^{\frac\pi2}\frac{\tan^{2n}\theta}{(1+\tan^2\theta)^{2+\gamma+2n}}\tan\theta\frac1{\cos^2\theta}\,d\theta\\
&=2\int_0^{\frac\pi2}\cos^{2n+2\gamma+1}\theta\sin^{2n+1}\theta\,d\theta\\
&=B(n+\gamma+1,n+1)=\frac{n!\Gamma(n+\gamma+1)}{\Gamma(2n+\gamma+2)}.
\end{align*}
Therefore,
$$\int_0^\infty \frac1{(1+z)^{2+\gamma}} \ _2F_1\left(1+\frac\gamma2,\frac32;3;\frac{4z}{(1+z)^2}\right)\,dz=\sum_{n=0}^\infty \frac{\Gamma\left(1+\frac\gamma2+n\right)}{\Gamma\left(1+\frac\gamma2\right)}\frac{\Gamma\left(\frac32+n\right)}{\Gamma\left(\frac32\right)}\frac{2\cdot 4^n}{(n+2)!}\frac{\Gamma(n+\gamma+1)}{\Gamma(2n+\gamma+2)}.$$
I can do a little more cancellation between the the numerator and the denominator. But I still don't know how to calculate the sum even after the simplication.
Any help would be appreciated!
Best Answer
If the series manipulations are giving you trouble or if you just would rather avoid them for whatever reason, it is in fact possible to do the calculation entirely with integrals.
Given $(a,b,c,z)\in\mathbb{R}^{4}$ such that $0<b<c\land z\le1$, Euler's integration formula for the Gauss hypergeometric function states
$$\int_{0}^{1}\mathrm{d}t\,\frac{t^{b-1}\left(1-t\right)^{c-b-1}}{\left(1-zt\right)^{a}}=\operatorname{B}{\left(b,c-b\right)}\,{_2F_1}{\left(a,b;c;z\right)};~~~\small{a+b-c<0\lor z<1}.$$
Suppose $0<\gamma<1$. We find
$$\begin{align} \mathcal{I}{\left(\gamma\right)} &=\int_{0}^{\infty}\mathrm{d}x\,\frac{1}{\left(1+x\right)^{2+\gamma}}\,{_2F_1}{\left(1+\frac{\gamma}{2},\frac32;3;\frac{4x}{\left(1+x\right)^{2}}\right)}\\ &=\int_{0}^{\infty}\mathrm{d}x\,\frac{1}{\left(1+x\right)^{2+\gamma}}\cdot\frac{1}{\operatorname{B}{\left(\frac32,\frac32\right)}}\int_{0}^{1}\mathrm{d}t\,\frac{\sqrt{t\left(1-t\right)}}{\left[1-\frac{4x}{\left(1+x\right)^{2}}t\right]^{1+\frac{\gamma}{2}}}\\ &=\frac{1}{\operatorname{B}{\left(\frac32,\frac32\right)}}\int_{0}^{\infty}\mathrm{d}x\int_{0}^{1}\mathrm{d}t\,\frac{1}{\left(1+x\right)^{2+\gamma}}\cdot\frac{\sqrt{t\left(1-t\right)}}{\left[1-\frac{4x}{\left(1+x\right)^{2}}t\right]^{1+\frac{\gamma}{2}}}\\ &=\frac{8}{\pi}\int_{0}^{\infty}\mathrm{d}x\int_{0}^{1}\mathrm{d}t\,\frac{1}{\left[\left(1+x\right)^{2}\right]^{1+\frac{\gamma}{2}}}\cdot\frac{\sqrt{t\left(1-t\right)}}{\left[1-\frac{4x}{\left(1+x\right)^{2}}t\right]^{1+\frac{\gamma}{2}}}\\ &=\frac{8}{\pi}\int_{0}^{\infty}\mathrm{d}x\int_{0}^{1}\mathrm{d}t\,\frac{\sqrt{t\left(1-t\right)}}{\left[\left(1+x\right)^{2}-4xt\right]^{1+\frac{\gamma}{2}}}\\ &=\frac{8}{\pi}\int_{0}^{\infty}\mathrm{d}x\int_{0}^{1}\mathrm{d}t\,\frac{\sqrt{t\left(1-t\right)}}{\left[x^{2}-2(2t-1)x+1\right]^{1+\frac{\gamma}{2}}}\\ &=\frac{2}{\pi}\int_{0}^{\infty}\mathrm{d}x\int_{-1}^{1}\mathrm{d}u\,\frac{\sqrt{1-u^{2}}}{\left(x^{2}-2ux+1\right)^{1+\frac{\gamma}{2}}};~~~\small{\left[t=\frac{u+1}{2}\right]}\\ &=\frac{2}{\pi}\int_{-1}^{1}\mathrm{d}u\int_{0}^{\infty}\mathrm{d}x\,\frac{\sqrt{1-u^{2}}}{\left(x^{2}-2ux+1\right)^{1+\frac{\gamma}{2}}}\\ &=\frac{1}{\pi}\int_{-1}^{1}\mathrm{d}u\int_{0}^{\infty}\mathrm{d}x\,\frac{\sqrt{1-u^{2}}}{\left(x^{2}-2ux+1\right)^{1+\frac{\gamma}{2}}}\\ &~~~~~+\frac{1}{\pi}\int_{-1}^{1}\mathrm{d}u\int_{-\infty}^{0}\mathrm{d}x\,\frac{\sqrt{1-u^{2}}}{\left(x^{2}-2ux+1\right)^{1+\frac{\gamma}{2}}};~~~\small{\left[(x,u)\mapsto(-x,-u)\right]}\\ &=\frac{1}{\pi}\int_{-1}^{1}\mathrm{d}u\int_{-\infty}^{\infty}\mathrm{d}x\,\frac{\sqrt{1-u^{2}}}{\left(x^{2}-2ux+1\right)^{1+\frac{\gamma}{2}}}\\ &=\frac{1}{\pi}\int_{-1}^{1}\mathrm{d}u\int_{-\infty}^{\infty}\mathrm{d}y\,\frac{\sqrt{1-u^{2}}}{\left(y^{2}+1-u^{2}\right)^{1+\frac{\gamma}{2}}};~~~\small{\left[x=y+u\right]}\\ &=\frac{1}{\pi}\int_{-1}^{1}\mathrm{d}u\int_{-\infty}^{\infty}\mathrm{d}z\,\frac{\left(1-u^{2}\right)}{\left[\left(1-u^{2}\right)z^{2}+1-u^{2}\right]^{1+\frac{\gamma}{2}}};~~~\small{\left[y=z\sqrt{1-u^{2}}\right]}\\ &=\frac{1}{\pi}\int_{-1}^{1}\mathrm{d}u\int_{-\infty}^{\infty}\mathrm{d}z\,\frac{1}{\left(1-u^{2}\right)^{\frac{\gamma}{2}}\left(1+z^{2}\right)^{1+\frac{\gamma}{2}}}\\ &=\frac{4}{\pi}\int_{0}^{1}\mathrm{d}u\int_{0}^{\infty}\mathrm{d}z\,\frac{1}{\left(1-u^{2}\right)^{\frac{\gamma}{2}}\left(1+z^{2}\right)^{1+\frac{\gamma}{2}}}\\ &=\frac{4}{\pi}\int_{0}^{1}\mathrm{d}v\,\frac{1}{2\sqrt{v}}\int_{0}^{\infty}\mathrm{d}w\,\frac{1}{2\sqrt{w}}\cdot\frac{1}{\left(1-v\right)^{\frac{\gamma}{2}}\left(1+w\right)^{1+\frac{\gamma}{2}}};~~~\small{\left[u=\sqrt{v}\land z=\sqrt{w}\right]}\\ &=\frac{1}{\pi}\int_{0}^{1}\mathrm{d}v\,v^{-1/2}\left(1-v\right)^{-\frac{\gamma}{2}}\int_{0}^{\infty}\mathrm{d}w\,\frac{\left(\frac{1}{1+w}\right)^{\frac32+\frac{\gamma}{2}}}{\left(\frac{w}{1+w}\right)^{1/2}}\\ &=\frac{1}{\pi}\int_{0}^{1}\mathrm{d}v\,v^{-1/2}\left(1-v\right)^{-\frac{\gamma}{2}}\int_{0}^{1}\mathrm{d}t\,\frac{\left(1-t\right)^{-\frac12+\frac{\gamma}{2}}}{t^{1/2}};~~~\small{\left[w=\frac{t}{1-t}\right]}\\ &=\frac{1}{\pi}\int_{0}^{1}\mathrm{d}v\,v^{1/2-1}\left(1-v\right)^{1-\frac{\gamma}{2}-1}\int_{0}^{1}\mathrm{d}t\,t^{1/2-1}\left(1-t\right)^{\frac12+\frac{\gamma}{2}-1}\\ &=\frac{1}{\pi}\operatorname{B}{\left(\frac12,1-\frac{\gamma}{2}\right)}\,\operatorname{B}{\left(\frac12,\frac12+\frac{\gamma}{2}\right)}.\blacksquare\\ \end{align}$$
Cheers. =)