$ \int_0^\infty (1+t^2)^{-s} (1+it)^{s'} 2t \; d t.$ The following integral bothers me since weeks:
$$
\int_{0}^{\infty}\left(\,1 + t^{2}\,\right)^{\large -s}\,
\left(\,1 + \,\mathrm{i}t\,\right)^{\large s'}\,2t\,\mathrm{d}t
$$ 
Has any body a suggestion for this integral ?.
$\Re s > 0$ sufficiently large and $s'$ an integer, but I doubt that restricting $s'$ to be an integer brings anymore insights. It is obviously analytic in variable $s$ and $s'$, so maybe even large enough real numbers are sufficient.
For $s'$ an intger, on can expand
$$
\sum_{r = 0}^{s'}\frac{s'!}{r!\left(\,s'- r\,\right)!}
\int_{0}^{\infty}\left(\,1 + t^{2}\,\right)^{-s}\,\left(\,\mathrm{i}t\,\right)^{r}\,2t\,\mathrm{d}t
$$
A variable transformation $x = t^{2}$ gives with the Beta function: 
$$
\mathrm{i}^{r}\int_{0}^{\infty}\left(\,1 + t\,\right)^{-s}\,t^{r/2}\,\mathrm{d}t =
\mathrm{i}^{r}\,\,\mathrm{B}\left(\,s - {r \over 2} - 1,{r \over 2} + 1\,\right).
$$
For my application, I can get rid of the odd $r$'s, obtaining an alternating sum. I was hoping for some closed form for the sum. Actually, I am expecting one quotient of Gamma-factors, not a sum of quotients.
 A: Mathematica spits out lots of hypergeometric junk which I won't reproduce here unless requested. A nice way to visualise the solution might be the relevant Barnes integral equivalent (which is sort of a quotient of gamma factors), introduce a parameter $a$ \begin{equation}
I(a,s,s') = \int_0^\infty (1+t^2)^{-s} (1+ait)^{s'} 2t \; d t.
\end{equation}
take a Mellin transform with respect to $a$ 
$$
\mathcal{M}_a [I(a,s,s')](g) = \int_0^\infty 2t(1+t^2)^{-s} \underbrace{\int_0^\infty (1+ait)^{s'}\; a^{g-1} da}_{Q} \; dt
$$
where $$
Q=\frac{\Gamma(g)\Gamma(-g-s')}{(it)^{g}\Gamma(-s')}
$$
Then try $$
\mathcal{M}_a [I(a,s,s')](g) = \frac{\Gamma(g)\Gamma(-g-s')}{\Gamma(-s')}\int_0^\infty \frac{2t(1+t^2)^{-s}}{(it)^{g}}  \; dt
$$
giving $$
\mathcal{M}_a [I(a,s,s')](g) = \frac{\Gamma(g)\Gamma(-g-s')\Gamma(1-\frac{g}{2})\Gamma(\frac{g}{2}+s-1)}{i^g\Gamma(s)\Gamma(-s')}
$$
then take the inverse Mellin transform and remember to take $a \to 1$
$$
I(a,s,s')=\frac{1}{\Gamma(s)\Gamma(-s')}\frac{1}{2\pi i}\int_{-i\infty}^{i\infty} \Gamma(-g-s')\Gamma(1-\frac{g}{2})\Gamma(\frac{g}{2}+s-1)\Gamma(g)\frac{1}{(ia)^g}\;dg
$$
Let's compare to $$
_2F_1(a,b;c;z) =\frac{\Gamma(c)}{\Gamma(a)\Gamma(b)} \frac{1}{2\pi i} \int_{-i\infty}^{i\infty} \frac{\Gamma(a+s)\Gamma(b+s)\Gamma(-s)}{\Gamma(c+s)}(-z)^s\,ds
$$
