How to calclulate a derivate of a hypergeometric function w.r.t. one of its parameters? Is it possible to take a derivative of a hypergeometric function w.r.t. one of its parameters and express it in a closed form? 
I am particularly interested in this case:
$$\large\left[\frac{d}{da}{_2F_1}\left(1/2,\,a;\,3/2;\,-1\right)\right]_{a=2}$$
 A: In general the answer is no. 
In the case at hand, however, the parameters are special and this becomes possible. One can use, for instance, the standard integral representation of the hypergeometric function to show that
\begin{align}
_2F_1\left(\frac12,a,\frac32,-1\right)=\frac12\int_0^1\frac{dt}{\sqrt{t}\left(1+t\right)^{a}},
\end{align}
which in turn yields
\begin{align}
\mathcal{I}=\left[\frac{d}{da} {}_2F_1\left(\frac12,a,\frac32,-1\right)\right]_{a=2}=
-\frac12\int_0^1\frac{\ln\left(1+t\right)dt}{\sqrt{t}\left(1+t\right)^{2}}.
\end{align}
The last integral can be expressed in terms of dilogarithms (e.g. after the change of variables $t=s^2$):
\begin{align}
\mathcal{I}&=-\int_0^1\frac{\ln\left(1+s^2\right)ds}{\left(1+s^2\right)^2}=\\&=
\frac{\pi}{8}\left[1-3\ln 2 +\ln\left(2+\sqrt{2}\right)\right]-\frac{1+\ln 2}{4}+\Im\left(\operatorname{Li}_2\left(-e^{i\pi/4}\right)-\operatorname{Li}_2\left(1-e^{i\pi/4}\right)\right)=\\
&=\frac{\pi\left(1-2\ln 2\right)}{8}-\frac{1+\ln 2}{4}+\frac12 \Im\operatorname{Li}_2\left(i\right)=\\
&=\frac{G}{2}+\frac{\pi\left(1-2\ln 2\right)}{8}-\frac{1+\ln 2}{4},
\end{align}
where $G$ denotes the Catalan's constant.
