# 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: $$\large\left[\frac{d}{da}{_2F_1}\left(1/2,\,a;\,3/2;\,-1\right)\right]_{a=2}$$

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.