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


1 Answer 1


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.

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    $\begingroup$ Perfect as usual ! Cheers, $\endgroup$ Nov 2, 2013 at 15:20
  • $\begingroup$ Why in general, it is not possible ? For example, in this document, there is a list of rule mhtlab.uwaterloo.ca/courses/me755/web_chap7.pdf $\endgroup$ Nov 23, 2016 at 19:20
  • $\begingroup$ @optimalcontrol if you are referring to the rule at the top of the second page of the linked document, that is the derivative wrt to the argument of the function. The OP is asking for the derivative wrt to one of the parameters, in particular the second one. $\endgroup$
    – Luca Citi
    Jul 19, 2017 at 18:59

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