How to calculate the derivative of a Hypergeometric function?

Question

I am studying the derivative of the following Hypergeometric function, regarded as a function in the variable x, for x=0 $$ F_{2}^{1}(1/2,x/2,3/2,a) $$ where a is a generic constant. I know how to expand this function only if the variable is the last entry, but what can I do when the variable is the second one?

Answer 1

Gammatester and J.M. provided the appropriate answers in the general case (+1)
(i.e. : I fear they won't help much without being specific!)

Coincidentally I began working on your exact hypergeometric function saturday!
(the log-sine integrals involved are directly related to the central-binomial series I considered)

A way to evaluate the derivative relatively to one parameter is to start with Euler's integral representation of the hypergeometric function and compute a partial derivative of the integral : $$\tag{1}{}_2F_1\left(a,b;c;z\right)=\frac {\Gamma(c)}{\Gamma(b)\Gamma(c-b)}\int_0^1 \frac {t^{b-1}(1-t)^{c-b-1}}{(1-tz)^a}\,dt$$ But it will be more convenient to use our specific parameters $\;a=\frac x2,\;b=\frac 12,\;c=\frac 32\;$ (from the definition of ${}_2F_1\,$ the two first parameters $a$ and $b$ can be exchanged).
These parameters verify $\;c=b+1\,$ so that : \begin{align} {}_2F_1\left(a,b;b+1;z\right)&=b\int_0^1 t^{b-1}(1-tz)^{-a}\,dt\\ \tag{2}&=\frac b{z^b}\int_0^zu^{b-1}(1-u)^{-a}\,du\\ \tag{3}&=\frac b{z^b}B_z(b,1-a)\\ \end{align} with $\,B_z\,$ the incomplete beta function but we won't use it here.

The substitution of our specific $\,a\,$ and $\,b\,$ and of $\,u:=(\cos t)^2\;$ gives : $${}_2F_1\left(\frac x2,\frac 12;\frac 32;z\right)=-\frac 1{2\sqrt{z}}\int_{\arccos 0}^{\arccos \sqrt{z}}(\cos t)^{1-2}(\sin t)^{-x}\,2(\sin t)(\cos t)\,dt$$ I'll set $z:=b^2$ to avoid the square roots and define : $$\tag{4}F_b(x):={}_2F_1\left(\frac 12,\frac x2;\frac 32;b^2 \right)=\frac 1b\int_{\arccos b}^{\pi/2} (\sin t)^{1-x}\,dt$$ Since $\,\displaystyle (\sin t)^{1-x}=e^{(1-x)\ln(\sin t)}\;$ the derivative relatively to $x$ under the integral sign will be : $$\tag{5}F_b(x)':=\frac 1b\int_{\pi/2}^{\arccos b} (\sin t)^{1-x}\ln(\sin t)\,dt$$ In the case $\;x=0\;$ we obtain $\;\displaystyle F_b(0)':=\frac {I_b}b$ with $\,I_b\,$ integrable by parts : \begin{align} I_b&:=\int_{\pi/2}^{\arccos b} (\sin t)\ln(\sin t)\,dt\\ &=\left.-\cos(t)\ln(\sin t)\right|_{\pi/2}^{\arccos b} +\int_{\pi/2}^{\arccos b} (\cos t)\frac{\cos t}{\sin t}\,dt\\ &=\left.-\cos(t)\ln(\sin t)+\cos t+\ln\tan\frac t2\right|_{\pi/2}^{\arccos b}\\ &=-\frac b2\ln(1-b^2)+b+\frac 12\ln\frac {1-b}{1+b}\\ \end{align} The final result (with $\;b=\sqrt{a}\,$ for positive values of $\,a\,$) is : $$\tag{6}\boxed{F_b(0)':=1-\frac 12\ln(1-b^2)+\frac 1{2b}\ln\frac {1-b}{1+b}}$$ (the derivative becomes complex for $|b|>1\,$ and $\;\displaystyle\lim_{b\to\pm 1} F_b(0)'=1-\ln 2\;$)