Closed form for derivative $\frac{d}{d\beta}\,{_2F_1}\left(\frac13,\,\beta;\,\frac43;\,\frac89\right)\Big|_{\beta=\frac56}$

Question

As far as I know, there is no general way to evaluate derivatives of hypergeometric functions with respect to their parameters in a closed form, but for some particular cases it may be possible. I am interested in this case: $$\mathcal{D}=\frac{d}{d\beta}\,{_2F_1}\left(\frac13,\,\beta;\,\frac43;\,\frac89\right)\Bigg|_{\beta=\frac56}\tag1$$ Could you suggest how to evaluate $\mathcal{D}$ in a closed form, if possible?

Answer 1

Using Euler-type integral representation for the Gauss's function: $$ {}_2F_1(a,b; c; z) = \frac{\Gamma(c)}{\Gamma(b) \Gamma(c-b)} \int_0^1 u^{b-1} (1-u)^{c-b-1} (1-z u)^{-a} \mathrm{d}u $$ for $c = \tfrac{4}{3}$ and $b=\tfrac{1}{3}$, differentiating with respect to $a$ at $a=\tfrac{5}{6}$: $$ \left.\frac{\mathrm{d}}{\mathrm{d}a} {}_2F_1(a,\tfrac{1}{3}; \tfrac{4}{3}; \tfrac{8}{9}) \right|_{a =\tfrac{5}{6}} = -\frac{1}{3} \int_0^1 \frac{\log\left(1- \tfrac{8}{9} u\right)}{\left(1- \tfrac{8}{9} u\right)^{5/6} u^{2/3}} \mathrm{d}u \tag{1} $$ The integral can be evaluated using Mellin convolution techniques for two functions $G_1(u) = \log\left(1- \tfrac{8}{9} u\right) \mathbf{1}_{0<u<1}$, and $G_2(u) = \left(1- \tfrac{8}{9} u\right)^{-5/6} \mathbf{1}_{0<u<1}$.

Asking Mathematica to evaluate the integral $(1)$ the answer comes out in closed form:

9 3^(1/6) Gamma[7/6] Gamma[4/3] (Pi + Sqrt[3] Log[3])/(2 Sqrt[Pi]) - 
 2 3^(1/3) (3 HypergeometricPFQ[{1/6, 1/6, 2/3}, {7/6, 7/6}, 1/9] + 
    Hypergeometric2F1[1/6, 2/3, 7/6, 1/9] Log[3])