Closed form of a Hypergeometric Function ${}_2F_1$ at $z=-8$

Question

How this can be proved?

$$ E = {}_2F_1(-\frac{1}{2}, \frac{1}{3}, \frac{4}{3},-8) = \frac{6}{5} - \frac{\chi}{2} $$

where

$$ \chi = \frac{6\sqrt{\pi}}{5}\frac{\Gamma(\frac{1}{3})}{\Gamma(-\frac{1}{6})} $$

This is not resolved by WolframAlpha and is not in any of the lists in Wolfram Function n Site

I'm not too familiar with cases where the parameters are not integers and with different denominators, and also the $z$ values are not $\pm 1$

My attempts so far has been to search for a transformation to get an easier or solved case:

1. With the $z \rightarrow \frac{1}{1-z}$ transformation I got: $$ E = \frac{6}{5}\cdot {}_2F_1(-\frac{1}{2}, 1, \frac{1}{6},\frac{1}{9}) - \chi $$ that I thought was a good progress since now $b=1$.

2. I also found that

$$ {}_2F_1(-\frac{1}{2}, 1, \frac{1}{6},\frac{1}{9}) = \frac{5}{2} - \frac{4}{3} \cdot{}_2F_1(\frac{1}{2}, 1, \frac{1}{6},\frac{1}{9}) $$

and I was hoping to use the fact that $$ \big(-\frac{1}{2} \big)_n = \big(-\frac{1}{2} \big) \big(\frac{1}{2} \big)_{n-1} $$

But I couldn't find a way.

Answer 1

Set $a=1$, $b=-1/2$, $c=1/6$, and $z=1/9$. We wish to show that $$F(a,b;c;z)=1+{\sqrt\pi\over2}{\Gamma(1/3)\over\Gamma(-1/6)}.$$ By DLMF 15.5.16_5, we have $$\begin{align*} F(a,b;c;z) &=F(a-1,b;c;z)+{b\over c}zF(a,b+1;c+1;z)\\ &=F(0,b;c;z)+{b\over c}zF(1,1/2;7/6;1/9)\\ &=1+{b\over c}zF(1,1/2;7/6;1/9). \end{align*}$$ By DLMF 15.4.32 with parameter $1/2$, we have $$F(1,1/2;7/6;1/9)={\sqrt{3\pi}\over2}{\Gamma(7/6)\over\Gamma(2/3)}.$$ Thus $$F(a,b;c;z)=1-{1\over3}{\sqrt{3\pi}\over2}{\Gamma(7/6)\over\Gamma(2/3)},$$ so it remains to be shown that $$-{\sqrt3\over3}\Gamma(7/6)\Gamma(-1/6)=\Gamma(1/3)\Gamma(2/3).$$ But this follows from Euler’s reflection formula $$\Gamma(z)\Gamma(1-z)={\pi\over\sin\pi z},$$ so we are done.