I am trying to convert $B(13/3, 11/3)$ into a Gamma function. I was hoping to reduce to $\Gamma(1/3 )$ for which there is a fixed value, but of course one reduces to $(6930/243) \Gamma(2/3)$ and the other $(630/81)\Gamma(2/3)$.
Can anyone advise? Euler's reflection formula looks like it might work, but I can't get it.
$$\begin{align}B\left(\frac{13}{3}, \frac{11}{3}\right)& = \frac{\Gamma(\frac{13}{3}) \cdot \Gamma(\frac{11}{3})}{\Gamma(\frac{13}{3} + \frac{11}{3})} \\& = \frac{\Gamma(1 + \frac{10}{3}) \cdot \Gamma(1 + \frac{8}{3})}{\Gamma(\frac{24}{3})} \\& = \frac{\Gamma(1 + \frac{10}{3})\cdot \Gamma(1 + \frac{8}{3})}{\Gamma(8)} \\&= \frac{\frac{10}{3}\Gamma(\frac{10}{3})\cdot \frac83\Gamma(\frac{8}{3})}{7!}\\&= \frac{\frac{10}{3}\Gamma(1 + \frac{7}{3}) + \frac83\Gamma(1 + \frac{5}{3})}{5040} \\&= \frac{\frac{10}{3}\cdot \frac73\Gamma(\frac{7}{3}) \cdot\frac83\cdot \frac53\Gamma(\frac{5}{3})}{5040} \\&= \frac{\frac{10}{3}\cdot \frac73\Gamma(1 + \frac{4}{3}) \cdot \frac83\cdot \frac53\Gamma(1 + \frac{2}{3})}{5040} \\&= \frac{\frac{10}{3}\cdot \frac43 \frac73\Gamma(\frac{4}{3}) \cdot \frac83\cdot \frac53\cdot \frac23\Gamma( \frac{2}{3})}{5040} \\&= \frac{\frac{10}{3}\cdot \frac43 \cdot\frac73\Gamma(1 + \frac{1}{3}) \cdot \frac83\cdot \frac53\cdot \frac23\Gamma( \frac{2}{3})}{5040} \\&= \frac{\frac{10}{3}\cdot \frac43 \cdot\frac73\cdot\frac13\Gamma(\frac{1}{3}) \cdot \frac83\cdot \frac53\cdot \frac23\Gamma( \frac{2}{3})}{5040}\end{align}$$
Now put the value of $\Gamma(\frac13) \cdot \Gamma(\frac23) = \frac{2}{\sqrt{3}}\pi$ and you will be done!
Gamma-beta relationship has been used in the first step.
In the further steps, I've used the result $\Gamma( 1+ n) = n\Gamma(n)$