Show that $\frac{\Gamma(\frac 1 3)^2}{\Gamma(\frac 1 6)}=\frac{\sqrt {\pi}\sqrt[3] 2}{\sqrt 3}$.

Question

Show that $$\frac{\Gamma(\frac 1 3)^2}{\Gamma(\frac 1 6)}=\frac{\sqrt {\pi}\sqrt[3] 2}{\sqrt 3}$$

Since there's $\sqrt {\pi}$, I suspect I have to related it to $\Gamma(1/2)$. Please give me some idea.

Answer 1

It's the Legendre duplication formula

$$\Gamma (2z) = \frac{2^{2z-1}}{\sqrt{\pi}}\Gamma(z)\Gamma\left(z+\tfrac{1}{2}\right),$$

and Euler's reflection formula

$$\Gamma(z)\Gamma(1-z) = \frac{\pi}{\sin (\pi z)}.$$

Thus we have

$$\Gamma\left(\tfrac{1}{3}\right)^2 = \underbrace{\frac{2^{-2/3}}{\sqrt{\pi}}\Gamma\left(\tfrac{1}{6}\right)\Gamma\left(\tfrac{4}{6}\right)}_{\text{duplication } \Gamma\left(\tfrac{1}{3}\right)}\cdot\Gamma\left(\tfrac{1}{3}\right) = \frac{2^{-2/3}}{\sqrt{\pi}}\Gamma\left(\tfrac{1}{6}\right) \frac{\pi}{\sin \frac{\pi}{3}} = \frac{2^{1/3}\sqrt{\pi}}{\sqrt{3}} \cdot\Gamma\left(\tfrac{1}{6}\right).$$