How do I get $ \int_0^1 \frac{dz}{\sqrt{z(z - 1\,)(z+1\,)}} = \frac{\sqrt{\pi}}{2} \frac{\Gamma(\frac{3}{4})}{\Gamma(\frac{9}{4})}$?
Posted by cactus314, at math.stackexchange.com,
While reading physics papers I found a very interesting integral so I decided to write it down. Let $p(z) = z^ 3 - 3\Lambda^ 2…