<s>[INST] 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})}$? While reading physics papers I found a very interesting integral so I decided to write it down.  Let $p(z) = z^ 3 - 3\Lambda^ 2 z$ where $\Lambda$ could be any number.  If you want $\Lambda = 1$ and $p(z) = z^ 3 - 3z$.  Then
$$ \int_0^{\sqrt{3}\Lambda} \frac{dz}{\sqrt{p(z)}} = \int_0^{\sqrt{3}\Lambda} \frac{dz}{\sqrt{z(z - \sqrt{3}\Lambda\,)(z+\sqrt{3}\Lambda\,)}} = \frac{\sqrt{\pi}}{2} \frac{\Gamma(\frac{3}{4})}{\Gamma(\frac{9}{4})} (\sqrt{3}\Lambda)^{5/2}$$
The physics paper at least tell us $\int \propto \Lambda^{5/2}$ so we know the growth rate. It could be that $\Lambda = \frac{1}{\sqrt{3}}$ is even simpler than $\Lambda = 1$. 
$$  \int_0^1 \frac{dz}{\sqrt{z(z - 1\,)(z+1\,)}} = \frac{\sqrt{\pi}}{2} \frac{\Gamma(\frac{3}{4})}{\Gamma(\frac{9}{4})}$$
This should be connected to the Riemann surface $y^ 2 = z(z^2 -1)$.  And we are computing a period of the Riemann surface.

Contour integration is really important here.  Checking on Wolfram Alpha gives a different answer:
$$ \int_0^1 \frac{dz}{x(x^2-1)} = - 2i\sqrt{\pi}\,\frac{ \Gamma(\frac{5}{4})}{\Gamma(\frac{3}{4})}$$
I am guessing this is either the same number or Mathematica is choosing a different contour.  Physicists love contour integrals (taken from physics.stackexchange):

 [/INST] Since:
$$\int_{0}^{1}x^{\alpha-1}(1-x)^{\beta-1}\,dx = B(\alpha,\beta) = \frac{\Gamma(\alpha)\Gamma(\beta)}{\Gamma(\alpha+\beta)}\tag{1}$$
through a change of variable we get:
$$ \int_{0}^{1} z^{-1/2}(1-z^2)^{-1/2}\,dz = \frac{1}{2}\int_{0}^{1}z^{1/4}(1-z)^{-1/2}\,dz = \frac{\Gamma\left(\frac{5}{4}\right)\Gamma\left(\frac{1}{2}\right)}{2\cdot \Gamma\left(\frac{7}{4}\right)}\tag{2} $$
and by using the reflection formulas for the $\Gamma$ function that simplifies to:

$$ \int_{0}^{1} \frac{dz}{\sqrt{z(1-z)(1+z)}} = \color{red}{\frac{\Gamma\left(\frac{1}{4}\right)^2}{2\sqrt{2\pi}}}.\tag{3}$$

</s>