Proving $\Gamma (\frac{1}{2}) = \sqrt{\pi}$ There are already proofs out there, however, the problem I am working on requires doing the proof by using the gamma function with $y = \sqrt{2x}$ to show that $\Gamma(\frac{1}{2}) = \sqrt{\pi}$.
Here is what I have so far:
$\Gamma(\frac{1}{2}) = \frac{1}{\sqrt{\beta}}\int_0^\infty x^{-\frac{1}{2}}e^{-\frac{x}{\beta}} \, dx.$
Substituting $x = \frac{y^2}{2}\, dx = y\,dy$, we get $\Gamma(\frac{1}{2}) = \frac{\sqrt{2}}{\sqrt{\beta}}\int_0^\infty e^{-\frac{y^2}{2\beta}} \, dy$.
At this point, the book pulled some trick that I don't understand.

Can anyone explain to me what the book did above? Thanks.

It's using the Gaussian integral or the fact that the area under normal density is 1. Since you have the statistics tag, I'm guessing the latter should be familiar already.
