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.
The key point is just that, through a change of variable in the integral defining the $\Gamma$ function: $$ \Gamma\left(\frac{1}{2}\right) = \int_{-\infty}^{+\infty}e^{-x^2}\,dx $$ but the integral in the RHS is well-known, it can be computed by using Fubini's theorem and polar coordinates. An alternative proof: $$ \int_{0}^{1}\sqrt{1-x^2}\,dx = \frac{\pi}{4} $$ since the LHS is the area of a quarter-circle. If we set $x=\sqrt{z}$ we have: $$ \frac{1}{2}\int_{0}^{1}z^{-1/2}(1-z)^{1/2}\,dz = \frac{1}{2}\,B\left(\frac{1}{2},\frac{3}{2}\right) = \frac{\Gamma\left(\frac{1}{2}\right)^2}{4}$$ by the properties of the Euler beta function and the identity $\Gamma(\alpha+1)=\alpha\,\Gamma(\alpha)$.