I want to prove that $\Gamma(\frac{1}{2})=\sqrt{\pi}$ using the probability density function given by ($X$ is a random random variable)
$$f_{X}(x)=\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}$$
From where $x\in \mathbb{R}$
Note that $\Gamma(1/2) = \int_{0}^{\infty}t^{-1/2}e^{-t}dt$ and if we take $t=\frac{z^2}{2}$ we have that
$$\Gamma(1/2)=\int_{0}^{\infty}\sqrt{2}e^{\frac{-z^{2}}{2}}dz=2\sqrt{\pi}\int_{0}^{\infty}\frac{1}{\sqrt{2 \pi}}e^{\frac{-z^{2}}{2}} dz$$
Apparently this last integral should give $\frac{1}{2}$, note that the integrand is just $f_{X}(x)$ should this influence the calculation of the integral?
but it is not clear to me why, any suggestions? help? thank you!
First recall that the support of the Standard Normal Distribution is $(-\infty,+\infty)$ and also integrating a PDF over it's entire support is equal to 1 since it's the entire area under the PDF.
So in this case we are integrating only on half the support and considering that the Standard Normal Distribution is symmetric. So the result of the last integral is only $\frac 12$
\begin{align} \Gamma(1/2)&=2\sqrt{\pi}\int_{0}^{\infty}\frac{1}{\sqrt{2 \pi}}e^{\frac{-z^{2}}{2}} dz\\ &=2\sqrt{\pi}\cdot{\frac12}\\ &=\sqrt \pi \end{align}
TL;DR It's because it is half the CDF of Standard Normal Distribution over it's entire support.