How to find the following Integral I am unable to get the following integral. I know the basics of integration. I have tried looking it up but to no avail.
$\int_0^\infty x^{-\frac{1}{2}}e^{-\frac{x}{2}}\,dx$
Thanks for the help.
 A: I assume the integral you are after is
$$I = \int_0^{\infty} \dfrac{e^{-x/2}}{\sqrt{x}}dx$$
Let $\sqrt{x} = t$, we then have $x=t^2 \implies dx = 2t dt$. Hence, we get that
$$I = \int_0^{\infty} \dfrac{e^{-t^2/2} 2tdt}{t} = 2 \int_0^{\infty}e^{-t^2/2}dt = 2 \times \sqrt{\dfrac{\pi}2} = \sqrt{2 \pi}$$
A: Take $x=w^{2}$ so that$\,dx=2wdw$ then we get:
$$\int_{0}^{\infty}\frac{1}{w}e^{-\frac{w^{2}}{2}}(2w)dw=2\int_{0}^{\infty}e^{-(\frac{w}{\sqrt{2}})^{2}}dw$$
Taking $u=\frac{w}{\sqrt{2}}$ so that $\,du=\frac{1}{\sqrt{2}}dw$ gives:
$$2\sqrt{2}\int_{0}^{\infty}e^{-u^{2}}du=2\sqrt{2}\frac{\sqrt{\pi}}{2}=\color{blue}{\sqrt{2\pi}}$$.
If you wish to look up information on integrals like this go here
A: $$
\int_{0}^{\infty}x^{-1/2}{\rm e}^{-x/2}\,{\rm d}x
=
\sqrt{\vphantom{\large A}2\,}\int_{0}^{\infty}x^{1/2\ -\ 1}\,{\rm e}^{-x}
\,{\rm d}x
=
\sqrt{\vphantom{\large A}2\,}
\quad
\overbrace{\ \Gamma\left(1 \over 2\right)\ }^{\sqrt{\vphantom{\large A}\pi\,}}
=
\color{#ff0000}{\large\sqrt{\vphantom{\Large A}2\pi\,}}
$$
