How to calculate the definite integral $\int\limits_0^\infty \sqrt{x} e^{-x} dx = \frac{\sqrt{\pi}}{2}$

Question

How to do the below integral:

$$\int_0^\infty \sqrt{x} e^{-x} dx = \frac{\sqrt{\pi}}{2} $$

I am guessing it somehow relates to

$$ \int_{-\infty}^\infty e^{-x^2} dx = \sqrt{\pi} $$

Answer 1

Change variable:

$$t^2:=x\implies 2tdt=dx\implies $$

$$I:=\int_0^\infty\sqrt x\,e^{-x}dx=\int_0^\infty te^{-t^2}2t\,dt=2\int_0^\infty t^2e^{-t^2}dt$$

and now by parts:

$$\begin{cases}u=t&u'=1\\v'=te^{-t^2}&v=-\frac12e^{-t^2}\end{cases}\implies$$

$$I=\overbrace{\left.-te^{-t^2}\right|_0^\infty}^{=0}+\int_0^\infty e^{-t^2}dt=\frac12\sqrt\pi$$