Can anyone explain me this definite integral I got from a probability problem:
$\mathbf E[X] = 2\int_0^\infty x^2e^{-x^2} dx$
$[x^2 = z]$
$=\int_0^\infty z^{{3\over2}-1}e^{-z} dz$
$= \Gamma({3\over2})$
$= {{\sqrt{\pi}}\over2}$
What I want to understand is from the first line to the third line.
How did $2x^2$ become $z^{{3\over2}-1}$ with $x^2 = z$?
(Please ignore if there are grammatical errors).