I'm going working through mathematical statistics by Larsen, and he provides the example of the expected value of the pdf for Maxwells infamous perfect gas equation:
$f_s(s) = 4\sqrt{\frac{a^3}{\pi}}s^2e^{-as^2}$
I'm unsure of how he derives the integral substitution for the equation here:
$2m\sqrt{\frac{a^3}{\pi}}\int_0^\infty s^4 e^{-as^2}ds$
after taking the sbustitution $t=as^2$
he gets:
$E(W) = \frac{m}{a\sqrt{\pi}}\int_0^\infty t^{\frac{3}{2}}e^-t\space dt$
Perhaps I'm a bit rusty with my substitution but I couldn't seem to derive his answer after implementing the substitution, any help on how it's properly implemented is much appreciated!
Step by step...
$$2m\sqrt{\frac{a^3}{\pi}}\int_0^{\infty}s^4e^{-as^2}ds$$
setting $t=as^2$
that means also
$$dt=2asds$$
you can rewrite you integral in the following form
$$m\frac{a^{3/2}}{\sqrt{\pi}}\int_0^{\infty}\frac{(as^2)^{3/2}}{a^{3/2}}e^{-as^2}\frac{2asds}{a}$$
and simplifying...
$$\frac{m}{a\sqrt{\pi}}\int_0^{\infty}t^{3/2}e^{-t}dt=\frac{m}{a\sqrt{\pi}}\Gamma\left(\frac{5}{2}\right)=\frac{3m}{4a}$$