# Thread: improper integration by parts

1. ## improper integration by parts

I need to prove the following:
$\int^\infty_0 x^2e^{-x^2}=\frac{1}{2}\int^\infty_{0} e^{-x^2}$
Im having trouble integrating this, although I do know, based on the Gamma function, that $\frac{1}{2}\int^\infty_{0} e^{-x^2}=\frac{\sqrt{\pi}}{4}$
So really what I need to show is that $\lim_{b\to \infty}\int^b_0 x^2e^{-x^2}=\frac{\sqrt{\pi}}{4}$
But when I try to integrate by parts, Im getting even more confusing terms, can someone help me out?

2. ## Question

When you're doing integration by parts, what are you letting u equal and v equal in the formula?

3. u = x^2, dv= $\int^\infty_0 e^{-x^2}dx$

4. Try integration by parts with

$u = x$
$dv = x e^{-x^2} \, dx$.

5. so... I did the integration by parts according to these suggestions, and what I came up with, supposing I did the integration correctly, is:
$\frac{1}{4}\lim_{b\to\infty}[b-3be^{-b^2}]$
I need this to equal $\frac{\sqrt{\pi}}{4}$...
So either I am missing something or the integration didnt go as it should have.
Any feedback? thanks

6. I don't know why you need to worry about the actual value- if you let u= x, $dv= xe^{-x^2}$, integration by parts puts $\int_0^\infty e^{-x^2}dx$ in your lap! I can only conclude that you have done the integration by parts incorrectly. Please show all your work.