Prove that
$\displaystyle \int_{0}^\infty x^2 e^{{-x}^{2}}=\frac{1}{2} \int_{0}^\infty e^{{-x}^{2}}$
and both integrals converge.
Any help on this? I tried doing integration by parts for the second one but couldn't reach an answer.
Prove that
$\displaystyle \int_{0}^\infty x^2 e^{{-x}^{2}}=\frac{1}{2} \int_{0}^\infty e^{{-x}^{2}}$
and both integrals converge.
Any help on this? I tried doing integration by parts for the second one but couldn't reach an answer.
That thread was really helpful!
But now I'm trying to solve
$\displaystyle \int_0^\infty x^2e^{-x^2}dx$ by using the same method, and it's not working out so nicely. I've set it up the same way and converted it into polar, and now I'm at
$\displaystyle \int_0^\infty\int_0^{2\pi}r^5cos^2\theta sin^2\theta e^{-r^2}d\theta dr$ and I have a feeling that's not simple to evaluate... :-(
Is there a simpler way to do this?