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.

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- Feb 7th 2010, 01:16 PMpaupsersProve these integrals are equal
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. - Feb 7th 2010, 01:38 PMJhevon
see post #2 here

- Feb 7th 2010, 02:23 PMpaupsers
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? - Feb 7th 2010, 02:37 PMpaupsers
Actually, it can be done using integration by parts. Thanks for the link, still!