TITLE: Need help with the integral $\int_{0}^\infty e^{-x^{2}}x^{2n+1}dx $
QUESTION [6 upvotes]: Problem: Prove that $$I = \int_{0}^\infty  e^{-x^{2}}x^{2n+1}dx = \frac{n!}{2}     $$
Source: A problem I found on an integral test. The problem bugged me for long and I did end up leaving it on the exam. Back home I decided to tackle it again. Here's my go on it.
My try: I have never encountered such a problem before, not even in my assignments and workbooks. Here's how I tried it
We can first take an indefinite integral and then we can work up towards a reduction formula.We can plug in the limits later. I don't know if it's any good but I did write this up on the exam: $$I_{2n+1} = \int{e^{x^{-2}}}{x^{2n+1}}dx$$
On applying Integration by parts:
$$I_{2n+1} = e^{-x^{2}}\frac{x^{2n+2}}{2n+2} +\frac{2}{2n+2} \int{e^{-x^{2}}}{x^{2n+3}}dx$$
or $$I_{2n+1} = e^{-x^{2}}\frac{x^{2n+2}}{2n+2} +\frac{I_{2n+3}}{n+1} $$
or
$$I_{2n+1} = {e^{-x^{2}}\over 2}\Bigl(\frac{x^{2n+2}}{n+1}\Bigl) +\frac{I_{2n+3}}{n+1} $$
What can I do next? Am I doing it incorrectly? Is this the wrong way? Please help me as I can't stop thinking about this problem! Thanks!
Edit 1: As suggested (already tried by me) lets put $u = -x^2$
$$ du = -2xdx$$
plug the above expression in the problem
$$I_{n} = -{1 \over 2}\int{e^{u}}{u^{n}}du$$
$$I_{n} = -{1 \over 2}e^{-u}\frac{u^{n+1}}{n+1} + \frac{I_(n+1)}{n+1}$$
Certainly helps. Now what?
Edit 2: Use the gamma function to get the result. Now unfortunately I didn't read it for the test that's why I didn't get it. Thank you all for pointing out the right direction!

REPLY [2 votes]: Let us start from your initial formula
$$ I_{2n+1} = \int{e^{x^{-2}}}{x^{2n+1}}dx $$
and define the definite integral
$$ J_{2n+1} = \int_0^{\infty}{e^{x^{-2}}}{x^{2n+1}}dx$$
If we take the formula you got via integration by parts:
$$I_{2n+1} = e^{-x^{2}}\frac{x^{2n+2}}{2n+2} +\frac{2}{2n+2} \int{e^{-x^{2}}}{x^{2n+3}}dx$$
and evaluate it at it's limits, it reduces to
$$J_{2n+1} = \frac{1}{n+1} \int_0^{\infty}{e^{-x^{2}}}{x^{2n+3}}dx=\frac{J_{n+3}}{n+1}.$$
Your formula now follows from evaluating $J_1$ and induction.  For this purpose, it is useful to rewrite the equation as $J_{2k+1}=kJ_{2(k-1)+1}$.