= Odd Gaussian moments on a half-line
{title2=$\int_0^\infty x^{2m+1}e^{-\alpha x^2}dx=m!/(2\alpha^{m+1})$}
For $\alpha>0$ and a nonnegative integer $m$, substitute $u=\alpha x^2$. Then $x^{2m+1}dx=u^mdu/(2\alpha^{m+1})$, so the integral reduces to the <gamma function> value $\Gamma(m+1)=m!$. This proves the formula. The corresponding whole-line integral is zero by odd symmetry, and both integrals converge absolutely. A half-line integral of an odd function need not vanish.
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