Odd Gaussian moments on a half-line (${\int }_0^{\infty }{x}^{2m+1}{e}^{-\alpha x^2}dx=m!/(2{\alpha }^{m+1})$)

For $\alpha >0$ and a nonnegative integer $m$, substitute $u=\alpha x^2$. Then $x^{2m+1}dx=u^{m}du/(2{\alpha }^{m+1})$, 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.