Gaussian moment recurrence ($I_n=\frac{n-1}{2}{I}_{n-2},I_n={\int }_0^{\infty }{x}^n{e}^{-x^2}dx$)

For integers $n\ge 2$, integration by parts givesThe boundary term vanishes because $n-1>0$ at zero and Gaussian decay dominates every power at infinity. The initial values are $I_0=\sqrt{\pi }/2$, the half-line Gaussian integral, and $I_1=1/2$ by the substitution rule $t=x^2$. More generally, for real $n>-1$, that substitution gives $I_n=\frac{1}{2}\Gamma ((n+1)/2)$, linking the recurrence to the Gamma function.

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.

For integer $n\ge 2$, integration by parts gives $J_n=(n-1)J_{n-2}/2$. The boundary term $-x^{n-1}{e}^{-x^2}/2$ vanishes at both infinities because Gaussian decay dominates every power. The full-line initial values are $J_0=\sqrt{\pi }$ and $J_1=0$, the latter by odd symmetry. HenceUnlike the half-line Gaussian moment recurrence, full-line odd moments vanish and even moments are twice their half-line values. The domain of integration must be retained when transferring a recurrence between the two settings.