= Solution
Use <integration by parts> in $I_{n+1}$, writing the integrand as $x^{2n+1}(xe^{-x^2})$. Since an <antiderivative> of $xe^{-x^2}$ is $-e^{-x^2}/2$,
$$
\begin{aligned}
I_{n+1}
&=\left[-\frac12x^{2n+1}e^{-x^2}\right]_{-\infty}^{\infty}
+\frac{2n+1}{2}\int_{-\infty}^{\infty}x^{2n}e^{-x^2}\,dx\\
&=\boxed{\left(n+\frac12\right)I_n}.
\end{aligned}
$$
The boundary term vanishes because the Gaussian <exponential function> decays faster than any fixed power. These <improper integrals> are absolutely convergent, justifying the limiting <integration by parts>.
Starting from the permitted <Gaussian integral> $I_0=\sqrt\pi$, the <Gaussian moment recurrence> gives $I_1=\sqrt\pi/2$ and then
$$
\boxed{I_2=\frac34\sqrt\pi,\qquad I_3=\frac{15}{8}\sqrt\pi.}
$$
More generally, iteration gives $I_n=(2n-1)!!\sqrt\pi/2^n$, using the <double factorial> with $(-1)!!=1$.
Back to article page
