Looking for a generic analytical solution to the following integral:
$\int^{+\infty}_{-\infty} x^k H_n(x) e^{-x^2} dx$
in terms of $k$ and $n$, there's probably a $\sqrt{\pi}$ in there somewhere. Where $k$ and $n$ are integers and $H_n$ is the $n^{\text{th}}$ order Hermit polynomial. It seems so simple so I think there should be one. Anyone know of a paper or source that has the analytical solution? Cheers Guys!
Plug in the definition of $H_n$ to get $$\int_{-\infty}^{\infty} (-1)^n x^k \frac{d^n}{dx^n}\left(e^{-x^2}\right)\,dx$$ which is begging for integration by parts: $$\int_{-\infty}^{\infty} (-1)^n x^k \frac{d^n}{dx^n}\left(e^{-x^2}\right)\,dx = -\int_{-\infty}^{\infty} (-1)^n kx^{k-1} \frac{d^{n-1}}{dx^{n-1}}\left(e^{-x^2}\right)\,dx.$$ So if $k<n$, you get zero. Otherwise, the integral is equal to $$k! \int_{-\infty}^{\infty} x^{k-n} e^{-x^2}\,dx$$ which is again zero if $k-n$ is odd. Otherwise, letting $u=x^2$, we get $$k!\int_0^\infty u^{(k-n-1)/2} e^{-u}\,du = k!\,\Gamma\left(\frac{k-n}{2}+\frac{1}{2}\right) = \frac{k!(k-n)!\sqrt{\pi}}{4^n\left(\frac{k-n}{2}\right)!}$$