An unusual integral involving Hermite polynomials

Question

I want a closed and nicely form for the following integral

$$ \int\limits_{-\infty}^{\infty} e^{-\frac{-x^2}{2}} He_n(x)He_m(a(x-b)), $$

but I didn't find this guy inside the Ryzhik's book.

Answer 1

This was a tricky one. It requires both the Hermite multiplication theorem $$ \operatorname{He}_n(ax) = n!\sum_{k=0}^{\left\lfloor \frac{n}{2} \right\rfloor }\frac{1}{2^kk!(n-2k)!}a^{n-2k}\left(a^2-1\right)^k \operatorname{He}_{n-2k}(x) $$ and the shift $$ \operatorname{He}_m(x+y)=\sum_{k=0}^m {m \choose k}y^{m-k}\operatorname{He}_k(x). $$ Using these two, the integral becomes $$ I_{nm}=\sum_{k=0}^m{m \choose k}(-ab)^{m-k}k!\sum_{l=0}^{\left\lfloor \frac{k}{2} \right\rfloor }\frac{a^{k-2l}(a^2-1)^l}{2^ll!(k-2l)!}\int_{-\infty}^\infty\operatorname{He}_n(x)\operatorname{He}_{k-2l}(x)e^{-\frac{x^2}{2}}dx. $$ Care must be taken when attempting to apply orthogonality $$ \int_{-\infty}^\infty\operatorname{He}_n(x)\operatorname{He}_{k-2l}(x)e^{-\frac{x^2}{2}}dx=\sqrt{2\pi}n!\delta_{n,k-2l}. $$ Note that the maximum $k-2l$ is $m$ when $k=m,l=0$, so $m\ge n$ for this integral to be non-zero. Furthermore, when $l=0$ we get the highest Hermite polynomial combination $\operatorname{He}_n(x)\operatorname{He}_k(x)$ so $k$ must be at least $n$, hence we can start the sum at $n$: $$ I_{nm}=\sum_{k=n}^m{m \choose k}(-ab)^{m-k}k!\sum_{l=0}^{\left\lfloor \frac{k}{2} \right\rfloor }\frac{a^{k-2l}(a^2-1)^l}{2^ll!(k-2l)!}\int_{-\infty}^\infty\operatorname{He}_n(x)\operatorname{He}_{k-2l}(x)e^{-\frac{x^2}{2}}dx. $$ Starting the sum here also has the side-effect of imposing that the integral is zero when $n<m$. Now we can apply orthogonality and set $n=k-2l$, but note that this will not always have integer solution for $l$, which can be seen by looking at the parity of this condition $\operatorname{mod}(n - k + 2l,2) = 0$, or $\operatorname{mod}(n,2) = \operatorname{mod}(k,2)$, the summation index k must have the same parity as $n$.

We will denote the parity of $n$ as $p=\operatorname{mod}(n,2)$, as it will appear in the solution. We could break this into two cases: even $n$ and odd $n$, and you may argue that the answer looks nicer, but we will continue to keep it general, at the expense of some notational burden.

To keep only the terms that will result in a non-zero integral, we must either choose $n = 2s$ or $n=2s+1$. To be general, we take $n=2s+p$. Simultaneously we use the orthogonality and get the final answer: $$ I_{nm}=\sqrt{2\pi}a^n\sum_{s=\left\lfloor \frac{n}{2} \right\rfloor}^{\left\lfloor \frac{m-p}{2} \right\rfloor }{m \choose 2s+p}\frac{(2s+p)!}{2^{s-\left\lfloor \frac{n}{2} \right\rfloor}\left(s-\left\lfloor \frac{n}{2} \right\rfloor\right)}(-ab)^{m-2s-p}(a^2-1)^{s- \left\lfloor \frac{n}{2} \right\rfloor} $$