General solution for an integral with Hermite polynomials?

Question

I'm currently studying quantum mechanics and thus dealing with Hermite polynomials.

By looking on Wikipedia I found this identity for Hermite polynomials:

$$\int\limits_{-\infty}^{\infty} H_m(x)H_n(x)e^{-x^2} \ dx=\sqrt{\pi}2^nn!\delta_{mn}$$

While doing a problem, after a little bit of calculations, I came to the following integral: $$\int\limits_{-\infty}^{\infty} H_m(x)\cdot x\cdot H_n(x)e^{-x^2} \ dx$$

By just looking at this integral, I figured that it has to be zero when $m-n=\pm2k, k=1,2,3...$ (cause the expression would be uneven and thus the integral would be zero).

I thought I could make the same argument by saying the integral is nonzero whene $m-n=\pm(2k+1)$. While this is true for $k=0$ (so $m-n=\pm1$), I turns out that this seems to be not true for $k>0$. For example I calculated: $$\int\limits_{-\infty}^{\infty} H_0(x)\cdot x\cdot H_3(x)e^{-x^2} \ dx=\int\limits_{-\infty}^{\infty} e^{-x^2}x(8x^3-12x)dx=0$$

I tried to understand why this is the case by using the recursion formula of the Hermite polynomials and plugging it into the integral but didn't succeed...

I wanted to ask if there is another already existing general identity for my integral (like the one on Wikipedia). If not it would be awesome if someone could help me out here and explain why the argumentation breaks down for $k>0$... Thanks!

Answer 1

Any polynomial of degree $\leqslant n$ is a linear combination of $H_k(x)$ with $0\leqslant k\leqslant n$. In particular, there exist numbers $a_{m,k}$ such that $xH_m(x)=\sum_{k=0}^{m+1}a_{m,k}H_k(x)$. And, if $m+1<n$, then this implies $\int_{-\infty}^\infty xH_m(x)H_n(x)e^{-x^2}\,dx=0$ by the orthogonality of the Hermite polynomials. (To handle $m+1=n$, one just has to compute $a_{m,m+1}$ by comparing the coefficients of $x^{m+1}$.)

Answer 2

The answer's already been accepted, but expanding on metamorphy's comment, an explicit formula for the integral of interest is

$$\int_{-\infty}^\infty \exp{(-x^2)} \ x H_m(x)H_n(x) \ dx = \sqrt{\pi} \ 2^n n! \sum_{j=0}^n \binom{m}{n-j}\frac{2^{-j}}{j!} \delta_{1,m-n+2j} $$

This is derivable from the orthogonality relationship and the explicit linearization formula $$ H_m(x)H_n(x) = 2^n n!\sum_{j=0}^n \binom{m}{n-j}\frac{2^{-j}}{j!} H_{m-n+2j}(x) $$ since $x = \tfrac{1}{2}H_1(x).$