Integral involving Hermite polynomials

Question

Is there any way to calculate the following integral ?

$\int_{-\infty}^{+\infty} H_n(x)H_m(x)e^{-\alpha x^2} d x$ with $\alpha > 0$ and $(H_n,H_m)_{w = e^{-x^2}} = 2^n n! \sqrt{\pi}\delta_{nm}$

(without using Gauss-Hermite quadrature)

Answer 1

We want to evaluate \begin{eqnarray*} J_{n,m}=\int_{-\infty}^{\infty} e^{-\alpha x^2} H_n(x) H_m(x) dx. \end{eqnarray*} The Hermite polynomials are defined by the generating function \begin{eqnarray*} \sum_{n=0}^{\infty} H_n(z) \frac{t^n}{n!} =\exp( 2tz-t^2). \end{eqnarray*} Now we define \begin{eqnarray*} I(t_1,t_2)&=&\sum_{n=0}^{\infty} \frac{t_1^n}{n!} \sum_{m=0}^{\infty} \frac{t_2^m}{m!} \int_{-\infty}^{\infty} e^{-\alpha x^2} H_n(x) H_m(x) dx \\ &=& \int_{-\infty}^{\infty} \exp( -\alpha x^2+2t_1x-t_1^2+2t_2x-t_2^2) dx \\ &=& \int_{-\infty}^{\infty} \exp \left( -\alpha \left(x-\frac{t_1+t_2}{\alpha} \right)^2 -t_1^2-t_2^2 +\frac{(t_1+t_2)^2}{\alpha}\right) dx \\ &=& \sqrt{\frac{\pi}{\alpha}} \exp \left( -t_1^2-t_2^2 +\frac{(t_1+t_2)^2}{\alpha}\right). \end{eqnarray*} It now becomes clear why $\alpha=1$ gives the orthogonality relation ($I = 2^n n! \delta_{n,m} \sqrt{\pi}$). In our case we require the coefficient of $t_1^n t_2^m$ in $I(t_1,t_2)$, in the case where $n,m$ are even ($(n,m) \rightarrow (2n,2m)$) we get \begin{eqnarray*} I(t_1,t_2)&=& \sqrt{\frac{\pi}{\alpha}} \sum_{j_1=0}^{\infty} \frac{(1/\alpha -1)^{j_1} t_1^{2j_1}}{j_1 !} \sum_{j=0}^{\infty} \frac{(2 t_1 t_2 )^{j}}{j !} \sum_{j_2=0}^{\infty} \frac{(1/\alpha -1)^{j_2} t_2^{2j_2}}{j_2 !} \end{eqnarray*} So $2j_1+j=2n$ and $2j_2+j=2m$ giving \begin{eqnarray*} J_{2n,2m} &=&(2n)!(2m)! \sqrt{\frac{\pi}{\alpha}} \sum_{j=0}^{\min(n,m)} \frac{2 ^{j}(1/\alpha -1)^{n+m-2j}}{(2j) ! (n-j)!(m-j)!} . \end{eqnarray*} $J$ will be zero in the case where one of $n,m$ is even and one is odd, I leave you to evaluate the case where they are both odd.