I'm attempting to find a closed form expression for $$\int_{-\infty}^{\infty}e^{-\frac{x^2\left(1+\lambda^2\right)}{2}}H_{n}(x)H_m(\lambda x)dx$$
where $H_n(x)$ are the physicist's hermite polynomials, but haven't had any luck. Anyone know of a way to compute this?
Denoting the integral by $I_{m,n}$, write the generating function \begin{align*}I(s,t)=&\sum_{m=0}^{\infty}\sum_{n=0}^{\infty}I_{m,n}\frac{s^m}{m!}\frac{t^n}{n!}=\\ =&\int_{-\infty}^{\infty}e^{-\frac{x^2(1+\lambda^2)}{2}} \underbrace{\left(\sum_{m=0}^{\infty}H_m(\lambda x)\frac{s^m}{m!}\right)}_{e^{2\lambda x s-s^2}} \underbrace{\left(\sum_{n=0}^{\infty}H_n( x)\frac{t^n}{n!}\right)}_{e^{2xt-t^2}}dx=\\ =&e^{-s^2-t^2}\int_{-\infty}^{\infty}e^{-\frac{x^2(1+\lambda^2)}{2}+2x(\lambda s+t)}dx=\\ =&\exp\left\{\frac{2(\lambda s+t)^2}{1+\lambda^2}-s^2-t^2\right\}\sqrt{\frac{2\pi}{1+\lambda^2}}. \end{align*} The rest should be clear. Of course, $I_{m,n}=0$ if $m,n$ are of different parity.