Fourier transform of a product of the Hermite polynomials I tried to find the second-order correction to eigenenergies of the quantum harmonic oscillator perturbed by a cosine potential: $V=2A\cos(Bx)=Ae^{iBx}+Ae^{-iBx}$. But I have no clue how to calculate an appearing integral, which is in fact the Fourier transform:
$$\int_{-\infty}^{+\infty}e^{\pm ikx}e^{-x^2}H_m(x)H_n(x)dx$$
where $H_n(x)=(-1)^ne^{x^2}\frac{d^n}{dx^n}e^{-x^2}$ is the physicists' Hermite polynomial and $k=\sqrt{\frac{\hbar}{m\omega}}B$ is just a constant.
I found an answer only for $m=n$ (here it is), but I'm not able to adapt it for arbitrary $m$ and $n$.
 A: Translating the integral into probabilists Hermite polynomials leads to
$$
I_{nm} = 2^{\frac{n+m-1}{2}}\int_{-\infty}^\infty e^{\pm i\frac{kx}{\sqrt{2}}}\operatorname{He}_n(x)\operatorname{He}_m(x)e^{-\frac{x^2}{2}}dx.
$$
To solve this integral, first linearize the Hermite polynomials
$$
\operatorname{He}_n(x)\operatorname{He}_m(x)=\sum_{l=0}^{\operatorname{min}(n,m)}{m \choose l}{n \choose l}l!\operatorname{He}_{n+m-2l}(x)
$$
and subsequently complete the square
$$
\pm ix\frac{k}{\sqrt{2}} - \frac{1}{2}x^2 = -\frac{1}{2}(x\pm ikx)+\frac{1}{4}k^2,
$$
this leads to the sum
$$
I_{nm}=2^{\frac{n+m-1}{2}}e^{-\frac{1}{4}k^2}\sum_{l=0}^{\operatorname{min}(n,m)}{m \choose l}{n \choose l}l!\int_{-\infty}^\infty \operatorname{He}_{n+m-2l}(x)e^{-\frac{1}{2}\left(x\pm i\frac{k}{\sqrt{2}}\right)}dx.
$$
The next step is to shift the integration variable $x\rightarrow x\pm i\frac{k}{\sqrt{2}}$ and Taylor expand the resulting shifted Hermite polynomial
$$
\operatorname{He}_{n+m-2l}\left(x \mp i\frac{k}{\sqrt{2}}\right)=\sum_{s=0}^{n+m-2l}{n+m-2l \choose s}\left(\mp i\frac{k}{\sqrt{2}}\right)^{n+m-2l-s}\operatorname{He}_{n+m-2l-s}(x)
$$
however by orthogonality we will only need the zeroth term, which integrates to $\sqrt{2\pi}$.
Therefore the integral is
$$
I_{nm}=2^{\frac{n+m}{2}}e^{-\frac{1}{4}k^2}\sqrt{\pi}\sum_{l=0}^{\operatorname{min}(n,m)}{m \choose l}{n \choose l}l!\left(\mp i\frac{k}{\sqrt{2}}\right)^{n+m-2l},
$$
which reduces to
$$
I_{nm}=\sqrt{\pi}e^{-\frac{1}{4}k^2}m!n!\sum_{l=0}^{\operatorname{min}(n,m)}\frac{2^l\left(\mp i\right)^{n+m-2l}}{l!(m-l)!(n-l)!}k^{n+m-2l}.
$$
Setting $n=m$ reproduces the result cited in the question Fourier transform of squared Gaussian Hermite polynomial
