I am looking for a simple expression (in terms of sums of polynomials for instance) of the integral
$$I = \int_{-\infty}^{+ \infty} dx \int_{-\infty}^{+ \infty} dy \, H_{k}(x+y) H_{l}(x-y) \, e^{-2x^2+ax-\lambda^2 y^2 +i b y+icxy} $$
where $H_k(x)$ is the $k$-th (physicists') Hermite polynomial and $a,b,c, \lambda \in \mathbb{R}$.
I have tried using the identities
$$H_k(x+y) = \sum_{s=0}^k {k \choose s} y^s \, H_{k-s}(x) \qquad \qquad \qquad (A) \\ H_k(x) H_l(x) = \sum_{s=0}^{\min(k,l)} s! {k \choose s} {l \choose s} H_{k+l-2s} (x) \qquad (B)$$
which leaves the above integral $I$ in the form
$$J=\int_{-\infty}^{+ \infty} dx \int_{-\infty}^{+ \infty} dy \, H_{m}(x) \, y^n \, e^{-2(x-a)^2 -y^2 +i b y +i c x y},$$
where $a,b,c \in \mathbb{R}$ are different than before. Now there are two possibilities:
- Integrate in $x$, i.e. computing
$$J_1 = \int_{-\infty}^{+ \infty} dx \, H_m(x) e^{-2(x-a-icy/4)^2},$$
for which I cannot find a simple expression (formula 7.374(8) or 7.374(10) in Gradshteyn & Ryzhik are similar, but I think there the offset in the exponential is real and mine is complex; formula 7.376(1) is also similar but not quite the same).
- Integrate in $y$, which yields another Hermite polynomial. After using $(A)$ again one arrives to an integral of the form
$$J_2 = \int_{-\infty}^{+ \infty} dx \, H_m(x) H_n(\lambda x) e^{-\alpha x^2 - \beta x}$$
where $\lambda, \alpha, \beta \in \mathbb{R}$. Again, I could not find a simple expression for this integral (there are similar expressions in Gradshteyn & Ryzhik, like 7.374(3) or 7.374(5), but I think they are all particular cases of $J_2$ above) .
Any ideas for any of the integrals $I$, $J$, $J_1$ or $J_2$?
I think $J_2$ is tractable by first completing the square to get $$ J_2=\frac{1}{\sqrt{\alpha}}e^{-\frac{\beta^2}{4\alpha}}\int_{-\infty}^\infty H_n\left(\frac{y}{\sqrt{\alpha}} -\frac{\beta}{2\alpha} \right) H_n\left(\frac{\lambda y}{\sqrt{\alpha}} -\frac{\lambda\beta}{2\alpha} \right)e^{-y^2}dy $$ It will be messy, but you could then use Taylor to get rid of the shift terms, then use the "multiplication theorem" to rescale. You'll then have the correct orthogonality integral which will reduce the answer from four sums to three.