Integral of two Hermite polynomials, one shifted, and exponentials I have to calculate the integral
$I=\int_{-\infty}^{\infty} H_{m}(x) e^{-x^{2}/2} H_{n}(x-a) e^{\frac{(x-a)^{2}}{2}}dx$
where $H_{n}(x)$ are the physicist's Hermite polynomials. This integral exists in the Wolfram function site and it is equal to
$I=\sqrt{\pi}\sqrt{2^{n}n!}\sqrt{2^{m}m!} \sqrt{\frac{n!m!}{2^{m-n}}}(-a)^{m-n} e^{-a^{2}/4} \sum_{k=0}^{n} \frac{(-a^{2}/2)^{k}}{k!(n-k)!(k-n+m)!}$
Wolfram_functions
However, on the NIST handbook of mathematical functions, there is this integral
$J=\int_{-\infty}^{\infty} H_{m}(x) e^{-x^{2}} H_{n}(x-a) e^{(x-a)^{2}}dx= \sqrt{\pi} 2^{- \frac{1}{2} (m+n+1)} H_{m+n}(\frac{x}{\sqrt{2}})e^{-x^{2}/2}$ (A)
I would like to have an expression like in integral $J$, however, I have $e^{-x^{2}/2}$ instead of $e^{-x^{2}}$. Do you have any suggestion to obtain the expression (A) for my integral? Or do you know how the integral (B) is calculated so that I can try the derivation starting from my integral $I$?
Thank you!
 A: The Wolfram site has the following integral
$$
I_{nm}(a)=\int_{-\infty}^\infty e^{-\frac{1}{2}x^2}e^{-\frac{1}{2}(x-a)^2}H_n(x)H_m(x-a)dx
$$
so I will assume you have a sign error and prove this integral.
First complete the square $-\left(x-\frac{a}{2}\right)^2-\frac{1}{4}a^2 = -\frac{1}{2}x^2 -\frac{1}{2}(x-a)^2$ to get
$$
I_{nm}(a)=e^{-\frac{a^2}{4}}\int_{-\infty}^\infty e^{-\left(x-\frac{a}{2}\right)^2}H_n(x)H_m(x-a)dx
$$
then switch to probabilists Hermite polynomials
$$
I_{nm}(a)=2^{\frac{n+m}{2}}e^{-\frac{a^2}{4}}\int_{-\infty}^\infty e^{-\left(x-\frac{a}{2}\right)^2}\operatorname{He}_n(\sqrt{2}x)\operatorname{He}_m(\sqrt{2}x-\sqrt{2}a)dx.
$$
Perform the substitution $z=\sqrt{2}{x} - \frac{a}{\sqrt{2}}$ so the integral becomes
$$
I_{nm}(a)=2^{\frac{n+m}{2}}e^{-\frac{a^2}{4}}\int_{-\infty}^\infty e^{-\frac{x^2}{2}}\operatorname{He}_n\left(z+\frac{a}{\sqrt{x}}\right)\operatorname{He}_m\left(z-\frac{a}{\sqrt{x}}\right)dx.
$$
Now apply the Taylor expansion
$$
\operatorname{He}_\alpha(x+c)=\sum_{l=0}^\alpha {\alpha\choose l}c^l\operatorname{He}_{\alpha-l}(x)
$$
twice to get the double sum
$$
I_{nm}(a)=2^{\frac{n+m}{2}}e^{-\frac{a^2}{4}}\sum_{l=0}^n\sum_{k=0}^n {n \choose l}{m\choose k}\left(\frac{a}{\sqrt{2}}\right)^l\left(-\frac{a}{\sqrt{2}}\right)^k\int_{-\infty}^\infty e^{-\frac{1}{2}z^2}\operatorname{He}_{n-l}(z)\operatorname{He}_{m-k}(z)dz
$$
which is in a form that can be solved by orthogonality
$$
\int_{-\infty}^\infty e^{-\frac{1}{2}z^2}\operatorname{He}_{n-l}(z)\operatorname{He}_{m-k}(z)dz = \delta_{n-l,m-k}\sqrt{2\pi}(n-l)!.
$$
This constrains one of the sums, solve for $k=m-n+l$.  Knowing $k\ge 0$ implies $l\ge n-m$.  Therefore the integral becomes
$$
I_{nm}(a)=2^n\sqrt{\pi}n!m!e^{-\frac{b^2}{4}}(-a)^{m-n}\sum_{l=\operatorname{max}(n-m,0)}^n \frac{\left(-\frac{a^2}{2}\right)^l}{l!(n-l)!(m-n+l)!}
$$
