The integral with two Hermite polynomials I have the following integral,
$$\int_{-\infty}^{\infty}dy\,e^{iqy}e^{-y^2/2}H_{n}(a-y)H_{m}(b+y)$$
and I really need help to calculate it. All my attempts are useless. I have checked Gradstein & Ryzhik and Harry Bateman books but it was unsuccesful.
 A: The answer is expressible in a triple nested sum, so it isn't very pretty, though the sums are finite.  The derivation uses Gradshteyn 7.374.9, and with a rescale of the integration variable, appropriate choice of the formula's $\alpha$ and Hermite polynomial variables, one gets ($i=\sqrt{-1}=\exp{(i \pi/2})$ , not an index)
$$ \int_{-\infty}^\infty \exp{(-y^2/2)} \exp{(i\,q\,y)} H_j(y) H_k(y) dy = $$
$$ \sqrt{2 \pi} \exp{(-q^2/2)} \sum_{L=0}^{min(j,k)}2^LL!\binom{j}{L} \binom{k}{L} 
i^{j+k - 2L} H_{j+k-2L}(q) $$
Now use the Appel property for the expansion of $H_n(a-y)$ (see the wiki of Hermite polys)
$$ H_n(a-y)=(-1)^n\,H_n(y-a)=(-1)^n \sum_{j=0}^n\binom{n}{j}H_j(y)(-2a)^{n-j}$$
$$ H_m(b+y)= \sum_{k=0}^m\binom{m}{k}H_k(y)(2b)^{m-k}$$
Insert these last two sums into the first to get
$$ \int_{-\infty}^\infty \exp{(-y^2/2)} \exp{(i\,q\,y)}H_n(a-y)\, H_m(b+y)  dy = $$
$$ \sqrt{2 \pi} \exp{(-q^2/2)} (-1)^n \sum_{j=0}^n\binom{n}{j}(-2a)^{n-j} \sum_{k=0}^m\binom{m}{k}(2b)^{m-k}  $$
$$ \quad \sum_{L=0}^{min(j,k)}2^LL!\binom{j}{L} \binom{k}{L} 
i^{j+k - 2L} H_{j+k-2L}(q) $$
