Integrals involving Hermite Polynomials Could you please tell me, How to evaluate this integral which involve hermite polynomials, $\int_{-\infty}^\infty e^{-ax^2}x^{2q}H_m(x)H_n(x)\,dx=?$ where $H_n$ is the $n$-th Hermite polynomial (Physicist's version) and $q,\,m$ and $n$ are positive integers.
If $x^{2q}$ term were absent, I am able to perform the integral by writing the product of the Hermite polynomials into a single Hermite polynomial with a higher degree.
Can anybody give me a hint to perform this integral?
 A: You say you know how to do it when the $x^{2q}$ term is missing.  But the following website explains the effect of multiplying a Hermite polynomial by $x$, and you could just apply this rule $2q$ times.
http://en.wikipedia.org/wiki/Hermite_polynomials#Recursion_relation_2
I know you will end up with a bit of a mess.  But at least this should work to give concrete formulae in the case that $q$ is small.
A: As noted in the comments, the solution for arbitrary $q$
$$
I_{nm}(q)=\int_{-\infty}^\infty e^{-ax^2}x^{2q}H_n(x)H_m(x)dx
$$
are the derivatives of the integral
$$
I_{nm}(0)=\int_{-\infty}^\infty e^{-ax^2}H_n(x)H_m(x)dx
$$
with respect to $a$
$$
I_{nm}(q)=(-1)^q\frac{\partial^q}{\partial a^q}I_{nm}(0).
$$
The $q=0$ integral can be done by linearization and scaling.  First, transform to the probabilists Hermite polynomials and use linearization to obtain
$$
I_{nm}(0) = 2^{\frac{n+m}{2}}\sum_{k=0}^{\operatorname{min}(n,m)}{n\choose k}{m\choose k}k!\int_{-\infty}^\infty e^{-ay^2}He_{n+m-2k}(\sqrt{2}y)dy.
$$
Now change the integration variable to get the probabilists weighting function $\frac{x}{\sqrt{2}}=\sqrt{a}y$ leading to
$$
I_{nm}(0) = 2^{\frac{n+m-1}{2}}\frac{1}{\sqrt{a}}\sum_{k=0}^{\operatorname{min}(n,m)}{n\choose k}{m\choose k}k!\int_{-\infty}^\infty e^{-\frac{x^2}{2}}He_{n+m-2k}\left(\frac{x}{\sqrt{a}}\right)dy.
$$
Now use the scaling formula
$$
\operatorname{He}_{n+m-2k}(\gamma x) =(n+m-2k)!\sum_{s=0}^{\frac{n+m}{2}-k}\frac{1}{2^ss!(n-m-2k-2s)!}\gamma^{n-m-2k-2s}\left(\gamma^2-1\right)^s \operatorname{He}_{n-m-2k-2s}(x),
$$
the floor function on the upper limit of the sum is not necessary since this integral is zero if $n$ and $m$ have opposite parity.  Therefore the sum is always even.  Also, since this sum is integrated, only the maximum $s$ term is retained by orthogonality.  Therefore the integral becomes
$$
I_{nm}(0)=\sqrt{\frac{\pi}{a}}n!m!\sum_{k=0}^{\operatorname{min}(n,m)}\frac{2^k(n+m-2k)!}{k!(n-k)!(m-k)!\left(\frac{n+m}{2}-k\right)!}\left(\frac{1-a}{a}\right)^{\frac{n+m}{2}-k}
$$
This is actually the $b=0$ limit of the question Integration involving Hermite Polynomials through generating functions.  To see this, the parity has to be even $p=0$ and $b^{2a}$ is only $1$ when $a=0$, in which case this integral reduces to this one.
