Integral involving Hermite polynomial Is there any way to calculate the following integral ?
$$I =\int_{-\infty}^\infty{x^kH_n(x)^2 e^{-x^2}\,dx}$$
Where $$H_n=(-1)^ne^{x^2}\frac{ d^n (e^{-x^{2}})}{dx^n}$$
The only result I got was
$$I=\int_{-\infty}^{\infty}(x^kH_n)^{(n)}e^{-x^2}\,dx$$
But in doesn't help much.
Thank for your help!
 A: First, note this is only non-zero for even $k$, so I will enforce this by setting $k=2m$.
Second, I prefer to work with the probabilists Hermite polynomials
$$
\operatorname{He}_{\alpha}(x) = 2^{-\frac{\alpha}{2}}H_\alpha\left(\frac{x}{\sqrt{2}}\right)
$$
in which case the integral becomes
$$
I_{nm}=2^{n-m-\frac{1}{2}}\int_{-\infty}^\infty x^{2m}\operatorname{He}_n^2(x)e^{-\frac{x^2}{2}}dx.
$$
We then transform the polynomial via the inversion formula
$$
x^{2m} = (2m)!\sum_{k=0}^{m} \frac{1}{2^k k!(2m-2k)!} \operatorname{He}_{2m-2k}(y), 
$$
and the square of the Hermite polynomial via the linearization formula
$$
\operatorname{He}_\alpha(x)\operatorname{He}_\beta(x)=\sum_{k=0}^{\min(\alpha,\beta)}{\alpha \choose k}{\beta \choose k}k!\operatorname{He}_{\alpha+\beta-2k}(x)
$$
resulting in the double summation
$$
I_{nm}=2^{n-m-\frac{1}{2}}(2m)!\sum_{k=0}^n {n \choose k}^2k!\sum_{l=0}^{m} \frac{1}{2^ll!(2m-2l)!}\int_{-\infty}^\infty \operatorname{He}_{2m-2l}(x)\operatorname{He}_{2n-2k}(x)e^{-\frac{x^2}{2}}dx.
$$
Orthogonality tells us that $m-l=n-k$, so we can solve for $l=m-n+k$ but this also has to be greater than $0$ so the sum over k has a lower bound $k_{\textrm{min}}=\operatorname{max}(n-m,0)$.  The sum over $l$ gives only one value due to orthogonality
$$
\int_{-\infty}^\infty \operatorname{He}_{\alpha}(x)\operatorname{He}_{\beta}(x)e^{-\frac{x^2}{2}}dx=\sqrt{2\pi}\alpha!\delta_{\alpha\beta}
$$
resulting in the closed form expression for the integral
$$
I_{nm} = 4^{n-m}(2m)!\sqrt{\pi}\sum_{k=\operatorname{max}(n-m,0)}^n {n\choose k}^2 \frac{k!}{2^k(m-n+k)!}.
$$
