Calculate $ \int_{-\infty}^{\infty} x^4 H(x)^2 e^{-x^2} dx$ where $H(x)$ is a Hermite polynomial I need to calculate the following integral 

$$ \int_{-\infty}^{\infty} x^4 H(x)^2 e^{-x^2} dx.$$

where $H(x)$ is a Hermite polynomial.
I tried using the recurrence relation, but I don't get the answer.
 A: You can have the closed form for your integral 

$$\int_{-\infty}^{\infty} x^4 H_n(x)^2 e^{-x^2} dx =  \frac{3\sqrt{\pi}}{4}{2}^{n}\left(2
\,{n}^{2}+2\,n+1 \right)n!\,,\quad n=0,1,2,3,\dots\,. $$

Added: One can get a closed form formula for another integral involves the cubic Hermite

$$\int_{-\infty}^{\infty} x H_n(x)^3 e^{-x^2} dx= {\frac { 48\,(512)^n \Gamma  \left( n+\frac{3}{2} \right)^{3}}{\pi \, \left( n+1 \right) }}.\quad n=0,1,2,3,\dots\,.$$

A: This paper, "On the Squares of Hermite Polynomials", tells us that
$$
\displaystyle \sum_{n=1}^{\infty} \frac{[H_n(x)]^2z^n}{2^nn!} = \frac{\exp{(2x^2z/(1+z))}}{\sqrt{1-z^2}}.
$$
Multiplying by $x^4\mathrm{e}^{-x^2}$ and integrating gives that
$$
\displaystyle \int_{-\infty}^{\infty} \, \sum_{n=1}^{\infty} \frac{[H_n(x)]^2x^4\mathrm{e}^{-x^2}z^n}{2^nn!} \, \mathrm{d}x = \int_{-\infty}^{\infty} \frac{\exp{(2x^2z/(1+z)-x^2)}}{\sqrt{1-z^2}}x^4 \, \mathrm{d}x. 
$$
The integral on the right hand side is a classical integral from statistical physics, 
$$
\int_{-\infty}^{\infty} \frac{\exp{(2x^2z/(1+z)-x^2)}}{\sqrt{1-z^2}}x^4 \, \mathrm{d}x = \int_{-\infty}^{\infty} \frac{\exp{(-x^2(1-z)/(1+z))}}{\sqrt{1-z^2}}x^4 \, \mathrm{d}x = \frac{3\sqrt{\pi}(1+z)^2}{4(1-z)^3}.
$$
Note that
$$
\frac{(1+z)^2}{(1-z)^3} = \displaystyle \sum_{n=0}^{\infty} \ (2n^2+2n+1) \ z^n, \qquad \mbox{for} \ |z|<1,
$$
which gives us the result that @Mhenni Benghorbal mentioned (under the caveat that we can swap the summation and integration on the left hand side). 
Now, let's try to evaluate the more general case of
$$
F(n,m) = \int^{\infty}_{-\infty} \, [H_n(x)]^2x^{2m}\mathrm{e}^{-x^2} \, \mathrm{d}x.
$$
We repeat the above process, where on the right hand side we have to evaluate integrals of the form
$$
(1-z^2)^{-\frac{1}{2}}\int^{\infty}_{-\infty} \,x^{2m}\mathrm{e}^{-\left(\frac{1-z}{1+z}\right)x^2} \, \mathrm{d}x = \frac{(1+z)^{m}}{(1-z)^{m+1}}\Gamma\left(\frac{2m+1}{2}\right).
$$
Note that
$$
\frac{(1+z)^{m}}{(1-z)^{m+1}} = \displaystyle \sum_{n=0}^{\infty} c_nz^{n+1}, 
$$
where $c_0=1$, $c_1=2m$ and $(2+n)c_{n+2} -2mc_{n+1}=nc_n$ (I won't attempt to solve the recurrence relation - even solving for our $m=2$ case in Wolfram Alpha gave the result in terms of something called the Lerch Transcendental!). In any case, 
$$
F(n,m) = 2^nn!c_n(m)\Gamma\left(\frac{2m+1}{2}\right).
$$
For the special case $m=0$,
$$
F(n,0) = 2^nn!c_n(0)\Gamma\left(\frac{1}{2}\right) =  2^nn!\sqrt{\pi},
$$
which confirms a well known result (from Wikipedia), that
$$
\int_{-\infty}^\infty H_m(x) H_n(x)\, \mathrm{e}^{-x^2}\, \mathrm{d}x =  2^n n! \sqrt{ \pi}\delta_{nm}
$$
