A method for calculating this integral hermite polynomials I need proof this,
$\int_{-\infty}^{\infty}e^{-x^2}H_n^{2}(x)x^2dx=2^nn!\sqrt{\pi}(n+\frac{1}{2})$
This is the idea: Multiply $(1-t^2)^{-1/2}e^{2x^2t/(1+t)}=\underset{n=0}{\overset{\infty}\sum}\frac{H_n^{2}(x)}{2^nn!}t^n,~~~|t|<1,$ by $x^2$, integrate from $-\infty$ to $\infty$, and evaluate the integral in the left-hand side, calling the result $\varphi(t).$ Then expand $\varphi(t)$ in powers of $t$ and equate coefficients of identical powers of $t$ in both sides of the equation  so obtained.
I have problems in evaluating the integral, and thus expand in powers of $t$. $(H_n(x)) Hermite-Polynomials$
Any ideas? would be helpful!
 A: *

*Method 1: recurrence relations and symmetries of the Hermite polynomials


Hint: I would use the relations
$$H_{n+1}(x)=xH_n(x)-H'_n(x), $$
$$H'_n(x)=nH_{n-1}(x) $$
and an induction argument, as follows. Let us suppose the steps $k=1,\dots,n-1$ are true; we want to show that 
$$\int e^{-x^2}H_n^2(x)x^2dx=2^nn!\sqrt{\pi}(n+\frac{1}{2}). $$
Using the above relations the l.h.s is 
$$\int e^{-x^2}H_n^2(x)x^2dx=\int e^{-x^2}x^2(xH_{n-1}(x)-H'_{n-1}(x))^2dx=
\int e^{-x^2}x^4H^2_{n-1}(x)dx \\
-2\int e^{-x^2}x^3H_{n-1}(x)H^{'}_{n-1}(x)dx
+\int e^{-x^2}x^2(H^{'}_{n-1}(x))^2dx=\\
\int e^{-x^2}x^4H^2_{n-1}(x)dx -2(n-1)\int e^{-x^2}x^3H_{n-1}(x)H_{n-2}(x)dx
+\\(n-1)^2\int e^{-x^2}x^2(H_{n-2}(x))^2dx. ~(*)
$$
Let us discuss $(*)$: the first 2 terms can be reduced using integration by parts on the products 
$$e^{-x^2}x^4H^2_{n-1}(x)=-(-\frac{1}{2}(2x)e^{-x^2})x^3H_{n-1}(x),$$
$$e^{-x^2}x^3H_{n-1}(x)H_{n-2}(x)=-(-\frac{1}{2}(2x)e^{-x^2})x^2H_{n-1}(x)H_{n-2}(x), $$ 
while the last term can be evaluated by the induction hypothesis.
All we need is to remember that
$$\int e^{-x^2}x^{2q+1}H^2_{r}(x)dx=0,$$
$$\int e^{-x^2}x^{2q}H_{r}(x)H_{r-1}(x)dx=0,$$
for all $q\in\mathbb N$, $r\geq 1$ for symmetry. For the second equality we used 
$H_r$ odd/even $\Leftrightarrow$ $H_{r-1}$ even/odd, which follows directly from the definition of the Hermite polynomials.


*

*Method 2: generating function


Using the generating function suggested by the OP, one needs to integrate w.r.t $x$ on the whole real axis both sides of 
$$x^2e^{-x^2}e^{-\frac{t}{1+t}x^2}\frac{1}{\sqrt{1-t^2}}= \sum_{n=1}^\infty x^2e^{-x^2}\frac{H_n^2(x)}{2^nn!}t^n,$$
and expanding w.r.t. $t$. The integral on the l.h.s. is Gaussian and can be evaluated with standard techniques reducing  to the case
$$\int_{-\infty}^\infty x^2e^{-\alpha(t)x^2}dx=\frac{1}{2}\sqrt{\frac{\pi}{\alpha(t)^3}}, ~(1)$$
for $\alpha(t)>0$ and $|t|<1$.


*

*Generating function with $\alpha(t)=\frac{1-t}{1+t}$


In the case under exam, multiplying both sides of the generating function identity w.r.t the product $x^2 e^{-x^2}$ ($e^{-x^2}$ is the weight for the integral identities involving the Hermite polynomials) and integrating, we arrive at (the exchange between summation and integration is allowed by the properties of Hermite poly.)
$$\frac{1}{\sqrt{1-t^2}} \int_{-\infty}^\infty x^2 e^{-x^2} e^{-\frac{t}{1+t}x^2}dx= \sum_{n=1}^\infty \int_{-\infty}^\infty x^2e^{-x^2}\frac{H_n^2(x)}{2^nn!}t^n; $$
to arrive at the thesis we write (using (1))
$$g(t):=\frac{1}{\sqrt{1-t^2}}\int_{-\infty}^\infty x^2e^{-\frac{1-t}{1+t}x^2}dx=\frac{1}{\sqrt{1-t^2}}\frac{1}{2}\sqrt{\frac{\pi}{\left(\frac{1-t}{1+t}\right)^3}}=\frac{\sqrt{\pi}}{2}\frac{1+t}{(1-t)^2}$$
for $|t|<1$ and we expand w.r.t $t$ at $t=0$ the function $g(t)$, obtaining
$$g(t)=\frac{\sqrt{\pi}}{2}\left(1+3t+\frac{1}{2!}10t^2+O(t^3)\right)$$
(the higher terms are easily computed, as well). This gives the thesis. I hope it helps.
A: This can be proven by using the linearization of Hermite polynomials.  I will work in the probabilists in which case the integral becomes
$$
I_n = 2^{n-\frac{3}{2}}\int_{-\infty}^\infty \operatorname{He}_n^2(x)x^2e^{-\frac{1}{2}x^2}dx.
$$
Applying the linearization formula
$$
\operatorname{He}^2_n(x) = \sum_{k=0}^n {n\choose k}^2k!\operatorname{He}_{2n-2k}(x)
$$
leads to the integral
$$
I_n = 2^{n-\frac{3}{2}}\sum_{k=0}^n {n\choose k}^2k!\int_{-\infty}^\infty \operatorname{He}_{2n-2k}(x)x^2e^{-\frac{1}{2}x^2}dx.
$$
Using $x^2=\operatorname{He}_2(x)+\operatorname{He}_0(x)$ together with orthogonality implies that the sum only has two non-zero terms
$$
I_n=2^{n-\frac{3}{2}}\left\{{n\choose n-1}^2 2!\sqrt{2\pi} + n!\sqrt{2\pi}  \right\}
$$
which reduces to precisely what was to be shown
$$
I_n = 2^n n!\left(n+\frac{1}{2}\right)\sqrt{\pi}
$$
