Hermite Polynomials How can I prove that
$\int^{\infty}_{-\infty} x^2 e^{-x^2} H^2_n(x)\ dx=2^nn!(pi)^{1\over2} $
My idea:
$H_0(x)=1, \ H_2(x)=4x^2-2$
$4x^2=H_2(x)+2H_0(x)$
$x^2=$$1\over4$$(H_2(x)+2H_0(x))$
so
$\int^{\infty}_{-\infty} x^2 e^{-x^2} H^2_n(x)\ dx=$
=$\int^{\infty}_{-\infty}$ $1\over4$ $(H_2(x)+2H_0(x))\ e^{-x^2} H^2_n(x)\ dx$
=$1\over4$[$\int^{\infty}_{-\infty}$ $ e^{-x^2}H_2(x)\ H^2_n(x)\ dx$
+2$\int^{\infty}_{-\infty}$  $e^{-x^2}H_0(x) H^2_n(x)\ dx$]
we know $H_0(x)=1$ then
=$1\over4$[$\int^{\infty}_{-\infty}$ $ e^{-x^2}H_2(x)\ H^2_n(x)\ dx$
+2$\int^{\infty}_{-\infty}$  $e^{-x^2} H^2_n(x)\ dx$
=$1\over4$[$\int^{\infty}_{-\infty}$ $ e^{-x^2}H_2(x)\ H^2_n(x)\ dx$
+$ 2.{2^nn!(pi)^{1\over 2}}$]
=$1\over4$ $\int^{\infty}_{-\infty}$ $ e^{-x^2}H_2(x)\ H^2_n(x)\ dx$
+$1\over2$.$ {2^nn!(pi)^{1\over 2}}$
True ? and if it true How can I to Find
$\int^{\infty}_{-\infty} e^{-x^2}H_2(x)\ H^2_n(x)\ dx$ ?
 A: Your formula is wrong, an easy check is to substitute $n=0,1.$ You should get the well-known integrals
$$\int_{-\infty}^{\infty}x^2e^{-x^2}dx = \frac{1}{2}\sqrt{\pi}$$
$$\int_{-\infty}^{\infty}x^2\cdot 4x^2e^{-x^2}dx = 3\sqrt{\pi}$$
but your results are $\sqrt{\pi}, 2\sqrt{\pi}.$ The correct result is 

$$\int_{-\infty}^{\infty}x^2H_n^2(x)e^{-x^2}dx = \frac{1}{2}\sqrt{\pi}\,2^{n}n! (2n+1)$$

Here is the proof using $$||f(x)||=\int_{-\infty}^{\infty}f(x)^2e^{-x^2}dx,$$
the recurrence relation
$$H_{n+1}(x) = 2xH_n(x) -H_{n-1}(x),$$
and the properties of the inner product. From 
$$xH_n(x)=\frac{1}{2}H_{n+1}(x)+nH_{n-1}$$
we compute
$$||xH_n|| = \frac{1}{4}||H_{n+1}||+n^2||H_{n-1}||$$
$$ = \frac{1}{4}\sqrt{\pi}2^{n+1}(n+1)!+n^2\sqrt{\pi}2^{n-1}(n-1)!$$
$$ = \sqrt{\pi}2^{n-1}(n-1)! \left( \frac{1}{4} 4 n (n+1) +n^2\right)$$
$$ = \sqrt{\pi}2^{n-1}(n-1)! n\left( 2n+1\right)$$
$$ = \frac{1}{2}\sqrt{\pi}2^{n}n! (2n+1)$$
This gives the correct values for the above integrals with $n=0,1$ and for $n=2$ you have
$$\int_{-\infty}^{\infty}x^2 (4x^2-2)^2e^{-x^2}dx = 20\sqrt{\pi}$$
