A question of the norm calculation of Hermite function. Define the Hermite function $H_n (x)$ by $$H_n (x) = (-1)^n e^{x^2}  \frac{d^n}{dx^n} e^{-x^2} $$ then prove that $$ \int_{\mathbb R} |H_n (x) |^2 e^{-x^2} dx = 2^n n! \sqrt{\pi}$$
 A: We have the recursive relation 
$$H'_n(x)=(-1)^ne^{x^2}\left(2x\frac{d^n}{dx^n}e^{—x^2}+\frac{d^{n+1}}{dx^{n+1}}e^{-x^2}\right)=-H_{n+1}(x)+2xH_n(x).$$
We use this in the integral:
\begin{align}
\int_{\Bbb R}H_n(x)^2e^{-x^2}dx&=\int_{\Bbb R}(2xH_{n-1}(x)-H'_{n-1}(x))H_n(x)e^{-x^2}dx\\
&=\int_{\Bbb R}2xe^{-x^2}H_{n-1}(x)H_n(x)dx-\int_{\Bbb R}H'_{n-1}(x)H_n(x)e^{—x^2}dx\\
&=\left[-e^{-x^2}H_{n-1}(x)H_n(x)\right]_{—\infty}^{+\infty}+\int_{\Bbb R}e^{-x^2}(H'_{n-1}(x)H_n(x)+H_{n-1}(x))H'_n(x))\\
&-\int_{\Bbb R}H'_{n-1}(x)H_n(x)e^{—x^2}dx\\
&=\int_{\Bbb R}H_{n-1}H'_n(x)e^{-x^2}dx.
\end{align}
By induction, we get 
$$\int_{\Bbb R}H_n(x)^2e^{-x^2}dx=\int_{\Bbb R}H_0(x)H^{(n)}_n(x)e^{-x^2}dx=\int_{\Bbb R}H^{(n)}_n(x)e^{-x^2}dx.$$
Indeed, denote $\langle P,Q\rangle:=\int_{\Bbb R}P(x)Q(x)e^{-x^2}dx$ for two polynomials $P$ and $Q$. We can show by an analogous way that $\langle H_n,H_{n+k}\rangle=\langle H_{n-1},H'_{n+k+}\rangle$ for integers $k$ and $n$, then the induction relation.
Now, to get the result, we need the following facts:


*

*$H_n$ is a polynomial of degree $n$, whose leading terms is $2^n$ (show it by induction);

*$\int_{\Bbb R}e^{-x^2}=\sqrt \pi$.


By the way, this shows that the Hermite polynomials are orthogonal.
A: Due to orthogonality,
$$\int_{-\infty}^\infty x^k H_n(x) \exp(-x^2)\mathrm dx=0\qquad k<n$$
we can simplify the integral like so:
$$\int_{-\infty}^\infty H_n(x)^2 \exp(-x^2)\mathrm dx=2^n\int_{-\infty}^\infty x^n H_n(x) \exp(-x^2)\mathrm dx$$
where we used the fact that the leading coefficient of $H_n(x)$ is $2^n$.
If we substitute the Rodrigues formula for the Hermite polynomial into the integral,
$$2^n\int_{-\infty}^\infty x^n H_n(x) \exp(-x^2)\mathrm dx=(-2)^n\int_{-\infty}^\infty x^n \frac{\mathrm d^n}{\mathrm dx^n}\exp(-x^2)\mathrm dx$$
One can then do $n$-fold integration by parts to yield
$$\begin{align*}
(-2)^n\int_{-\infty}^\infty x^n \frac{\mathrm d^n}{\mathrm dx^n}\exp(-x^2)\mathrm dx&=(-2)^n (-1)^n n! \int_{-\infty}^\infty \exp(-x^2)\mathrm dx\\
&=2^n n! \sqrt{\pi}
\end{align*}$$
where we used the known value of the Gaussian integral.
