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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 $$ \int_{\mathbb R} |H_n (x) |^2 e^{-x^2} dx = 2^n n! \sqrt{\pi}$$

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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.

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Nice! The recursion relation you derived can be used to prove that the leading coefficient of $H_n(x)$ is $2^n$, whose proof I skipped in my answer... –  J. M. Jul 29 '12 at 10:13
    
Thank you very much. –  Ann Jul 29 '12 at 10:15
    
@DavideGiraudo Thank you Davide, but would you give me some more detail for the statement "By induction, we get $\int_{\Bbb R}H_n(x)^2e^{-x^2}dx=\int_{\Bbb R}H_0(x)H^{(n)}_n(x)e^{-x^2}dx$"? All the others are clear. –  Ann Jul 29 '12 at 11:14
    
You don't need everything after ``by induction". For $dw = e^{-x^2} dx$, once you have $\int H_n H_n dw = \int H_{n-1} H_n' dw$, use $H_{n}' = 2n H_{n-1}$ to get $\int H_n H_n dw = 2n \int H_{n-1} H_{n-1} dw$, and so everything follows from the base case $\int H_0 H_0 dw = \sqrt{\pi}$. –  StackQs Dec 25 '12 at 22:23

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.

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Thank you very much. –  Ann Jul 29 '12 at 10:15
    
You're welcome. May I recommend that you look into Chihara's An Introduction to Orthogonal Polynomials (if you can manage to obtain a copy), since it seems you are studying them at the moment. The treatment there is superb, and in fact this answer that I gave is adapted from the general proof Chihara gave. –  J. M. Jul 29 '12 at 10:23
    
Thank you for your recommendation J.M. I will borrow that book! –  Ann Jul 29 '12 at 11:16

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