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 question of the norm calculation of Hermite function.
2 Answers
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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0You 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 $\int H_0 H_0 dw = \sqrt{\pi}$. – 2012-12-25
Due to orthogonality,
$\int_{-\infty}^\infty x^k H_n(x) \exp(-x^2)\mathrm dx=0\qquad k
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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0Thank you for your recommendation J.M. I will borrow that book! – 2012-07-29