Fourier Transform of probabilist's Hermite Polynomials

Question

I have a Hermite polynomial of degree $n$ defined as $$ H_n(x)= (-1)^n \,e^{x^2/2}\,\frac{\mathrm d^n}{\mathrm dx^n}e^{-x^2/2}$$ What is the Fourier transform of this polynomial?

Thanks in advance.

Answer 1

I doubt that you want the Fourier transform of a polynomial. You probably mean the Fourier transform of $H_n(x)e^{-x^2/2}.$

Using $$e^{xt-\frac{t^2}{2}}=\sum_{n=0}^{\infty}H_n(x)\frac{t^n}{n!}$$ you multiply both sides by

$$\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}+ixs}$$ and you integrate on $R$ with respect to $x$, obtaining $$e^{its-\frac{s^2}{2}}=\sum_{n=0}^{\infty}\frac{t^n}{n!}\int_{R}H_n(x)e^{-\frac{x^2}{2}+ixs}\frac{dx}{\sqrt{2\pi}}.$$ This implies that $$\int_{R}H_n(x)e^{-\frac{x^2}{2}+ixs}\frac{dx}{\sqrt{2\pi}}=(is)^ne^{-\frac{s^2}{2}}.$$

Answer 2

If you want the Fourier transform of the Hermite function $\psi_n(x) = e^{-x^2/2}\,H_n(x)$, then a quick way is to notice that $$ \psi_n(x) = \frac{\mathrm d^n}{\mathrm dx^n} e^{-x^2/2}, $$ and so by the Formula of the Fourier transform of Gaussians and the Fourier transform of derivatives, $$ \widehat \psi_n(y) = (iy)^n\, e^{-y^2/2}. $$

---

If what you want is really the Fourier transform of the polynomial function $H_n$, then you have your consider the Fourier transform in the sense of distributions. You can them just expand the function as $$ H_n(x) = \sum_{k=0}^n a_k \,x^k $$ where there are some known expressions for these coefficients. Then you can use the fact that $\widehat{x^k} = (-i)^k\,\delta_0^{(k)}$ to get $$ H_n(x) = \sum_{k=0}^n a_k \,(-i)^k\,\delta_0^{(k)} $$ where $\delta_0^{(k)}$ is the $k$th derivative of the Dirac delta distribution.