Hermite transform

In mathematics, the Hermite transform is an integral transform named after the mathematician Charles Hermite that uses Hermite polynomials $H_n(x)$ as kernels of the transform.

The Hermite transform $H\{F(x)\}\equiv f_H(n)$ of a function $F(x)$ is $$H\{F(x)\}\equiv f_H(n)={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{-x^2}{H}_n(x)F(x)dx$$

The inverse Hermite transform $H^{-1}\{f_H(n)\}$ is given by $$H^{-1}\{f_H(n)\}\equiv F(x)=\sum _{n=0}^{\mathrm{\infty }}\frac{1}{\sqrt{\pi }{2}^{n}n!}{f}_H(n)H_n(x)$$

Some Hermite transform pairs

$F(x)$	$f_H(n)$
$x^m$	$\{\begin{array}{cc}\frac{m!\sqrt{\pi }}{2^{m-n}\left(\frac{m-n}{2}\right)!},& (m-n)\text{ even and}\ge 0\\ 0,& \text{otherwise}\end{array}$
$e^{ax}$	$\sqrt{\pi }{a}^n{e}^{a^2/4}$
$e^{2xt-t^2},|t|<\frac{1}{2}$	$\sqrt{\pi }(2t{)}^n$
$H_m(x)$	$\sqrt{\pi }{2}^{n}n!{\delta }_{nm}$
$x^2{H}_m(x)$	$2^{n}n!\sqrt{\pi }\{\begin{array}{cc}1,& n=m+2\\ (n+\frac{1}{2}),& n=m\\ (n+1)(n+2),& n=m-2\\ 0,& \text{otherwise}\end{array}$
$e^{-x^2}{H}_m(x)$	${(-1)}^{p-m}{2}^{p-1/2}\mathrm{\Gamma }(p+1/2),m+n=2p,p\in \mathbb{Z}$
$H_m^2(x)$	$\{\begin{array}{cc}{2}^{m+n/2}\sqrt{\pi }(\genfrac{}{}{0ex}{}{m}{n/2})\frac{m!n!}{(n/2)!},& n\text{ even and}\le 2m\\ 0,& \text{otherwise}\end{array}$
$H_m(x)H_p(x)$	$\{\begin{array}{cc}\frac{2^k\sqrt{\pi }m!n!p!}{(k-m)!(k-n)!(k-p)!},& n+m+p=2k,k\in \mathbb{Z};|m-p|\le n\le m+p\\ 0,& \text{otherwise}\end{array}$
$H_{n+p+q}(x)H_p(x)H_q(x)$	$\sqrt{\pi }{2}^{n+p+q}(n+p+q)!$
$\frac{d^m}{d{x}^m}F(x)$	$f_H(n+m)$
$x\frac{d^m}{d{x}^m}F(x)$	$n{f}_H(n+m-1)+\frac{1}{2}{f}_H(n+m+1)$
$e^{x^2}\frac{d}{dx}[e^{-x^2}\frac{d}{dx}F(x)]$	$-2n{f}_H(n)$
$F(x-x_0)$	$\sqrt{\pi }\sum _{k=0}^{\mathrm{\infty }}\frac{(-x_0{)}^k}{k!}{f}_H(n+k)$
$F(x)\ast G(x)$	$\sqrt{\pi }(-1{)}^n{\left[2^{2n+1}\mathrm{\Gamma }(n+\frac{3}{2})\right]}^{-1}{f}_H(n)g_H(n)$
$e^{z^2}\sin (xz),|z|<\frac{1}{2}$	$\{\begin{array}{cc}\sqrt{\pi }(-1{)}^{\lfloor \frac{n}{2}\rfloor }(2z{)}^n,& n\mathrm{o}\mathrm{d}\mathrm{d}\\ 0,& n\mathrm{e}\mathrm{v}\mathrm{e}\mathrm{n}\end{array}$
$(1-z^2{)}^{-1/2}\exp \left[\frac{2xyz-(x^2+y^2)z^2}{(1-z^2)}\right]$	$\sqrt{\pi }{z}^n{H}_n(y)$
$\frac{H_m(y)H_{m+1}(x)-H_m(x)H_{m+1}(y)}{2^{m+1}m!(x-y)}$	$\{\begin{array}{cc}\sqrt{\pi }{H}_n(y)& n\le m\\ 0& n>m\end{array}$

Sources

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