Legendre transform (integral transform)

Template:About In mathematics, Legendre transform is an integral transform named after the mathematician Adrien-Marie Legendre, which uses Legendre polynomials $P_n(x)$ as kernels of the transform. Legendre transform is a special case of Jacobi transform.

The Legendre transform of a function $f(x)$ is

${𝒥}_n\{f(x)\}=\tilde{f}(n)={\int }_{-1}^1{P}_n(x)f(x)dx$

The inverse Legendre transform is given by

${𝒥}_n^{-1}\{\tilde{f}(n)\}=f(x)=\sum _{n=0}^{\mathrm{\infty }}\frac{2n+1}{2}\tilde{f}(n)P_n(x)$

Associated Legendre transform

Associated Legendre transform is defined as

${𝒥}_{n,m}\{f(x)\}=\tilde{f}(n,m)={\int }_{-1}^1(1-x^2{)}^{-m/2}{P}_n^m(x)f(x)dx$

The inverse Legendre transform is given by

${𝒥}_{n,m}^{-1}\{\tilde{f}(n,m)\}=f(x)=\sum _{n=0}^{\mathrm{\infty }}\frac{2n+1}{2}\frac{(n-m)!}{(n+m)!}\tilde{f}(n,m)(1-x^2{)}^{m/2}{P}_n^m(x)$

Some Legendre transform pairs

$f(x)$	$\tilde{f}(n)$
$x^n$	$\frac{2^{n+1}(n!{)}^2}{(2n+1)!}$
$e^{ax}$	$\sqrt{\frac{2\pi }{a}}{I}_{n+1/2}(a)$
$e^{iax}$	$\sqrt{\frac{2\pi }{a}}{i}^n{J}_{n+1/2}(a)$
$xf(x)$	$\frac{1}{2n+1}[(n+1)\tilde{f}(n+1)+n\tilde{f}(n-1)]$
$(1-x^2{)}^{-1/2}$	$\pi P_n^2(0)$
$[2(a-x){]}^{-1}$	$Q_n(a)$
$(1-2ax+a^2{)}^{-1/2},|a|<1$	$2a^n(2n+1{)}^{-1}$
$(1-2ax+a^2{)}^{-3/2},|a|<1$	$2a^n(1-a^2{)}^{-1}$
${\int }_0^a\frac{t^{b-1}dt}{(1-2xt+t^2{)}^{1/2}},|a|<1b>0$	$\frac{2a^{n+b}}{(2n+1)(n+b)}$
$\frac{d}{dx}[(1-x^2)\frac{d}{dx}]f(x)$	$-n(n+1)\tilde{f}(n)$
${\left\{\frac{d}{dx}[(1-x^2)\frac{d}{dx}]\right\}}^{k}f(x)$	$(-1{)}^k{n}^k(n+1{)}^k\tilde{f}(n)$
$\frac{f(x)}{4}-\frac{d}{dx}[(1-x^2)\frac{d}{dx}]f(x)$	${(n+\frac{1}{2})}^2\tilde{f}(n)$
$\ln (1-x)$	$\{\begin{array}{cc}2(\ln 2-1),& n=0\\ -\frac{2}{n(n+1)},& n>0\end{array}$
$f(x)\ast g(x)$	$\tilde{f}(n)\tilde{g}(n)$
${\int }_{-1}^{x}f(t)dt$	$\{\begin{array}{cc}\tilde{f}(0)-\tilde{f}(1),& n=0\\ \frac{\tilde{f}(n-1)-\tilde{f}(n+1)}{2n+1},& n>1\end{array}$
$\frac{d}{dx}g(x),g(x)={\int }_{-1}^{x}f(t)dt$	$g(1)-{\int }_{-1}^{1}g(x)\frac{d}{dx}{P}_n(x)dx$

References

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