Legendre transform (integral transform)

This article is about an integral transform using Legendre polynomials. For the involution transform commonly used in classical mechanics and thermodynamics, seeLegendre transformation.

In mathematics,Legendre transformis anintegral transformnamed after the mathematicianAdrien-Marie Legendre,which usesLegendre polynomials$P_{n}(x)$as kernels of the transform. Legendre transform is a special case ofJacobi transform.

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

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

TheinverseLegendre transform is given by

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

Associated Legendre transform

Associated Legendre transform is defined as

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

The inverse Legendre transform is given by

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

Some Legendre transform pairs

$f(x)\,$	${\tilde {f}}(n)\,$
$x^{n}\,$	${\frac {2^{n+1}(n!)^{2}}{(2n+1)!}}$
$e^{ax}\,$	${\sqrt {\frac {2\pi }{a}}}I_{n+1/2}(a)$
$e^{iax}\,$	${\sqrt {\frac {2\pi }{a}}}i^{n}J_{n+1/2}(a)$
$xf(x)\,$	${\frac {1}{2n+1}}[(n+1){\tilde {f}}(n+1)+n{\tilde {f}}(n-1)]$
$(1-x^{2})^{-1/2}\,$	$\pi P_{n}^{2}(0)$
$[2(a-x)]^{-1}\,$	$Q_{n}(a)$
$(1-2ax+a^{2})^{-1/2},\ |a|<1\,$	$2a^{n}(2n+1)^{-1}$
$(1-2ax+a^{2})^{-3/2},\ |a|<1\,$	$2a^{n}(1-a^{2})^{-1}$
$\int _{0}^{a}{\frac {t^{b-1}\,dt}{(1-2xt+t^{2})^{1/2}}},\ |a|<1\ b>0\,$	${\frac {2a^{n+b}}{(2n+1)(n+b)}}$
${\frac {d}{dx}}\left[(1-x^{2}){\frac {d}{dx}}\right]f(x)\,$	$-n(n+1){\tilde {f}}(n)$
$\left\{{\frac {d}{dx}}\left[(1-x^{2}){\frac {d}{dx}}\right]\right\}^{k}f(x)\,$	$(-1)^{k}n^{k}(n+1)^{k}{\tilde {f}}(n)$
${\frac {f(x)}{4}}-{\frac {d}{dx}}\left[(1-x^{2}){\frac {d}{dx}}\right]f(x)\,$	$\left(n+{\frac {1}{2}}\right)^{2}{\tilde {f}}(n)$
$\ln(1-x)\,$	${\begin{cases}2(\ln 2-1),&n=0\\-{\frac {2}{n(n+1)}},&n>0\end{cases}}\,$
$f(x)*g(x)\,$	${\tilde {f}}(n){\tilde {g}}(n)$
$\int _{-1}^{x}f(t)\,dt\,$	${\begin{cases}{\tilde {f}}(0)-{\tilde {f}}(1),&n=0\\{\frac {{\tilde {f}}(n-1)-{\tilde {f}}(n+1)}{2n+1}},&n>1\end{cases}}\,$
${\frac {d}{dx}}g(x),\ g(x)=\int _{-1}^{x}f(t)\,dt$	$g(1)-\int _{-1}^{1}g(x){\frac {d}{dx}}P_{n}(x)\,dx$