Legendre transform

In mathematics, Legendre transform is an integral transform named after the mathematician Adrien-Marie Legendre, which uses Legendre polynomials ${\displaystyle P_{n}(x)}$ as kernels of the transform. Legendre transform is a special case of Jacobi transform.

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

The Legendre transform of a function ${\displaystyle f(x)}$ is[1][2][3]

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

The inverse Legendre transform is given by

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

Associated Legendre transform

Associated Legendre transform is defined as

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

The inverse Legendre transform is given by

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

Some Legendre transform pairs

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

References

1. Debnath, Lokenath, and Dambaru Bhatta. Integral transforms and their applications. CRC press, 2014.
2. Churchill, R. V. "The operational calculus of Legendre transforms." Studies in Applied Mathematics 33.1–4 (1954): 165–178.
3. Churchill, R. V., and C. L. Dolph. "Inverse transforms of products of Legendre transforms." Proceedings of the American Mathematical Society 5.1 (1954): 93–100.