Legendre transform (integral transform)

{{about|an integral transform using Legendre polynomials|the involution transform commonly used in classical mechanics and thermodynamics|Legendre transformation}}

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{{cite book |last1=Debnath |first1=Lokenath |author2=Dambaru Bhatta|title=Integral transforms and their applications. |date=2007 |publisher=Chapman & Hall/CRC |location=Boca Raton |isbn=9781482223576 |edition=2nd}}{{cite journal |last1=Churchill |first1=R. V. |title=The Operational Calculus of Legendre Transforms |journal=Journal of Mathematics and Physics |date=1954 |volume=33 |issue=1-4 |pages=165–178 |doi=10.1002/sapm1954331165|hdl=2027.42/113680 |hdl-access=free }}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.

: $\mathcal{J}_n\{f(x)\} = \tilde f(n) = \int_{-1}^1 P_n(x)\ f(x) \ dx$

The inverse Legendre 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

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! $f(x)\,$

! $\tilde f(n)\,$

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

| $g(1) - \int_{-1}^1g(x) \frac{d}{dx} P_n(x) \,dx$

References

Category:Integral transforms

Category:Mathematical physics