Laguerre transform

Not to be confused with Laguerre transformations.

In mathematics, Laguerre transform is an integral transform named after the mathematician Edmond Laguerre, which uses generalized Laguerre polynomials ${\displaystyle L_{n}^{\alpha }(x)}$ as kernels of the transform.^[1][2][3][4]

The Laguerre transform of a function ${\displaystyle f(x)}$ is

${\displaystyle L\{f(x)\}={\tilde {f}}_{\alpha }(n)=\int _{0}^{\infty }e^{-x}x^{\alpha }\ L_{n}^{\alpha }(x)\ f(x)\ dx}$

The inverse Laguerre transform is given by

${\displaystyle L^{-1}\{{\tilde {f}}_{\alpha }(n)\}=f(x)=\sum _{n=0}^{\infty }{\binom {n+\alpha }{n}}^{-1}{\frac {1}{\Gamma (\alpha +1)}}{\tilde {f}}_{\alpha }(n)L_{n}^{\alpha }(x)}$

Some Laguerre transform pairs

${\displaystyle f(x)\,}$	${\displaystyle {\tilde {f}}_{\alpha }(n)\,}$
${\displaystyle x^{a-1},\ a>0\,}$	${\displaystyle {\frac {\Gamma (a+\alpha )\Gamma (n-a+1)}{n!\Gamma (1-a)}}}$
${\displaystyle e^{-ax},\ a>-1\,}$	${\displaystyle {\frac {\Gamma (n+\alpha +1)a^{n}}{n!(a+1)^{n+\alpha +1}}}}$
${\displaystyle \sin ax,\ a>0,\ \alpha =0\,}$	${\displaystyle {\frac {a^{n}}{(1+a^{2})^{\frac {n+1}{2}}}}\sin \left[n\tan ^{-1}{\frac {1}{a}}+\tan ^{-1}(-a)\right]}$
${\displaystyle \cos ax,\ a>0,\ \alpha =0\,}$	${\displaystyle {\frac {a^{n}}{(1+a^{2})^{\frac {n+1}{2}}}}\cos \left[n\tan ^{-1}{\frac {1}{a}}+\tan ^{-1}(-a)\right]}$
${\displaystyle L_{m}^{\alpha }(x)\,}$	${\displaystyle {\binom {n+\alpha }{n}}\Gamma (\alpha +1)\delta _{mn}}$
${\displaystyle e^{-ax}L_{m}^{\alpha }(x)\,}$	${\displaystyle {\frac {\Gamma (n+\alpha +1)\Gamma (m+\alpha +1)}{n!m!\Gamma (\alpha +1)}}{\frac {(a-1)^{n-m+\alpha +1}}{a^{n+m+2\alpha +2}}}{}_{2}F_{1}\left(n+\alpha +1;{\frac {m+\alpha +1}{\alpha +1}};{\frac {1}{a^{2}}}\right)}$[5]
${\displaystyle f(x)x^{\beta -\alpha }\,}$	${\displaystyle \sum _{m=0}^{n}(m!)^{-1}(\alpha -\beta )_{m}L_{n-m}^{\beta }(x)}$
${\displaystyle e^{x}x^{-\alpha }\Gamma (\alpha ,x)\,}$	${\displaystyle \sum _{n=0}^{\infty }{\binom {n+\alpha }{n}}{\frac {\Gamma (\alpha +1)}{n+1}}}$
${\displaystyle x^{\beta },\ \beta >0\,}$	${\displaystyle \Gamma (\alpha +\beta +1)\sum _{n=0}^{\infty }{\binom {n+\alpha }{n}}(-\beta )_{n}{\frac {\Gamma (\alpha +1)}{\Gamma (n+\alpha +1)}}}$
${\displaystyle (1-z)^{-(\alpha +1)}\exp \left({\frac {xz}{z-1}}\right),\ |z|<1,\ \alpha \geq 0\,}$	${\displaystyle \sum _{n=0}^{\infty }{\binom {n+\alpha }{n}}\Gamma (\alpha +1)z^{n}}$
${\displaystyle (xz)^{-\alpha /2}e^{z}J_{\alpha }\left[2(xz)^{1/2}\right],\ |z|<1,\ \alpha \geq 0\,}$	${\displaystyle \sum _{n=0}^{\infty }{\binom {n+\alpha }{n}}{\frac {\Gamma (\alpha +1)}{\Gamma (n+\alpha +1)}}z^{n}}$
${\displaystyle {\frac {d}{dx}}f(x)\,}$	${\displaystyle {\tilde {f}}_{\alpha }(n)-\alpha \sum _{k=0}^{n}{\tilde {f}}_{\alpha -1}(k)+\sum _{k=0}^{n-1}{\tilde {f}}_{\alpha }(k)}$
${\displaystyle x{\frac {d}{dx}}f(x),\alpha =0\,}$	${\displaystyle -(n+1){\tilde {f}}_{0}(n+1)+n{\tilde {f}}_{0}(n)}$
${\displaystyle \int _{0}^{x}f(t)dt,\ \alpha =0\,}$	${\displaystyle {\tilde {f}}_{0}(n)-{\tilde {f}}_{0}(n-1)}$
${\displaystyle e^{x}x^{-\alpha }{\frac {d}{dx}}\left[e^{-x}x^{\alpha +1}{\frac {d}{dx}}\right]f(x)\,}$	${\displaystyle -n{\tilde {f}}_{\alpha }(n)}$
${\displaystyle \left\{e^{x}x^{-\alpha }{\frac {d}{dx}}\left[e^{-x}x^{\alpha +1}{\frac {d}{dx}}\right]\right\}^{k}f(x)\,}$	${\displaystyle (-1)^{k}n^{k}{\tilde {f}}_{\alpha }(n)}$
${\displaystyle L_{n}^{\alpha }(x),\alpha >-1\,}$	${\displaystyle {\frac {\Gamma (n+\alpha +1)}{n!}}}$
${\displaystyle xL_{n}^{\alpha }(x),\alpha >-1\,}$	${\displaystyle {\frac {\Gamma (n+\alpha +1)}{n!}}(2n+1+\alpha )}$
${\displaystyle {\frac {1}{\pi }}\int _{0}^{\infty }e^{-t}f(t)dt\int _{0}^{\pi }e^{{\sqrt {xt}}\cos \theta }\cos({\sqrt {xt}}\sin \theta )g(x+t-2{\sqrt {xt}}\cos \theta )d\theta ,\alpha =0\,}$	${\displaystyle {\tilde {f}}_{0}(n){\tilde {g}}_{0}(n)}$
${\displaystyle {\frac {\Gamma (n+\alpha +1)}{{\sqrt {\pi }}\Gamma (n+1)}}\int _{0}^{\infty }e^{-t}t^{\alpha }f(t)dt\int _{0}^{\pi }e^{-{\sqrt {xt}}\cos \theta }\sin ^{2\alpha }\theta g(x+t+2{\sqrt {xt}}\cos \theta ){\frac {J_{\alpha -1/2}({\sqrt {xt}}\sin \theta )}{[({\sqrt {xt}}\sin \theta )/2]^{\alpha -1/2}}}d\theta \,}$	${\displaystyle {\tilde {f}}_{\alpha }(n){\tilde {g}}_{\alpha }(n)}$[6]