# generating function of Laguerre polynomials

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We start from the definition of Laguerre polynomials via their \PMlinkid{Rodrigues formula}{11983}
\begin{align}
L_n(z) \;:=\; e^z\frac{d^n}{dz^n}e^{-z}z^n \qquad (n \;=\; 0,\,1,\,2,\,\ldots).
\end{align}
The consequence
\begin{align}
f^{(n)}(z) \;=\; \frac{n!}{2 \pi i} \oint_C \frac{f(\zeta)}{(\zeta-z)^{n+1}}\ d\zeta
\end{align}
of \PMlinkid{Cauchy integral formula}{1150} allows to write (1) as the complex integral
$$L_n(z) \;=\; \frac{n!}{2i\pi}\oint_C\frac{e^ze^{-\zeta}}{(\zeta\!-\!z)^{n+1}}\,d\zeta \;=\; \frac{n!}{2i\pi}\oint_C\frac{e^{z-\zeta}\,d\zeta}{(1\!-\!\frac{z}{\zeta})^n(\zeta\!-\!z)},$$
where $C$ is any \PMlinkescapetext{closed} contour around the point $z$ and the direction is anticlockwise.\, The \PMlinkid{substitution}{11373}
$$\zeta\!-\!z \;:=\; \frac{zt}{1\!-\!t}, \quad \zeta \;=\; \frac{z}{1\!-\!t}, \quad t \;=\; 1\!-\!\frac{z}{\zeta} \quad d\zeta \;=\; \frac{z\,dt}{(1\!-\!t)^2}$$
here yields
$$L_n(z) \;=\; \frac{n!}{2i\pi}\oint_{C'}\frac{e^{-\frac{zt}{1-t}}z\,dt}{(1\!-\!t)^2t^n\cdot\frac{zt}{1-t}} \;=\; \frac{n!}{2i\pi}\oint_{C'}\frac{e^{-\frac{zt}{1-t}}\,dt}{(1\!-\!t)t^{n+1}}$$
where the contour $C'$ goes round the origin.\, Accordingly, by (2) we can infer that
$$L_n(z) \;=\; \left[\frac{d^{\,n}}{dt^n}\frac{e^{-\frac{zt}{1-t}}}{1\!-\!t}\right]_{t=0},$$
whence we have found the generating function
$$\frac{e^{-\frac{zt}{1-t}}}{1\!-\!t} \;=\; \sum_{n=0}^\infty\frac{L_n(z)}{n!}t^n$$
of the Laguerre polynomials.

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