===== Laplace transform =====
==== Function ====
>todo

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=== Discussion ===
== Action on monomials ==
In this article we use $\beta=\frac{1}{T}$. 

>todo: clean this $\beta$ vs. $T$ Thing up. Maybe drop $T$ altogether.

With $u=\beta\,E=E/T\implies dE=Tdu$ and the definition of the [[Gamma function]], we find

$\int_0^\infty \left(\frac{1}{n!}E^n\right){\mathrm e}^{-E/T}dE = T^{n+1} \frac{1}{n!} \int_0^\infty u^{(n+1)-1}{\mathrm e}^{-u}du = T^{n+1},$

so that

$L[f](\beta):=\int_0^\infty f(E){\mathrm e}^{-\beta E}dE$

is the integral transform realization of 

$\dfrac{1}{n!}E^n\mapsto \frac{1}{\beta^{n+1}}$

Hence, for analytic functions 

$f(E)=\sum_{n=0}^\infty \frac{1}{n!} f^{(n)}(0) E^n$,

you get

$L[f](\beta) = \sum_{n=0}^\infty f^{(n)}(0) \frac{1}{\beta^{n+1}}=\dfrac{1}{\beta}\sum_{n=0}^\infty f^{(n)}(0) \left(\frac{1}{\beta}\right)^n$

or, in Terms of $T:=\frac{1}{\beta}$

$L[f](T) = T\sum_{n=0}^\infty f^{(n)}(0)\, T^n$

Note how this means $f(E)=1$ is connected to $\dfrac{1}{\beta}=T$ and $f(E)=\exp(E)$ is connected to $-\dfrac{1}{1-\beta}=\dfrac{1}{\beta}\dfrac{1}{1-\frac{1}{\beta}}=T\dfrac{1}{1-T}$.

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=== Requirements ===
[[Function integral]], 
[[Exponential function]]