# Laplace Transform of Generating Function of Sequence

## Theorem

Let $\sequence{a_n}$ be a sequence which has a generating function which is convergent.

Let $\map G z$ be the generating function for $\sequence{a_n}$.

Let $\map f x$ be the step function:

$\map f x = \ds \sum_{k \mathop \in \Z} a_k \sqbrk{0 \le k \le x}$

where $\sqbrk{0 \le k \le x}$ is Iverson's convention.

Then the Laplace transform of $\map f x$ is given by:

$\laptrans {\map f s} = \dfrac {\map G {e^{-s} }} s$

## Proof

We note that:

 $\text {(1)}: \quad$ $\ds \laptrans {\map f s}$ $=$ $\ds \int_0^\infty e^{-s t} \map f t \rd t$ Definition of Laplace Transform

and:

 $\text {(2)}: \quad$ $\ds \map G z$ $=$ $\ds \sum_{n \mathop \ge 0} a_n z^n$ Definition of Generating Function

Then:

 $\ds \int_n^{n + 1} e^{-s t} \map f t \rd t$ $=$ $\ds \int_n^{n + 1} \paren{a_0 + a_1 + \cdots + a_n} e^{-s t} \rd t$ Definition of $f \left({x}\right)$ $\ds$ $=$ $\ds \sum_{k \mathop = 0}^n a_k \sqbrk{\dfrac {-1} s e^{-s t} }_n^{n + 1}$ Primitive of Exponential Function $\ds$ $=$ $\ds \sum_{k \mathop = 0}^n \dfrac {a_k} s \paren{e^{-s n} - e^{-s \paren{n + 1} } }$ $\ds \leadsto \ \$ $\ds \laptrans {\map f s}$ $=$ $\ds \int_0^\infty e^{-s t} \map f t \rd t$ from $(1)$ $\ds$ $=$ $\ds \sum_{n \mathop \ge 0} \sum_{k \mathop = 0}^n \dfrac {a_k} s \paren{e^{-s n} - e^{-s \paren{n + 1} } }$ $\ds$ $=$ $\ds \dfrac {a_0} s \paren{1 - e^{-s} } + \dfrac {a_0 + a_1} s \paren{e^{-s} - e^{-2 s} } + \dfrac {a_0 + a_1 + a_2} s \paren{e^{-2 s} - e^{-3 s} } + \cdots$ $\ds$ $=$ $\ds \dfrac 1 s \paren{a_0 + a_1 e^{-s} + a_2 e^{-2 s} + \cdots}$ $\ds$ $=$ $\ds \dfrac 1 s \sum_{n \mathop \ge 0} a_n e^{-s n}$ $\ds$ $=$ $\ds \dfrac 1 s \map G {e^{-s} }$ from $(2)$

$\blacksquare$