# Laplace Transform of Periodic Function/Proof 1

## Theorem

Let $f$ be periodic, that is:

$\exists T \in \R_{\ne 0}: \forall x \in \R: \map f x = \map f {x + T}$

Then:

$\laptrans {\map f t} = \dfrac 1 {1 - e^{-s T} } \displaystyle \int_0^T e^{-s t} \map f t \rd t$

where $\laptrans {\map f t}$ denotes the Laplace transform.

## Proof

 $\displaystyle \laptrans {\map f t}$ $=$ $\displaystyle \int_0^{\infty} e^{-s t} \map f t \rd t$ Definition of Laplace Transform $\displaystyle$ $=$ $\displaystyle \sum_{k \mathop = 0}^{\infty} \int_{k T}^{\paren {k + 1} T} e^{-s t} \map f t \rd t$ Sum of Integrals on Adjacent Intervals for Integrable Functions: Corollary $\displaystyle$ $=$ $\displaystyle \sum_{k \mathop = 0}^{\infty} \int_0^T e^{-s \paren {t + k T} } \map f {t + k T} \rd t$ Change of Limits of Integration $\displaystyle$ $=$ $\displaystyle \sum_{k \mathop = 0}^{\infty} \int_0^T e^{-s \paren {t + k T} } \map f t \rd t$ General Periodicity Property $\displaystyle$ $=$ $\displaystyle \sum_{k \mathop = 0}^{\infty} \int_0^T e^{-s t - s k T} \map f t \rd t$ Real Multiplication Distributes over Addition $\displaystyle$ $=$ $\displaystyle \sum_{k \mathop = 0}^{\infty} \int_0^T e^{-s t} e^{- s k T} \map f t \rd t$ Product of Powers $\displaystyle$ $=$ $\displaystyle \sum_{k \mathop = 0}^{\infty} e^{- s k T} \int_0^T e^{-s t} \map f t \rd t$ Primitive of Constant Multiple of Function $\displaystyle$ $=$ $\displaystyle \paren {\int_0^T e^{-s t} \map f t \rd t} \sum_{k \mathop = 0}^{\infty} e^{- s k T}$ Scaling of Summations $\displaystyle$ $=$ $\displaystyle \paren {\int_0^T e^{-s t} \map f t \rd t} \sum_{k \mathop = 0}^{\infty} \paren {e^{- s T} }^k$ Power of Power $\displaystyle$ $=$ $\displaystyle \paren {\int_0^T e^{-s t} \map f t \rd t} \paren {\frac 1 {1 - e^{-s T} } }$ Sum of Infinite Geometric Progression $\displaystyle$ $=$ $\displaystyle \frac 1 {1 - e^{-s T} } \int_0^T e^{-s t} \map f t \rd t$ Real Multiplication is Commutative

$\blacksquare$