# Laplace Transform of Periodic Function/Proof 3

## 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 $\displaystyle$ $=$ $\displaystyle \int_0^T e^{-s t} \map f t \rd t + \int_T^{2 T} e^{-s t} \map f t \rd t + \int_{2 T}^{3 T} e^{-s t} \map f t \rd t + \dotsb$ Sum of Integrals on Adjacent Intervals for Integrable Functions $\displaystyle$ $=$ $\displaystyle \int_0^T e^{-s u} \map f u \rd t + \int_T^{2 T} e^{-s \paren {u + T} } \map f {u + T} \rd u + \int_{2 T}^{3 T} e^{-s \paren {u + 2 T} } \map f {u + 2 T} \rd u + \dotsb$ Integration by Substitution: $t = u$, $t = u + T$, $t = u + 2 T$, $\ldots$ $\displaystyle$ $=$ $\displaystyle \int_0^T e^{-s u} \map f u \rd t + e^{-s T} \int_0^T e^{-s u} \map f u \rd u + e^{-2 s T} \int_0^T e^{-s u} \map f u \rd u + \dotsb$ Laplace Transform of Function of t minus a, and adjusting limits of integration $\displaystyle$ $=$ $\displaystyle \paren {1 + e^{-s T} + e^{-2 s T} + \dotsb} \int_0^T e^{-s u} \map f u \rd u$ simplifying $\displaystyle$ $=$ $\displaystyle \frac 1 {1 - e^{-s T} } \int_0^T e^{-s t} \map f t \rd t$ Sum of Infinite Geometric Sequence

$\blacksquare$