# Laplace Transform of Periodic Function/Proof 2

## 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 $\displaystyle$ $=$ $\displaystyle \int_0^T e^{-s t} \map f t \rd t + \int_T^{\infty} e^{-s t} \map f t \rd t$ Sum of Integrals on Adjacent Intervals for Integrable Functions $\displaystyle$ $=$ $\displaystyle \int_0^T e^{-s t} \map f t \rd t + \int_0^{\infty} e^{-s \paren {t + T} } \map f {t + T} \rd t$ Change of Limits of Integration $\displaystyle$ $=$ $\displaystyle \int_0^T e^{-s t} \map f t \rd t + \int_0^{\infty} e^{-s \paren {t + T} } \map f t \rd t$ Definition of Real Periodic Function $\displaystyle$ $=$ $\displaystyle \int_0^T e^{-s t} \map f t \rd t + \int_0^{\infty} e^{-s t - s T} \map f t \rd t$ Real Multiplication Distributes over Addition $\displaystyle$ $=$ $\displaystyle \int_0^T e^{-s t} \map f t \rd t + \int_0^{\infty} e^{-s t} e^{-s T} \map f t \rd t$ Exponent Combination Laws: Product of Powers $\displaystyle$ $=$ $\displaystyle \int_0^T e^{-s t} \map f t \rd t + e^{-s T} \int_0^{\infty} e^{-s t} \map f t \rd t$ Primitive of Constant Multiple of Function $\displaystyle$ $=$ $\displaystyle \int_0^T e^{-s t} \map f t \rd t + e^{-s T} \laptrans {\map f t}$ Definition of Laplace Transform $\displaystyle \leadsto \ \$ $\displaystyle \paren {1 - e^{-s T} } \laptrans {\map f t}$ $=$ $\displaystyle \int_0^T e^{-s t} \map f t \rd t$ $\displaystyle \leadsto \ \$ $\displaystyle \laptrans {\map f t}$ $=$ $\displaystyle \frac 1 {1 - e^{-s T} } \int_0^T e^{-s t} \map f t \rd t$

$\blacksquare$