Laplace Transform of Periodic Function/Proof 1

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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

$\blacksquare$