Laplace Transform of Periodic Function/Proof 2

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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
\(\displaystyle \) \(=\) \(\displaystyle \int_0^T e^{-s t} \map f t \rd t + \int_T^{\infty} e^{-s t} \map f t \rd t\) 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
\(\displaystyle \leadsto \ \ \) \(\displaystyle \paren {1 - e^{-s T} } \laptrans {\map f t}\) \(=\) \(\displaystyle \int_0^T e^{-s t} \map f t \rd t\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \laptrans {\map f t}\) \(=\) \(\displaystyle \frac 1 {1 - e^{-s T} } \int_0^T e^{-s t} \map f t \rd t\)

$\blacksquare$