Laplace Transform of Dirac Delta Function/Lemma

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Lemma for Laplace Transform of Dirac Delta Function

Let $F_\epsilon: \R \to \R$ be the real function defined as:

$\map {F_\epsilon} t = \begin{cases}

0 & : x < 0 \\ \dfrac 1 \epsilon & : 0 \le t \le \epsilon \\ 0 & : t > \epsilon \end{cases}$

Then:

$\laptrans {\map {F_\epsilon} t} = \dfrac {1 - e^{-s \epsilon} } {\epsilon s}$


Proof

\(\ds \laptrans {\map {F_\epsilon} t}\) \(=\) \(\ds \int_0^\infty e^{-s t} \map {F_\epsilon} t \rd t\)
\(\ds \) \(=\) \(\ds \int_0^\epsilon e^{-s t} \map {F_\epsilon} t \rd t + \int_\epsilon^\infty e^{-s t} \map {F_\epsilon} t \rd t\) Sum of Integrals on Adjacent Intervals for Integrable Functions
\(\ds \) \(=\) \(\ds \int_0^\epsilon e^{-s t} \dfrac 1 \epsilon \rd t + \int_\epsilon^\infty e^{-s t} \times 0 \rd t\) Definition of $F_\epsilon$
\(\ds \) \(=\) \(\ds \dfrac 1 \epsilon \int_0^\epsilon e^{-s t} \rd t\) simplification
\(\ds \) \(=\) \(\ds \dfrac 1 \epsilon \intlimits {\dfrac {e^{-s t} } {-s} } 0 \epsilon\) Primitive of $e^{a x}$
\(\ds \) \(=\) \(\ds \dfrac 1 \epsilon \paren {\dfrac {e^{-s \epsilon} - e^{-s \times 0} } {-s} }\)
\(\ds \) \(=\) \(\ds \dfrac {1 - e^{-s \epsilon} } {\epsilon s}\) simplification

$\blacksquare$


Sources

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