# Laplace Transform of Dirac Delta Function/Lemma

## 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

$\blacksquare$