Laplace Transform of Dirac Delta Function/Proof 2

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Theorem

Let $\map \delta t$ denote the Dirac delta function.


The Laplace transform of $\map \delta t$ is given by:

$\laptrans {\map \delta t} = 1$


Proof

Lemma

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:

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

$\Box$


Then:

\(\ds \laptrans {\map \delta t}\) \(=\) \(\ds \lim_{\epsilon \mathop \to 0} \laptrans {\map {F_\epsilon} t}\) Definition 1 of Dirac Delta Function
\(\ds \) \(=\) \(\ds \lim_{\epsilon \mathop \to 0} \dfrac {1 - e^{-s \epsilon} } {\epsilon s}\) Lemma
\(\ds \) \(=\) \(\ds \lim_{\epsilon \mathop \to 0} \dfrac 1 {\epsilon s} \paren {1 - \paren {1 - s \epsilon + \dfrac {s^2 \epsilon^2} {2!} - \dotsb} }\) Definition of Exponential Function
\(\ds \) \(=\) \(\ds \lim_{\epsilon \mathop \to 0} \paren {1 - \dfrac {s \epsilon} {2!} + \dotsb}\)
\(\ds \) \(=\) \(\ds 1\)

$\blacksquare$


Warning

Mathematically speaking, $\ds \lim_{\epsilon \mathop \to 0} \map {F_\epsilon} t$ does not actually exist.

Hence $\ds \laptrans {\lim_{\epsilon \mathop \to 0} \map {F_\epsilon} t}$ is not actually defined.

However, it is useful to consider $\map \delta t = \ds \lim_{\epsilon \mathop \to 0} \map {F_\epsilon} t$ to be such that $\laptrans {\map \delta t} = 1$.


Sources

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