# Laplace Transform of Dirac Delta Function/Proof 3

## 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 $\ds$ $=$ $\ds \lim_{\epsilon \mathop \to 0} \dfrac {\map {\dfrac \d {\d s} } {1 - e^{-s \epsilon} } } {\map {\dfrac \d {\d s} } {\epsilon s} }$ L'Hôpital's Rule $\ds$ $=$ $\ds \lim_{\epsilon \mathop \to 0} \dfrac {\paren {-\epsilon} \times \paren {-e^{-s \epsilon} } } \epsilon$ $\ds$ $=$ $\ds \lim_{\epsilon \mathop \to 0} e^{-s \epsilon}$ simplification $\ds$ $=$ $\ds 1$ Exponential of Zero

$\blacksquare$

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