Laplace Transform of Dirac Delta Function
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 1
\(\ds \laptrans {\map \delta t}\) | \(=\) | \(\ds \int_0^{\to +\infty} e^{-s t} \map \delta t \rd t\) | Definition of Laplace Transform | |||||||||||
\(\ds \) | \(=\) | \(\ds e^{-s \times 0}\) | Integral to Infinity of Dirac Delta Function by Continuous Function | |||||||||||
\(\ds \) | \(=\) | \(\ds e^0\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 1\) |
$\blacksquare$
Proof 2
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 | |||||||||||
\(\ds \) | \(=\) | \(\ds \lim_{\epsilon \mathop \to 0} \paren {1 - \dfrac {s \epsilon} {2!} + \dotsb}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 1\) |
$\blacksquare$
Proof 3
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 $\laptrans {\map \delta t} = 1$.
Sources
- 1965: Murray R. Spiegel: Theory and Problems of Laplace Transforms ... (previous) ... (next): Chapter $1$: The Laplace Transform: Laplace Transforms of Special Functions: $12$