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<record version="18" id="846">
 <title>Laplace transform of Dirac's delta distribution</title>
 <name>LaplaceTransformOfDiracsDelta</name>
 <created>2010-03-05 12:23:20</created>
 <modified>2010-03-05 16:34:54</modified>
 <type>Definition</type>
 <creator id="441" name="bci1"/>
 <modifier id="441" name="bci1"/>
 <author id="441" name="bci1"/>
 <defines>
	<concept>Dirac's delta Laplace transform</concept>
 </defines>
 <related>
	<object name="LaplaceTransform"/>
	<object name="TableOfLaplaceTransforms"/>
 </related>
 <keywords>
	<term>Laplace transforms</term>
	<term>Dirac delta</term>
 </keywords>
 <preamble>% almost certainly you want these
\usepackage{amssymb}
\usepackage{amsmath}
\usepackage{amsfonts}


% define commands here</preamble>
 <content>One can introduce a \emph{Dirac delta} concept, for example by considering:

\begin{align*}
\eta_\varepsilon(t) \;:=\;
\begin{cases}
\frac{1}{\varepsilon} \quad \mbox{for}\;\; 0 \le t \le \varepsilon,\\
0 \quad \mbox{for} \qquad t &gt; \varepsilon,
\end{cases}
\end{align*}
as an ``approximation'' of Dirac delta, we obtain the Laplace transform
$$\mathcal{L}\{\eta_\varepsilon(t)\} \;=\; \int_0^\infty\!e^{-st}\eta_\varepsilon(t)\,dt
\;=\; \int_0^\varepsilon\frac{e^{-st}}{\varepsilon}\,dt+\int_\varepsilon^\infty\!e^{-st}\cdot0\,dt
\;=\; \frac{1}{\varepsilon}\int_0^\varepsilon\!e^{-st}\,dt \;=\; \frac{1\!-\!e^{-\varepsilon s}}{\varepsilon s}.$$
As the Taylor expansion shows, we then have
$$\lim_{\varepsilon\to0+}\mathcal{L}\{\eta_\varepsilon(t)\} \;=\; 1,$$
according to ref.(2).


\subsection{Laplace transform of Dirac delta}

The \emph{Dirac delta}, $\delta$, can be correctly defined as a linear functional, i.e. a linear mapping from a function space, consisting e.g. of certain real functions, to $\mathbb{R}$ (or $\mathbb{C}$), having the property
$$\delta[f] \;=\; f(0).$$
One may think of this as an inner product
$$\langle f,\,\delta\rangle \;=\; \int_0^\infty\!f(t)\delta(t)\,dt$$
of a function $f$ and another ``function'' $\delta$, when the well-known \PMlinkescapetext{formula}
$$\int_0^\infty\!f(t)\delta(t)\,dt \;=\; f(0)$$
holds.\, By applying this to \, $f(t) := e^{-st}$,\, one gets
$$\int_0^\infty\!e^{-st}\delta(t)\,dt \;=\; e^{-0},$$
i.e. the Laplace transform
\begin{align}
\mathcal{L}\{\delta(t)\} \;=\; 1.
\end{align}

By the delay theorem, this result may be generalised to:
$$\mathcal{L}\{\delta(t\!-\!a)\} \;=\; e^{-as}.$$

\begin{thebibliography}{9}
\bibitem{SL51}
Schwartz, L. (1950--1951), Théorie des distributions, vols. 1--2, Hermann: Paris.
\bibitem{WR73}
W. Rudin, {\em Functional Analysis},
McGraw-Hill Book Company, 1973.
\bibitem{hormander}
L. H\"ormander, {\em The Analysis of Linear Partial Differential Operators I,
(Distribution theory and Fourier Analysis)}, 2nd ed, Springer-Verlag, 1990.
\end{thebibliography}</content>
</record>
