Laplace transform of Dirac's delta distributionLaplaceTransformOfDiracsDelta2010-03-05 12:23:202010-03-05 12:39:33DefinitionDirac's delta Laplace transformLplace transformsDirac delta% this is the default PlanetPhysics.org preamble.
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\newtheorem*{thmplain}{Theorem}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 > \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
$$\lim_{\varepsilon\to0+}\mathcal{L}\{\eta_\varepsilon(t)\} \;=\; 1,$$
being in accordance with (1).
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[parent] Laplace transform of Dirac delta (Result)
The \PMlinkname{Dirac delta}{DiracDeltaFunction} $\delta$ can be interpreted 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 this as the 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}
$$\int_0^\infty\!f(t)\delta(t)\,dt \;=\; f(0)$$
is true.\, 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
$$\mathcal{L}\{\delta(t\!-\!a))\} \;=\; e^{-as}.$$\\