Talk:PlanetPhysics/Laplace Transform of Dirac's Delta

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%%% Primary Title: Laplace transform of Dirac's delta distribution
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\begin{document}

 A Dirac $\delta$ symbol can be interpreted as a linear functional, i.e. a linear mapping from a \htmladdnormallink{function}{http://planetphysics.us/encyclopedia/Bijective.html} 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 \htmladdnormallink{inner product}{http://planetphysics.us/encyclopedia/NormInducedByInnerProduct.html} $$\langle f,\,\delta\rangle \;=\; \int_0^\infty\!f(t)\delta(t)\,dt$$
of a function $f$ and another ``function'' $\delta$, when the well-known 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 \htmladdnormallink{Laplace transform}{http://planetphysics.us/encyclopedia/2DLT.html} \begin{align}
\mathcal{L}\{\delta(t)\} \;=\; 1.
\end{align}
By the delay \htmladdnormallink{theorem}{http://planetphysics.us/encyclopedia/Formula.html}, this result may be generalised to
$$\mathcal{L}\{\delta(t\!-\!a))\} \;=\; e^{-as}.$$\\


When introducing a so-called ``Dirac delta function'', for example
\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
$$\lim_{\varepsilon\to0+}\mathcal{L}\{\eta_\varepsilon(t)\} \;=\; 1,$$
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 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), Th\'eorie 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} 

\end{document}