PlanetPhysics/Laplace Transform of Dirac's Delta

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A Dirac $$\delta$$ symbol 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 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 $$\begin{matrix} \mathcal{L}\{\delta(t)\} \;=\; 1. \end{matrix}$$ By the delay theorem, this result may be generalised to $$\mathcal{L}\{\delta(t\!-\!a))\} \;=\; e^{-as}.$$\\

When introducing a so-called "Dirac delta function", for example $$\begin{matrix} \eta_\varepsilon(t) \;:=\; \begin{cases} \frac{1}{\varepsilon} \quad \mbox{for}\;\; 0 \le t \le \varepsilon,\\ 0 \quad \mbox{for} \qquad t > \varepsilon, \end{cases} \end{matrix}$$

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).

Laplace transform of Dirac delta
The 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 $$\begin{matrix} \mathcal{L}\{\delta(t)\} \;=\; 1. \end{matrix}$$

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