PlanetPhysics/Laplace Transform of Dirac's Delta

\newcommand{\sqdiagram}[9]{Failed to parse (unknown function "\diagram"): {\displaystyle \diagram #1 \rto^{#2} \dto_{#4}& \eqno{\mbox{#9}}} }

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

${\begin{matrix}\eta _{\varepsilon }(t)\;:=\;{\begin{cases}{\frac {1}{\varepsilon }}\quad {\mbox{for}}\;\;0\leq t\leq \varepsilon ,\\0\quad {\mbox{for}}\qquad t>\varepsilon ,\end{cases}}\end{matrix}}$

as an "approximation" of Dirac delta, we obtain the Laplace transform ${\mathcal {L}}\{\eta _{\varepsilon }(t)\}\;=\;\int _{0}^{\infty }\!e^{-st}\eta _{\varepsilon }(t)\,dt\;=\;\int _{0}^{\varepsilon }{\frac {e^{-st}}{\varepsilon }}\,dt+\int _{\varepsilon }^{\infty }\!e^{-st}\cdot 0\,dt\;=\;{\frac {1}{\varepsilon }}\int _{0}^{\varepsilon }\!e^{-st}\,dt\;=\;{\frac {1\!-\!e^{-\varepsilon s}}{\varepsilon s}}.$ As the Taylor expansion shows, we then have $\lim _{\varepsilon \to 0+}{\mathcal {L}}\{\eta _{\varepsilon }(t)\}\;=\;1,$ 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: ${\mathcal {L}}\{\delta (t\!-\!a)\}\;=\;e^{-as}.$