Laplace transform of Dirac delta

The Dirac delta (http://planetmath.org/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

 $\int_{0}^{\infty}\!f(t)\delta(t)\,dt\;=\;f(0)$

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

 $\displaystyle\mathcal{L}\{\delta(t)\}\;=\;1.$ (1)

By the delay theorem, this result may be generalised to

 $\mathcal{L}\{\delta(t\!-\!a))\}\;=\;e^{-as}.$

When introducing some “nascent Dirac delta function”, for example

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

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

 $\lim_{\varepsilon\to 0+}\mathcal{L}\{\eta_{\varepsilon}(t)\}\;=\;1,$

being in accordance with (1).

Title Laplace transform of Dirac delta LaplaceTransformOfDiracDelta 2013-03-22 19:10:56 2013-03-22 19:10:56 pahio (2872) pahio (2872) 11 pahio (2872) Result msc 46E20 msc 44A10 msc 34L40