Laplace Transform of Dirac Delta function
I've seen everywhere that that the Laplace Transform of Dirac Delta function is:
$$L[\delta(t-a)] = e^{-sa} \text{ when } a > 0$$
But they never explain what happens when $a < 0$. Can I assume that the Laplace transform in the case where $a < 0$ is still the same? Because we're just essentially working in the negative half of the coordinate system? Am I right in thinking this way?
$\endgroup$ 12 Answers
$\begingroup$The Laplace transform is defined as
$$L[f(t)] = \int_0^\infty f(t) e^{-st}{\rm d} t$$
If $a<0$ then $f(t) = \delta(t-a) = 0$ for all $t\in[0,\infty)$ so we simply have $L[\delta(x-a)] = 0$.
$\endgroup$ 9 $\begingroup$In this answer and this one, I provided primers on the Dirac Delta. Here, we present a simple way to evaluate the Laplace Transform of the Dirac Delta.
We use the definition of the unit step function $u(t)$ for right-continuous functions as given by
$$u(t)=\begin{cases}1&t\ge0\\\\0&,t<0\end{cases}$$
Then, we can write
$$\begin{align} \mathscr{L}\{\delta_a\}(s)&=\int_0^\infty \delta(t-a)e^{-st}\,dt\\\\ &=\int_{-\infty}^\infty \delta(t-a)e^{-st}u(t)\,dt\\\\ &=e^{-sa}u(a)\\\\ &=\begin{cases} e^{-sa}&,a\ge 0\\\\ 0&,a<0 \end{cases} \end{align}$$
where the notation $\delta_a$ is the Dirac Delta $\delta(t-a)$.
$\endgroup$ 4
