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?
Best Answer
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 $L[\delta(x-a)] = 0$.