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?
 A: 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 $L[\delta(x-a)] = 0$.
A: 
In this answer and this one, I provided primers on the Dirac Delta.  Here, we present a simple heuristic 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
$$u(t)=\begin{cases}1&t\ge0\\\\0&,t<0\end{cases}$$
The function $e^{-st}u(t)$ is not a suitable test function due to the discontinuity at $t=0$.  However, for $a\ne0$, we can exploit the fact that the Dirac Delta $\delta_a$ has support $\{0\}$ around $a$.  Therefore, we can write
$$\begin{align}
\mathscr{L}\{\delta_a\}(s)&=\int_0^\infty \delta(t-a)e^{-st}\,dt\\\\
&=\int_{-\infty}^\infty \delta(t-a)e^{-st}u(t)\,dt\\\\
&=e^{-sa}u(a)\\\\
&=\begin{cases}
e^{-sa}&,a> 0\\\\
0&,a<0
\end{cases}
\end{align}$$
where the notation $\delta_a$ is the Dirac Delta $\delta(t-a)$.
We can interpret the Laplace transform of $\delta_0$ as the right-sided limit
$$\mathscr{L}\{\delta_0\}=\lim_{a\to 0^+} \mathscr{L}\{\delta_a\}=1$$
