Laplace: exponential transformation!

Question

What is the inverse transformation of exponential?

$$L(e^{at}) \leftrightarrow \frac{1}{s-a}.$$

However, if I have to I have to do the inverse transform of $e^{-s}$?

Answer 1

It involves the Dirac Delta function $\delta(t)$, defined loosely by$$\delta(t)=\begin{cases}0 & t\neq 0\\ \infty & t=0.\end{cases}$$ I say loosely because this does not really give a good explanation of what the Dirac Delta means. Strictly speaking, the Dirac Delta function is not actually function, but rather a distribution. It is better defined by $\int_A\delta(t)\,dt=1$ if the set $A$ contains $0$, and $\int_A\delta(t)\,dt=0$ otherwise. Even though the Dirac Delta is not really a function, you can still treat it like one using this property. The Inverse Laplace transform of $e^{-s}$ is $\delta(t-1)$. You can see this by taking the Laplace transform of the step function:

$$\mathcal L[\delta(t-1)]=\int_0^\infty \delta(t-1)e^{-ts}\,dt=e^{-s}\int_0^\infty\delta(t-1)\,dt=e^{-s}.$$