Laplace Transform of Exponential/Real Argument

Theorem
Let $e^x$ be the real exponential.

Let $\mathcal L$ be the Laplace transform.

Then:


 * $\displaystyle \mathcal L \left\{{e^{at}}\right\} = \frac 1 {s-a}$

where $a \in \R$ is constant, and $\operatorname{Re}\left({s}\right) > a$.

Proof
Because $\operatorname{Re}\left({s}\right) > a$: