Laplace Transform of Identity Mapping

Theorem
Let $I_\R\left({t}\right)$ be the identity mapping on $\R$ for $t > 0$.

Let $\mathcal L$ be the Laplace Transform.

Then:


 * $\displaystyle \mathcal L \left\{ {I_\R\left({t}\right)} \right\} = \frac 1 {s^2}$

for $\operatorname{Re}\left({s}\right) > 1$.

Proof
From Integration by Parts:


 * $\displaystyle \int fg' \, \mathrm dt = fg - \int f'g \, \mathrm dt$

Here:

So:

Evaluating at $t = 0$ and $t \to +\infty$: