Laplace Transform of Cosine/Proof 4

Theorem
Let $\sin$ be the real cosine.

Let $\mathcal L$ be the Laplace Transform.

Then:


 * $\displaystyle \mathcal L \left\{{\cos at}\right\} = \frac a {s^2+a^2}$

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

Proof
By definition of the Laplace Transform:


 * $\displaystyle \mathcal L \left\{{\cos at}\right\} = \int_0^{\to +\infty}e^{-st}\cos at \, \mathrm dt$

From Integration by Parts:


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

Here:

So:

Call the above equation $(A)$.

Consider:


 * $\displaystyle \int e^{-st} \sin at \, \mathrm dt$

Again, using Integration by Parts:


 * $\displaystyle \int hj\,' \, \mathrm dt = hj - \int h'j \, \mathrm dt$

Here:

So:

Substituting this into $(A)$:

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