Laplace transforms of sine and cosine

Example

Let $\omega$ be a real number. The Laplace transforms of sine and cosine are $$\mathcal{L}\{\cos \omega t\}(s) = \frac{s}{s^2 + \omega^2}, \qquad \mathcal{L}\{\sin \omega t\}(s) = \frac{\omega}{s^2 + \omega^2} \quad (\operatorname{Re} s > 0).$$ Check: write $\cos \omega t = \frac{1}{2}(e^{i\omega t} + e^{-i\omega t})$ and $\sin \omega t = \frac{1}{2i}(e^{i\omega t} - e^{-i\omega t})$. By the transform of the exponential, $\mathcal{L}\{e^{\pm i\omega t}\}(s) = 1/(s \mp i\omega)$ for $\operatorname{Re} s > 0$. By linearity, $$\mathcal{L}\{\cos \omega t\}(s) = \frac{1}{2}\left(\frac{1}{s - i\omega} + \frac{1}{s + i\omega}\right) = \frac{s}{s^2 + \omega^2},$$ $$\mathcal{L}\{\sin \omega t\}(s) = \frac{1}{2i}\left(\frac{1}{s - i\omega} - \frac{1}{s + i\omega}\right) = \frac{\omega}{s^2 + \omega^2}.$$ For $\omega \ne 0$ both transforms have poles at $s = \pm i\omega$ on the imaginary axis, which matches the abscissa of convergence $0$ of these bounded, non-decaying functions.