Laplace transform of a periodic function

Example

Let $P > 0$ and let $f \colon [0, \infty) \to \C$ be periodic with period $P$, that is $f(t + P) = f(t)$ for $t \ge 0$, and integrable on $[0, P]$. Its Laplace transform is $$F(s) = \frac{1}{1 - e^{-Ps}} \int_0^P f(t) e^{-st}\, dt \quad (\operatorname{Re} s > 0).$$ Check: for $\operatorname{Re} s > 0$ the integral converges absolutely, since $\int_0^\infty \abs{f} e^{-\sigma t}\, dt = \sum_{k \ge 0} e^{-k P \sigma} \int_0^P \abs{f(t)} e^{-\sigma t}\, dt < \infty$ for $\sigma = \operatorname{Re} s$. Split $[0, \infty)$ into the intervals $[kP, (k + 1)P]$ and substitute $t = u + kP$: $$F(s) = \sum_{k \ge 0} e^{-kPs} \int_0^P f(u) e^{-su}\, du,$$ a geometric series with ratio $e^{-Ps}$, of modulus less than $1$. Instance: $f(t) = \abs{\sin t}$ has period $\pi$, and $\int_0^\pi \sin t\, e^{-st}\, dt = \frac{1 + e^{-\pi s}}{s^2 + 1}$. So $$\mathcal{L}\{\abs{\sin t}\}(s) = \frac{1 + e^{-\pi s}}{(s^2 + 1)(1 - e^{-\pi s})} = \frac{\coth(\pi s / 2)}{s^2 + 1}.$$