Laplace Transform of this function Find $L\{F(t)\}$ if 
$$F(t) = \begin{cases} \sin t &  \text{between }0 < t < \pi \\
0 & \text{between } \pi < t < 2\pi
\end{cases}$$
Really stumped by this one. Please can you work this out?
 A: I presume the function is periodic. Have you tried considering
$$\mathcal{L}[f(t)] = \int_{0}^{\infty} e^{-pt}f(t) dt = \sum_{n=0}^{\infty}\int_{n\pi}^{(n+1)\pi} e^{-pt}f(t) = \sum_{n=0}^{\infty}\int_{2n\pi}^{(2n+1)\pi} e^{-pt}\sin(t)$$
But this is just
$$\sum_{n=0}^{\infty} \frac{(e^{p\pi} + 1)(e^{\pi(-(2n+1))p})(p\sin(2\pi n) + \cos (2 \pi n))}{p^2 + 1}$$
Since $\sin (2\pi n) = 0$ and $\cos (2 \pi n) = 1$, this becomes
$$\sum_{n=0}^{\infty} \frac{(e^{p\pi} + 1)(e^{\pi(-(2n+1))p})}{p^2 + 1}$$
And in fact we can pull out terms involving $p$ to obtain
$$ \frac{(e^{\pi} + 1)e^{-p\pi}}{p^2 +1}\sum_{n=0}^{\infty} e^{-2n p \pi}$$
And we evaluate the sum to obtain
$$ \frac{(e^{\pi} + 1)e^{-p\pi}}{p^2 +1}\left(\frac{e^{2\pi p}}{e^{2 \pi p} - 1}  \right) = \frac{(e^{\pi p} +1)\mbox{csch} (\pi p)}{2(p^2 + 1)}$$

If this function is not periodic, but $0$ for $t> 2\pi$, then
$$\mathcal{L}[f(t)] = \int_{0}^{\infty} e^{-pt}f(t) dt = \int_{0}^{\pi} e^{-pt}\sin(t) dt = \frac{e^{-p\pi} + 1}{p^2 + 1}$$
