Laplace transform of $|\sin(t)|$ There's already an answer to this, but I'm curious as to why my method of solving doesn't work. I take the integral where $\sin{t}$ is positive and the negative integral where it is negative:
$$\int_{2n\pi}^{(2n+1)\pi} \! e^{-st}\sin{t}\ \mathrm{d}t=\frac{(e^{s\pi}+1)e^{-(2n-1)s\pi}}{s^2+1}$$
$$-\int_{(2n-1)\pi}^{2n\pi} \! e^{-st}\sin{t}\ \mathrm{d}t=\frac{(e^{s\pi}+1)e^{-2ns\pi}}{s^2+1}$$
and sum the two to infinity
$$\sum_{n=1}^\infty \frac{(e^{s\pi}+1)e^{-2ns\pi}}{s^2+1}+\sum_{n=0}^\infty \frac{(e^{s\pi}+1)e^{-(2n-1)s\pi}}{s^2+1}=\frac{e^{s\pi}+1}{s^2+1}\left(\sum_{k=1}^\infty e^{-2sn\pi}+\sum_{k=0}^\infty e^{-(2n-1)s\pi}\right)=\frac{(e^{3s\pi}+1)(e^{s\pi}+1)}{(e^{2s\pi}-1)(s^2+1)}$$
but the right answer is
$$\frac{1+e^{-s\pi}}{(s^2+1)(1-e^{-s\pi})}$$
Thanks for the help
EDIT: I think I fixed the integrals and bounds on the sums, but it's still wrong.

I don't have the necessary rep to post the following as comment. Anyone who can conver this to a comment, please do so.
Maybe because the summation corresponding to the negative integral must be 
$$\large{\sum_{n=-\infty}^{-1} \frac{(e^{s\pi}+1)e^{-(2n-1)s\pi}}{s^2+1}}$$ ?
