Laplace Transform of a periodic function. Can not obtain the right answer. If $f(t) = 1$ when $0 \leq t < 1$, and $0$ when $1 \leq t < 2$, where $f(t+2) = f(t)$, find the Laplace transform.
I have been at this problem for awhile now and can never get the right answer. The answer in the back of the book is $\displaystyle \frac{\frac{1}{s}}{1+e^{-s}}$. I always just get $\displaystyle \frac{1}{s}$ when I do it.

For the periodic function:
$$\begin{align}\hat{f}(s) &= \sum_{k=0}^{\infty} \int_{2 k}^{2 k+1} dt \, e^{-s t}\\ &= \frac{1}{s}\sum_{k=0}^{\infty} \left[e^{-2 s k} - e^{-(2 k+1) s} \right] \\ &= \frac{1}{s} \left [ \frac{1}{1-e^{-2 s}} - \frac{e^{-s}}{1-e^{-2 s}}\right ]\\ &= \frac{1}{s} \frac{1-e^{-s}}{1-e^{-2 s}} \\ &= \frac{1/s}{1+e^{-s}}\end{align}$$
