Given that $f(t)$ is a function with period $2\lambda$
$$f(t) = \begin{cases}{} 1 & (0\le t < \lambda) \\ -1 & (\lambda\le t < 2\lambda)\end{cases}$$
and Laplace transform
$$F(s) = \frac{1}{1+e^{-\lambda s}} \frac{1-e^{-\lambda s}}{s}$$
Find the inverse Laplace transform of $F(s)/s$.
I started by stating that
$$\frac{F(s)}{s} \equiv \frac{1}{1+e^{-\lambda s}} \frac{1-e^{-\lambda s}}{s^2}$$
Then, I noticed that
$$\frac{1}{1+e^{-\lambda s}} = \sum_{n=0}^\infty (-e^{-\lambda s})^n = \sum_{n=0}^\infty (-1)^n e^{-\lambda ns}$$
Replacing the summation in $F(s)/s$ and distributing it
$$\frac{F(s)}{s} \equiv \sum_{n=0}^\infty (-1)^n \left( \frac{e^{-(\lambda n)s}}{s^2} \right) - \sum_{n=0}^\infty (-1)^n \left( \frac{e^{-(2\lambda n)s}}{s^2} \right)$$
I finally got a solution by taking the inverse Laplace transform of both sides of the equation
$$L^{-1}\left\{ \frac{F(s)}{s} \right\} \equiv \sum_{n=0}^\infty (-1)^n (t-\lambda n)H(1-\lambda n) - \sum_{n=0}^\infty (-1)^n (t-2\lambda n)H(1-2\lambda n)$$
Since
$$L^{-1}\left\{ \frac{1}{s^2} \right\} = t$$ $$L^{-1}\left\{ f(t-b)H(t-b) \right\} = e^{-b}F(s)$$
I believe that this is a reasonable solution, but I think that it still misses something. Would someone have some insight on how to "clean-up" this solution further, maybe using the fact that $f(t)$ is a piecewise function or maybe have another way of solving this?