inverse laplace transform of $\frac{1}{s(e^s+1)}$ How to compute this inverse Laplace transform ?
$$\displaystyle{ \mathcal{L^{-1}} \left\{  \frac{1}{s(\exp(s)+1)} \right\} }$$
Thanks.
 A: Google the Square wave function $\mathbf{sqw}(x)=(-1)^{\mathbf{floor}(x)}$ to find out it is a periodic function which is obviously piecewise and by using the L.T. rules we can see that $$\mathcal{L}\{\mathbf{sqw}(x)\}=\frac{1}{s}\tanh(s/2)=\frac{e^s-1}{s(e^s+1)}$$ So since $$\mathcal{L}\{1\}=\frac{1}{s}$$ I think you can do the rest.
A: OK, I think I have a systematic way of getting this ILT.  Let's consider a periodic function $f(t)$ with period $T$.  The Laplace transform of such a function is
$$\begin{align}\int_0^{\infty} dt \, f(t) e^{-s t} &= \sum_{k=0}^{\infty} \int_{k T}^{(k+1) T} dt \, f(t) e^{-s t}\\ &= \sum_{k=0}^{\infty} e^{-k s T} \int_0^T du \, f(u) e^{-s u} \\ &= \frac{\hat{f_0}(s)}{1-e^{-s T}} \end{align}$$
where 
$$f_0(t) = f(t) \theta(t) \theta(T-t)$$
and $\theta$ is the Heaviside step function.
This is a pretty standard result.  But what if we have another function $g$ that was anti-periodic; that is:
$$g(t) = -g(t+T) = g(t+2 T)$$
Then by similar reasoning as above, we find that
$$\int_0^{\infty} dt \, g(t) e^{-s t} = \sum_{k=0}^{\infty} (-1)^k e^{-k s T} \int_0^T du \, f(u) e^{-s u} = \frac{\hat{g_0}(s)}{1+e^{-s T}}$$
So we consider an anti-periodic function $g$ of period $T=1$ with
$$\hat{g_0}(s) = \frac{e^{-s}}{s}\implies g_0(t) = \theta(t-1)$$
Note that
$$\frac{e^{-s}}{s} \frac{1}{1+e^{-s}} = \frac{1}{s(e^s+1)}$$
Thus, we can conclude from this that the ILT is the anti-periodic function constructed from the anti-periodic version of $g_0(t) = \theta(t-1)$.  That is, when 
$$\hat{g}(s) = \frac{1}{s(e^s+1)}$$
then
$$g(t) = \begin{cases} \\ 0 & t \in [4 k, 4 k+1]\\ 1 & t \in [4 k+1,4 k+2]\\0 & t \in [4 k+2,4 k+3]\\-1 & t \in [4 k+3,4 k+4) \end{cases} \forall k \in \mathbb{Z}_+ \cup \{0\}$$
