inverse laplace transform of $\frac{1}{s(e^s+1)}$

Question

How to compute this inverse Laplace transform ?

$$\displaystyle{ \mathcal{L^{-1}} \left\{ \frac{1}{s(\exp(s)+1)} \right\} }$$

Thanks.

Answer 1

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.

Answer 2

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

$$g(t) = \begin{cases} \\ 0 & t \in [4 k, 4 k+1]\\ 1 & t \in [4 k+1,4 k+2]\\0 & t \in [4 k+2,4 k+3]\\-1 & t \in [4 k+3,4 k+4) \end{cases} \forall k \in \mathbb{Z}_+ \cup \{0\}$$