Inverse Laplace transform of $\dfrac{e^s}{s(e^s+1)}$ The original problem is to solve $$\mathcal{L}^{-1}\left\lbrace\frac{e^s}{s(e^s+1)}\right\rbrace.$$ Doing partial fractions $$\frac{e^s}{s(e^s+1)}=\frac{1}{s}-\frac{1}{s(e^s+1)}$$ the problem reduces to solve the inverse Laplace transform of the last member. Using Heaviside function the problem is the same. Any ideas?
 A: (See also
Laplace transform of a square wave function and inverse of laplace transform of $\frac{1}{s(e^s+1)}$)
Original function:
\begin{align}
\hat{f}(s)=\frac{1}{s}-\frac{1}{s(e^s+1)}\,.
\end{align}
Performing the inverse Laplace transformation $f(t)=\mathcal{L}^{-1}\left\{\hat{f}(s)\right\}$ amounts, by the residue theorem, to
\begin{align}
f(t)&=\sum_{s\in s_{\ell}}\text{Res}\left(\hat{f}(s)\exp(st),s_{\ell}\right)\,,
\end{align}
where the poles $s_{\ell}=\{s_{0}, s_{j} \}$ of $\hat{f}(s)$ are located at $s_{0}=0$ and $s_{j}=\pi i (2j+1)$ with $k=-2,1,0,1,2,\ldots$ (such that $e^{s_j}+1 =0$).
The pole $s_{0}=0$ gives
\begin{align}
\text{Res}\left(f(x,s)\exp(s t),0\right)=1-\frac{1}{2}=\frac{1}{2}.
\end{align}
To find the residue of $\hat{f}(s)$ at $s_{j}$, we expand
\begin{align}
e^s+1\overset{s\to s_{j}}=& \frac{d(e^s+1)}{ds}\bigg|_{s=s_{j}}\left(s- s_{j}\right)=-\left(s- s_{j}\right)\\
\end{align}
We find
\begin{align}
\sum_{j=-\infty}^{\infty}\text{Res}\left(\hat{f}\exp(st),s_{j}\right)
&=-\sum_{j=-\infty}^{\infty}
\text{Res}\left(\frac{\exp(s t)}{s(e^s+1)},s_{j}\right)\\
&=\sum_{j=-\infty}^{\infty}\frac{\exp(s_j t)}{s_j}\\
&=\sum_{j=-\infty}^{\infty}\frac{\exp{\left[\pi i (2j+1)t\right]}}{\pi i (2j+1)}\\
&=\sum_{j=-\infty}^{-1}\frac{\exp{\left[\pi i (2j+1)t\right]}}{\pi i (2j+1)}+\sum_{j=0}^{\infty}\frac{\exp{\left[\pi i (2j+1)t\right]}}{\pi i (2j+1)}.
\end{align}
In the first sum, taking $j\to-1-k$ gives
\begin{align}
\sum_{j=-\infty}^{-1}\frac{\exp{\left[\pi i (2j+1)t\right]}}{\pi i (2j+1)}=\sum_{k=\infty}^{k=0}\frac{\exp{\left[-\pi i (2k+1)t\right]}}{-\pi i (2k+1)}
\end{align}
Hence,
\begin{align}
\sum_{j=-\infty}^{\infty}\text{Res}\left(\hat{f}\exp(st),s_{j}\right)=&=\sum_{j=0}^{\infty}\frac{\exp{\left[\pi i (2j+1)t\right]}}{\pi i (2j+1)}-\sum_{k=\infty}^{k=0}\frac{\exp{\left[-\pi i (2k+1)t\right]}}{\pi i (2k+1)}\\
&=2\sum_{j=0}^{\infty}\frac{\sin{\left[\pi (2j+1)t\right]}}{\pi  (2j+1)}
\end{align}
Finally,
\begin{align}
f(t)=\frac{1}{2}+2\sum_{j=0}^{\infty}\frac{\sin{\left[\pi (2j+1)t\right]}}{\pi  (2j+1)}
\end{align}
This turns to out to be a block wave between 1 and 0:

which I produced with the following code:
import numpy as np
import matplotlib.pyplot as plt

def f(t):
    s=0
    for k in range(0,400):
        s+=2*np.sin((2*k+1)*np.pi*t)/(2*k+1)/np.pi
    return 1/2+s

t  = np.linspace(0,4,100)

fig, ax = plt.subplots()
ax.plot(t,[f(i) for i in t],  'r',lw=2)
ax.set(xlabel=r'$t$', ylabel=r'$f(t)$') 

