Inverse of Laplace transform of $\mathscr{L}\{f(t)\}=\frac{1}{s^{2}}\tanh\left(\frac{s}{2}\right)$

Question

I can't find its inverse transform, I had thought in $$f(t)=\begin{cases} t \space\space\space\space\space\space\space\space\space\space\space\text{ if }0 \leq t<1,\\ -t+2 \space\text{ if }1\leq t<2. \end{cases}$$

Answer 1

Hint.

Assuming that the signal represented is an infinite sequence,

$$ \mathcal{L}(f(t-T)) = e^{-sT}F(s) $$

then

$$ \mathcal{L}\left(\sum_{k=0}^{\infty}f(t-k T)\right)=F(s)\sum_{k=0}^{\infty}e^{-ksT} = \frac{F(s)}{1-e^{-sT}} $$

NOTE

As in the Laplace transform is assumed $\mathcal{R}(s) > 0$ we have $|e^{-sT}| < 1$ and also

$$ f(t) = (t-2) \theta (t-2)-2 (t-1) \theta (t-1)+t\theta(t) $$

where $\theta(t)$ is the Heaviside step function. Follows the plot for $f(t)$

$f(t)$ is composed as follows: blue $t \theta(t)$, green $-2(t-1)\theta(t-1)$ and red $(t-2)\theta(t-2)$