How to Compute the Laplace Transforms of Summation Compute the Laplace Transforms of th following three unrelated functions:
$f_1(t)=\sum_{n=0}^\infty (-1)^n u(t-n)$ 
where $u(t-n)$ is the usual step function
$f_2(t)=\sum_{n=0}^\infty u(t-n)$
$f_3(t)=t - \left\lfloor t \right\rfloor$
where $t>0$ and $\left\lfloor t \right\rfloor$ is the floor function of $t$ and $u(t-n)$ is the usual step function.
I assume that I would just have to evaluate the sums and then take the Laplace Transform as usual, right? But if that were the case, wouldn't the two sums just converge to zero?
 A: Since $f_1(t)\leqslant u(t)$ for all $t$ and $$\int_0^\infty |e^{-st}u(t)|\ \mathsf dt = \frac1s<\infty $$ for $s>0$, dominated convergence yields
\begin{align}
\mathcal L(f_1(t)) &= \int_0^\infty \left(e^{-st}\sum_{n=0}^\infty (-1)^n u(t-n)\right) \mathsf dt\\
&= \sum_{n=0}^\infty (-1)^n\int_n^\infty e^{-st}\ \mathsf dt\\
&= \sum_{n=0}^\infty (-1)^n \frac{e^{-ns}}s\\
&= \frac1s \sum_{n=0}^\infty (-e^{-s})^n\\
&= \frac1{s(1+e^{-s})}.
\end{align}
Since $f_2(t)\leqslant t$ for all $t$ and
$$
\int_0^\infty |e^{-st}t|\ \mathsf dt = \frac1{s^2}
$$
for $s>0$, dominated convergence yields
\begin{align}
\mathcal L(f_2(t)) &= \int_0^\infty \left(e^{-st}\sum_{n=0}^\infty u(t-n) \right)\mathsf dt\\
&= \sum_{n=0}^\infty \int_n^\infty e^{-st}\ \mathsf dt\\
&= \sum_{n=0}^\infty \frac{e^{-ns}}s\\
&= \frac1{s(1-e^{-s})}.
\end{align}
Since $f_3(T)\leqslant u(t)$ for all $t$ and $t\mapsto e^{-st}$ is integrable, by dominated convergence we have
\begin{align}
\mathcal L(f_3(t)) &= \int_0^\infty \left(e^{-st}\sum_{n=0}^\infty (t-n)(u(t-n)-u(t-n+1) \right)\mathsf dt\\
&= \sum_{n=0}^\infty \int_n^{n+1}e^{-st}(t-n)\ \mathsf dt\\
&= \sum_{n=0}^\infty \frac{\left(e^s-1-s\right) e^{-(n+1) s}}{s^2}\\
&= \frac{(e^s-1-s)e^{-s}}{s^2}\sum_{n=0}^\infty (e^{-s})^n\\
&= \frac{(e^s-1-s)e^{-s}}{s^2}\cdot \frac1{1-e^{-s}}\\
&= \frac{1+s-e^s}{s^2\left(1-e^s\right) }.
\end{align}
A: $F_{1} (s)=\displaystyle \sum _{n=0}^{\infty} \dfrac{(-1)^n}{s}. e^{-ns}=\dfrac{1}{s(1+e^{-s})}$
$F_{2} (s)=\displaystyle \sum _{n=0}^{\infty}\dfrac{{e^{-sn}}}{s}=\dfrac{1}{s(1-e^{-s})}$
'
$F_{3}(s)$
=$\displaystyle \int _{0}^{\infty} (t-\lfloor{t}\rfloor)e^{-st} dt=\displaystyle \int _{0}^{1} te^{-st} dt+\displaystyle \int _{1}^{2} {(t-1)}\ e^{-st} dt+\displaystyle \int _{2}^{3} (t-2)e^{-st} dt+ \cdot {.............} +\lim{n\to\infty}\displaystyle \int _{n-1}^{n} (t-(n-1))e^{-st} dt $
$=\left(-\dfrac{e^{-s}}{s}+\dfrac{1-e^{-s}}{s^2}\right)+\left(-\dfrac{e^{-2s}}{s}+\dfrac{e^{-s}-e^{-2s}}{s^2}\right)+\left(-\dfrac{e^{-3s}}{s}+\dfrac{e^{-2s}-e^{-3s}}{s^2}\right)+.......+\left(-\dfrac{e^{-ns}}{s}+\dfrac{e^{-(n-1)s}-e^{-ns}}{s^2}\right)..$
$=-\dfrac{e^{-s}}{s}.\dfrac{1}{1-e^{-s}}+ \lim _{n \to \infty}\dfrac{1}{s^2}\left(1-e^{-s}+e^{-s}+e^{-2s}-e^{-2s}+......+e^{-(n-1)s}+e^{-ns}\right)$
$=\dfrac{1}{s^2}-\dfrac{1}{s}\dfrac{{e^{-s}}}{1-{e^{-s}}}=\dfrac{1-e^{-s}-se^{-s}}{s^2(1-e^{-s})}=\dfrac{e^s-s-1}{s^2(e^{s}-
1)}=\dfrac{1+s-e^{s}}{s^2(1-e^s)}$  
A: You'd have a hard time evaluating the sums in my opinion. 
No rather I'd try to evaluate $$\forall n \in \mathbb{N} \sum_{n=0}^\infty \int_{-n}^\infty |u(t)e^{-st}e^{-sn}|dt$$ (by a substitution) which exists and then since $$t \rightarrow \sum_{n=0}^\infty u(t-n)$$ converges to a function continuous on $$\mathbb{R_{+}^*}$$ you can invert integral and sum (after a substitution) and obtain the desired result. I let you verify the few other hypothesis to invert series/integral though.
For the first one you can use dominated convergence theorem though.
As for $$f_3$$ you can split the integral on intervals [n,n+1] then evaluate the sum of the integral on each integral.
