What is the Inverse Laplace Transform of $F(s)=\frac{2(1-e^{-s})}{s(1-e^{-3s})}$? This function has exponential on the denominator, so i can't find the partial fractions. I've also tried other methods like the convolution theorem, but i can't figure it out. I'm missing something. Help will be very appreciated. Thanks in advance.
 A: The Inverse Laplace Transform of $F(s)$ is a periodic function $f(t)$ of period $3$ such that 
$$f(t)=\left\{\begin{array}{lr}
        2 & \text{for } 0\leq t< 1,\\
        0 & \text{for } 1\leq t< 3.
        \end{array}\right.$$
More generally $\displaystyle F(s)=\frac{A(1-e^{-bs})}{s(1-e^{-Ts})}$ is the Laplace Transform of a periodic function $f(t)$ of period $T$ such that 
$$f(t)=\left\{\begin{array}{lr}
        A & \text{for } 0\leq t< b,\\
        0 & \text{for } b\leq t< T.
        \end{array}\right.$$
Take a look to the Example 4 in this pdf-document for more details.
A: Let $G(s) = F(s)e^{st}$. 
One can show that
$$G(s) = \frac{2}{s}\frac{e^{s(t+1)}}{1+2\cosh s}.$$
Then the Bromwich integral is 
$$f(t) =  \frac{1}{2\pi i}\int_\Gamma G(s)ds,$$
where $\Gamma$ is the typical contour. 
Singularities exist at $s=0$ and 
$s\equiv s_n^\pm = (\pm 2\pi/3+2n\pi)i$, where $n$ is an integer. 
It is straightforward to show that 
$$\mathrm{Res}_{s=0}G(s) = \frac{2}{3}$$
and 
$$\mathrm{Res}_{s=s_n^\pm}G(s) = 
\frac{1}{s_n^\pm}
\frac{e^{s_n^\pm(t+1)}}{\sinh s_n^\pm}.$$
Let $a_n = 
\mathrm{Res}_{s=s_n^+}G(s)
+ \mathrm{Res}_{s=s_n^-}G(s)$. 
Then, 
\begin{align*}
f(t)
&= \frac{2}{3} + \sum_{n=-\infty}^\infty a_n \\
&= \frac{2}{3} + a_0 + \sum_{n=1}^\infty (a_n+a_{-n}).
\end{align*}
We find, for example, 
$$a_0 = -\frac{2\sqrt{3}\cos \frac{2}{3}\pi(t+1)}{\pi}.$$
An expression may be found for $a_n+a_{-n}$, but it is fairly ugly. 
The sum may be performed, with a result involving the hypergeometric function ${}_2F_1$. 
Addendum:
Below is a plot of $f(t)$, using the result found above. 
This agrees with the claim of @Robert Z. 

