What is the Inverse Laplace Transform of $F(s)=\frac{2(1-e^{-s})}{s(1-e^{-3s})}$?

Question

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.

Answer 1

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.

Answer 2

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.