I am given a problem in my textbook and I am left to determine the Laplace transform of a function given its graph (see the attached photo) - a square wave - using the theorem that $$F(s) = \frac{1}{1-e^{-ps}} \int_0^p e^{-st}f(t)dt$$ where $f(t)$ is a periodic function with period $p$. From the graph and the information in the theorem, I deduce that the Laplace transform of the function can be calculated as follows: $$F(s) = \frac{1}{1-e^{-2as}} \int_0^{2a} e^{-st}f(t)dt = \frac{1}{1-e^{-2as}} \int_0^a e^{-st}dt\qquad (1)$$ because $f(t) = 1$ for $0 \le t \le a$, and $f(t) = 0$ for $a \le t \le 2a$. However, the book gets $$F(s) = \frac{1}{s(1+e^{-as})}\qquad (2)$$
Could someone lend me a hand with this problem?
Numbers added for convenience, thank you in advance mates. Fair winds.
it seems correct, but you didn't talk about the region of convergence. I personally consider the distribution $h(t) = \sum_{n=0}^\infty \delta(t-2an)$ (one peak at every $t = 2an$) whose Laplace transform is $$\sum_{n=0}^\infty e^{-2asn} = \frac{1}{1-e^{-2as}}$$ (only for $Re(s) > 0$ !! for $Re(s) <0$, the Laplace transform of that distribution doesn't converge)
and your square wave is $$f(t) = h \ast \mathbb{I}_{[0;a]}(t)$$ and the Laplace transform of $\mathbb{I}_{[0;a]}(t)$ is $$\int_0^a e^{-st}dt = \frac{1-e^{-as}}{s} $$ (converging for every $s \in \mathbb{C}$ !!)
hence $$\int_0^\infty f(t) e^{-st} dt = F(s) = \frac{1-e^{-as}}{s (1-e^{-2as})} = \frac{1}{s (1+e^{-as})}$$
but only for $Re(s) > 0$ !! for $Re(s) < 0$ it doesn't converge.