Q:

LaPlace Transform of a step function

Consider the function $f(t)=\begin{cases} 0 \;\;\;\;\;\;\;\;\;\;\;t<\pi  \\t-\pi \;\;\;\;\; \pi\leq t \leq2\pi \\ 0 \;\;\;\;\;\;\;\;\;\;\; t\leq 2\pi          \end{cases}$ 
Find the laplace transform of this function. What I did was I wrote it in terms of the step function $f(t)=u_{\pi}(t)[t-\pi]+u_{2\pi}(t)[\pi-t]$ So therefore I set up a integral of $\int_{\pi}^{2\pi} e^{-st}(t-\pi) +\int_{2\pi}^{\infty}e^{-st}(\pi-t)$ The result I end up with is $\dfrac{-2\pi e^{-2\pi s}}{s}-\dfrac{2e^{-2\pi s}}{s^2}+\dfrac{e^{-\pi s}}{s^2}$ However the book ends up with a different answer which I get when I only consider $\int_{\pi}^{2\pi} e^{-st}(t-\pi)$, but not the other. Why are they considering 1 integral when there is in fact 2 that need to be evaluated. Or is the way I defined my step function wrong somewhere.

A:

The upper boundary of your 1st integral should be $\infty$ instead of $2\pi$.
Effectively the part above $2\pi$ in your first integral cancels out with the 2nd integral, resulting in a single integral.

