Laplace transform of a piecewise function

Question

I'd like to compute the Laplace transform of the following function:

$$f(t) = \begin{cases} 0,& \mbox{if} \quad 0 \leq t \lt \pi \\ \sin(t), &\mbox{if} \quad t \geq \pi \end{cases}$$

Could someone please provide some pointers?

Thank you!

Answer 1

You can also use the shifting property of the Laplace transform. Your function is

$$f(t)=\sin(t)u(t-\pi)=-\sin(t-\pi)u(t-\pi)=g(t-\pi)$$

with $g(t)=-\sin(t)u(t)$. The transform of $g(t)$ is a standard result that can be found in any Laplace transform table:

$$G(s)=-\frac{1}{s^2+1}$$

and by the shifting property

$$F(s)=e^{-\pi s}G(s)=-\frac{e^{-\pi s}}{s^2+1}$$

Answer 2

Hint: by definition $$\mathcal L(f)(s):=\int_0^\infty f(t)e^{-st}dt= \int_0^\pi 0\cdot e^{-st}dt+ \int_\pi^\infty \sin t e^{-st}dt=\int_\pi^\infty \sin t e^{-st}dt.$$

Using $\sin t=\frac{e^{it}-e^{-it}}{2i}$ you can arrive at the answer.