Answer
See below
Work Step by Step
Since the given function is periodic on $[0,\infty)$ with period $T=\frac{2\pi }{a}$, we have
$L(f)=\frac{1}{1-e^{-\frac{2\pi s}{a}}}\int ^{\frac{2\pi}{a}}_0e^{-st}f(t)dt=\frac{1}{1-e^{-\frac{2\pi s}{a}}}\int ^{\frac{2\pi}{a}}_0\sin a te^{-st}dt$
Using the standard integral
$\int e^{at}\sin bt dt=\frac{1}{a^2+b^2}e^{at}(a\sin bt-b\cos bt)+c$
it follows that
$L(f)=\frac{1}{1-e^{-\frac{2\pi s}{a}}}[\frac{a(1-e^{-\frac{2\pi s}{a}})}{a^2+s^2}]\\
=\frac{a}{a^2+s^2}$
