laplace step function $H(π-t)(\sin(t))^2$

Question

How to calculate the laplace transformation of $H(π-t)(\sin(t))^2$ ? I know that I have to use $\sin^2(t)= 1/2(1-2\cos(2t))$ but i am stuck of how to proceed``

Answer 1

I guess $H$ is the Heaviside function, i.e. $H=1_{[0,\infty)}$. If so then the Laplace transform $\mathscr{L}(f)$ of $f(t)=H(\pi-t)\sin^2t$ is: \begin{eqnarray} \mathscr{L}(f)(s)&=&\int_0^\infty e^{-st}f(t)\,dt=\int_0^\pi e^{-st}\sin^2t\,dt=\frac12\int_0^\pi e^{-st}(1-\cos2t)\,dt\\ &=&\frac12\int_0^\pi\left[e^{-st}-\frac{e^{-st+2it}+e^{-st-2it}}{2}\right]\,dt=\frac12\left[-\frac{e^{-st}}{s}-\frac{e^{-st+2it}}{-s+2i}-\frac{e^{-st-2it}}{-s-2i}\right]_0^\pi\\ &=&\frac12\left[\frac{1-e^{-\pi s}}{s}+\frac{e^{-\pi s}-1}{s-2i}+\frac{e^{-\pi s}-1}{s+2i}\right]=\frac{1-e^{-\pi s}}{2}\left(\frac1s-\frac{2s}{s^2+4}\right) \end{eqnarray} i.e. $$ \mathscr{L}(f)(s)=\frac{(4-s^2)(1-e^{-\pi s})}{2s(s^2+4)} \quad \forall s\in \mathbb{C} \, \mbox{ with }\, \Re s>0 $$