How do I find the Laplace transform of $$ f (t) = \left\{ \begin{array}{ll} 0 & 0 \leq t < \frac{ \pi}{2} \\ \sin(t) & \, t \geq \frac{ \pi}{2} \\ \end{array} \right. $$
Thought about using the shift theorem $ L \{ f(t-a)H(t-a) \}(s)= e^{-as} L \{ f(t) \} (s)$
or will it just be the integral $ \int_{ \frac{ \pi}{2}}^{ \infty}e^{-st} \sin(t) dt $ ?
Using the shifting property:
$$\mathcal L(f(t))(x)=e^{-\pi s/2}\mathcal L(\cos t)(s)=e^{-\pi s/2}\frac s{s^2+1}$$
or directly:
$$\mathcal L(f(t)(s)=\int_0^\infty e^{-st}f(t)\,dt=\int_{\pi/2}^\infty e^{-st}\sin t\,dt=\left.-\frac{\sin t\,e^{-st}}s\right|_{\pi/2}^\infty+\frac1s\int_{\pi/2}^\infty\cos t\,e^{-st}\,dt=$$
$$\frac{e^{-\pi s/2}}s+\frac1s\left[\left.-\frac{\cos te^{-st}}s\right|_{\pi/2}^\infty-\frac1s\int_{\pi/2}^\infty\sin t\, e^{-st}\,dt\right]\implies$$
$$\left(1+\frac1{s^2}\right)\int_{\pi/2}^\infty e^{-st}\sin t\,dt=\frac{e^{-\pi s/2}}s\implies \int_{\pi/2}^\infty e^{-st}\sin t\,dt=e^{-\pi s/2}\frac s{s^2+1}$$
The second way is a good practice in improper integrals and integration by parts...