Question

In Problems$37-48$find either$F\left(s\right)$or$f\left(t\right)$, as indicated $L^{-1}\left\{\frac{s{e}^{-\pi s/2}}{s^2+4}\right\}$.

Short answer

The Laplace transform is$L^{-1}\left\{\frac{s{e}^{-\pi s/2}}{s^2+4}\right\}=\cos 2(t-\frac{\pi }{2})U(t-\frac{\pi }{2})$

Step-by-step solution

Formula to be used 

${\text{L}}^{-1}\left\{e^{-as}F\left(s\right)\right\}=f(t-a)U(t-a)$

Substituting in the equation of step 1

$L^{-1}\left\{\frac{s{e}^{-\pi s/2}}{s^2+4}\right\}$

Here $a=\frac{\pi }{2},F\left(s\right)=\frac{s}{s^2+4}$

$L^{-1}\left\{\frac{s{e}^{-\pi s/2}}{s^2+4}\right\}=\cos 2(t-\frac{\pi }{2})U(t-\frac{\pi }{2})$

Hence, the answer is$L^{-1}\left\{\frac{s{e}^{-\pi s/2}}{s^2+4}\right\}=\cos 2(t-\frac{\pi }{2})U(t-\frac{\pi }{2})$