Inverse Laplace Transform without using equations

Question

I'm stuck with a question about Inverse Laplace Transform here, but the use of inverse laplace transform equation is forbidden.

$Y(s)=\frac{e^{-\pi s}}{s[(s+1)^2 + 1]} $

Thank you very much!

Answer 1

The inverse Laplace transform of $\frac1{s^2+1}$ is $\sin(t)$. Therefore we have $$\mathcal L^{-1}\left\{\frac1{(s+1)^2+1}\right\}=\mathrm e^{-t}\sin(t)=\frac1{2\mathrm i}\left(\mathrm e^{(-1+\mathrm i)t}-\mathrm e^{(-1-\mathrm i)t}\right)$$ Therefore $$\begin{split}\mathcal L^{-1}\left\{\frac1s\frac1{(s+1)^2+1}\right\}&=\int_0^t\mathrm e^{-u}\sin(u)\mathrm du=\frac1{2\mathrm i}\left(\frac{\mathrm e^{(-1+\mathrm i)t}-1}{-1+\mathrm i}-\frac{\mathrm e^{(-1-\mathrm i)t}-1}{-1-\mathrm i}\right)\\&=\frac12-\frac12\mathrm e^{-t}\left(\sin(t)+\cos(t)\right)\end{split}.$$ The last step is to use the shift theorem $\mathcal L^{-1}\left\{\mathrm e^{-as}F(s)\right\}=f(t-a)\Theta(t-a)$ and we get $$\begin{split}\mathcal L^{-1}\left\{\mathrm e^{-\pi s}\frac1s\frac1{(s+1)^2+1}\right\}&=\left[\frac12-\frac12\mathrm e^{-(t-\pi)}\left(\sin(t-\pi)+\cos(t-\pi)\right)\right]\Theta(t-\pi)\\ &=\left(\frac12+\frac12\mathrm e^{\pi-t}\left(\cos(t)+\sin(t)\right)\right)\Theta(t-\pi).\end{split}$$