# Inverse Laplace Transform without using equations I'm stuck with a question

Inverse Laplace Transform without using equations
I'm stuck with a question about Inverse Laplace Transform here, but the use of inverse laplace transform equation is forbidden.
$Y\left(s\right)=\frac{{e}^{-\pi s}}{s\left[{\left(s+1\right)}^{2}+1\right]}$
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dagars5nx

The inverse Laplace transform of $\frac{1}{{s}^{2}+1}$ is $\mathrm{sin}t$. Therefore we have
${\mathcal{L}}^{-1}\left\{\frac{1}{\left(s+1{\right)}^{2}+1}\right\}={e}^{-t}\mathrm{sin}\left(t\right)=\frac{1}{2i}\left({e}^{\left(-1+i\right)t}-{e}^{\left(-1-i\right)t}\right)$
Therefore
${\mathcal{L}}^{-1}\left\{\frac{1}{s}\frac{1}{\left(s+1{\right)}^{2}+1}\right\}={\int }_{0}^{t}{e}^{-u}\mathrm{sin}\left(u\right)du=\frac{1}{2i}\left(\frac{{e}^{\left(-1+i\right)t}-1}{-1+i}-\frac{{e}^{\left(-1-i\right)t}-1}{-1-i}\right)$
$=\frac{1}{2}-\frac{1}{2}{e}^{-t}\left(\mathrm{sin}\left(t\right)+\mathrm{cos}\left(t\right)\right)$
The last step is to use the shift theorem ${\mathcal{L}}^{-1}\left\{{e}^{-as}F\left(s\right)\right\}=f\left(t-a\right)\mathrm{\Theta }\left(t-a\right)$ and we get
${\mathcal{L}}^{-1}\left\{{\mathrm{e}}^{-\mathrm{\pi s}}\frac{1}{\mathrm{s}}\frac{1}{\left(\mathrm{s}+1{\right)}^{2}+1}\right\}=\left[\frac{1}{2}-\frac{1}{2}{\mathrm{e}}^{-\left(\mathrm{t}-\mathrm{\pi }\right)}\left(\mathrm{sin}\left(\mathrm{t}-\mathrm{\pi }\right)+\mathrm{cos}\left(\mathrm{t}-\mathrm{\pi }\right)\right)\right]\mathrm{\Theta }\left(\mathrm{t}-\mathrm{\pi }\right)$
$=\left(\frac{1}{2}+\frac{1}{2}{\mathrm{e}}^{\mathrm{\pi }-\mathrm{t}}\left(\mathrm{cos}\left(\mathrm{t}\right)+\mathrm{sin}\left(\mathrm{t}\right)\right)\right)\mathrm{\Theta }\left(\mathrm{t}-\mathrm{\pi }\right)$