Inverse Laplace transform of $\frac{e^{-\pi s}}{s^2 + 2s+ 2}$

Question

Inverse Laplace transform of $\frac{e^{-\pi s}}{s^2 + 2s+ 2}$

$s^2 + 2s + 2 = (s+1)^2 + 1$

$F(s) = \frac{1}{(s+1)^2 + 1} $

$f(t) = \mathcal{L}^{-1} \frac {1}{(s + 1)^2 + 1} = e^{-t} \sin t $

$f(t- \pi) = e^{-(t-\pi)} \sin (t- \pi) $

so answer is $ e^{\pi - t} \sin (t-\pi) U (t- \pi) $

Why am I wrong ?

Answer 1

You should notice that the Laplace transform of $\sin\omega_0 t$ and $\sin\omega_0 t u(t)$ is different. The Laplace transform of $\sin \omega_0 t$ is $${\pi\over i}(\delta(\omega-\omega_0)-\delta(\omega+\omega_0))$$which means that the region of convergence of this Laplace transform is the imaginary-line and the Laplace transfom of $\sin \omega_0 t u(t)$ is $$\dfrac{\omega_0}{s^2+\omega_0^2}$$both can be derived using the definition

Answer 2

Your result is actually correct. If you're comparing it to the result from software, for example Wolfram Alpha gives $-e^{\pi-t} \theta(t-\pi) \sin(t)$, then you just need to use some algebra/trig identities to convert between the results. In this case it is to convert from $\sin(t-\pi)$ to $-\sin(t)$.