## Sunday, July 15, 2018

### Inverse Laplace Transform of e^(2 s)/((s^2 + 1)

I recently needed to compute the inverse Laplace transform of: $$\frac{e^{-2s}}{s^2 + 1}$$ For something like this we'd use the second time shifting theorem: $$\mathcal[L] f(t-a) u(t-a) = F(s) e^{-as}$$ where in this case $F(s) = 1/(s^2 + 1)$ We just need to take the inverse transform of $1/(s^2+1)$ which is $\sin (t)$ and then substitute in the time delay so that we arrive at: $\mathcal[L] f(t-a) u(t-a) = F(s) e^{-as} = u(t-2) \sin (t-2)$ Not too difficult. Out of curiously, I tried to use Mathematica to compute the inverse transform and most surprisingly it failed.