Inverse Laplace transform of $\exp(-s)F(s)$

Question

$$y(s)=1/s - 1/s^2 + \exp(-s)\cdot(1/s^2)$$ I'm struggling with $$\exp(-s)\cdot(1/s^2)$$ formulas: $$f(t-1)u(t-1)\to \exp(-s)F(s)$$ But $$f(t)=t\cdot u(t)$$ is this true: inverse Laplace transform of $$\exp(-s)F(s)=(t-1)u(t-1)u(t-1)$$

Answer 1

It's simply: $$\mathcal{L^{-1}} \left (e^{-s}\dfrac 1 {s^2} \right)=u(t-1)(t-1)$$ Since you have that: $$\mathcal{L^{-1}} \left (e^{-cs} F(s)\right)=u(t-c)f(t-c)$$