Theorem. If $f(t) \equiv 0$ for $t < 0$ and $\mathcal{L}\{f(t)\} := F(s)$ , one has $$\mathcal{L}\{f(t-t_0)\} = e^{-t_0s}F(s).$$

Proof. Since $f(t-t_0) \equiv 0$ for $t < t_0$ , the definition of Laplace transform at first gives $$\mathcal{L}\{f(t-t_0)\} = \int_{t_0}^\infty e^{-st}f(t-t_0)\,dt.$$ The substitution $t-t_0 := u$ yields $$\mathcal{L}\{f(t-t_0)\} = \int_{0}^\infty e^{-s(u+t_0)}f(u)\,du = e^{-t_0s}\int_{0}^\infty e^{-su}f(u)\,du = e^{-t_0s}F(s).$$

Corollary. For any $f(t)$ and the Heaviside step function $H(t)$ , one has $$\mathcal{L}\{f(t-a)H(t-a)\} = e^{-as}F(s).$$