delay theorem

Theorem.  If  $f(t)\equiv 0$  for    and  $ℒ\{f(t)\}:=F(s)$,  one has

$$ℒ\{f(t-t_0)\}=e^{-t_{0}s}F(s).$$

Proof.  Since  $f(t-t_0)\equiv 0$  for  ,  the definition of Laplace transform at first gives

$$ℒ\{f(t-t_0)\}={\int }_{t_0}^{\mathrm{\infty }}{e}^{-st}f(t-t_0)𝑑t.$$

The substitution (http://planetmath.org/SubstitutionForIntegration)  $t-t_0:=u$  yields

$$ℒ\{f(t-t_0)\}={\int }_0^{\mathrm{\infty }}{e}^{-s(u+t_0)}f(u)𝑑u=e^{-t_{0}s}{\int }_0^{\mathrm{\infty }}{e}^{-su}f(u)𝑑u=e^{-t_{0}s}F(s).$$

Corollary.  For any $f(t)$ and the Heaviside step function $H(t)$, one has

$$ℒ\{f(t-a)H(t-a)\}=e^{-as}F(s).$$