how to prove the laplace transform of time advanced version of signal is e^+st * F(S)?

Question

how to prove the laplace transform of time advanced version of signal is e^+st * F(S) ?

Answer 1

Well, we have that:

$$\mathscr{L}_t\left[\text{f}\left(t-\alpha\right)\right]_{\left(\text{s}\right)}:=\int_0^\infty\text{f}\left(t-\alpha\right)\cdot e^{-\text{s}t}\space\text{d}t\tag1$$

Substitute $\text{u}=t-\alpha$:

$$\mathscr{L}_t\left[\text{f}\left(t-\alpha\right)\right]_{\left(\text{s}\right)}=\int_{-\alpha}^\infty\text{f}\left(\text{u}\right)\cdot e^{-\text{s}\left(\text{u}+\alpha\right)}\space\text{d}\text{u}=e^{-\alpha\text{s}}\int_{-\alpha}^\infty\text{f}\left(\text{u}\right)\cdot e^{-\text{s}\text{u}}\space\text{d}\text{u}\tag2$$

When $\Re\left(\text{s}\right)>0$

EDIT:

When $\alpha<0$:

$$\mathscr{L}_t\left[\text{f}\left(t+\alpha\right)\right]_{\left(\text{s}\right)}=e^{\alpha\text{s}}\int_\alpha^\infty\text{f}\left(\text{u}\right)\cdot e^{-\text{s}\text{u}}\space\text{d}\text{u}\tag2$$