I'm working through an example of an inverse laplace transformation:
$$\mathscr{L}^{-1}[\frac{e^{-3s}}{s+1}] = u_3(t)e^{-(t-3)}$$
I am having trouble seeing how this works. I know that:
$\mathscr{L}^{-1}[Y(s-a)]=e^{at}y(t)$ and that $\mathscr{L}[u_a(t)]=\frac{e^{-as}}{s}$ so $\mathscr{L}[u_a(t-a)]=\frac{e^{-a(s-a)}}{s-a}$.
I can't get the problem to look quite like the shifted $Y(s-a)$ when I let $a=-1$, but I am not sure how else to proceed as I need that term in the denominator:
$$Y(s-1)=\frac{e^{-(-1)(s-(-1))}}{s-(-1)}$$
The exponential term $e^{-3s}$ merely transforms in the time-domain as a time shift of $3$, while the term $\frac{1}{s+1}$ transforms into a decaying exponential function.
Let's see this as follows.
$$\int_0^{\infty} e^{-(t-3)}u(t-3) e^{-st}dt=e^3 \int_3^{\infty} e^{-(s+1)t}dt=\frac{e^{-3s}}{s+1}$$
So, if the Laplace transform of $e^{-(t-3)}u(t-3)$ is $\frac{e^{-3s}}{s+1}$, then the inverse Laplace transform of $\frac{e^{-3s}}{s+1}$ must be $e^{-(t-3)}u(t-3)$.