Question

Showing why

Laplace{g(t)U(t-a)} = e^(-as) Laplace{g(t+a)}
is an alternative form of
Laplace{f(t-a)U(t-a)} =e^(-as) F(s) ?

Answer 1

Are you asking for a clarification, or a formal proof?

Answer 2

I think you can see how to get the results by using the definition of the transform. Consider some function \(f(t)~u(t-a)\), where \(u(t-a)\) denotes the Heaviside step function shifted by \(a\).
\[\large\begin{align*}\mathcal{L}\left\{f(t)~u(t-a)\right\}&=\int_0^\infty e^{-st}f(t)~u(t-a)~dt\\
&=\int_a^\infty e^{-st}f(t)~dt\\
&=\int_a^\infty e^{-s(t-a+a)}f(t-a+a)~dt\\
&=\int_0^\infty e^{-s(r+a)}f(r+a)~dr &\text{substituting }r=t-a\\
&=e^{-as}\int_0^\infty e^{-rs}f(r+a)~dr\\
&=e^{-as}\mathcal{L}\left\{f(r+a)\right\}\end{align*}\]
Now for a bit of a hand-wavey explanation: \(r\) is a variable that can easily be replaced by another. Using \(t\) in its stead will give you the desired result.

Answer 3

Sorry, the cut-off part is a substitution, \(r=t-a\).

The second formula can be established similarly, straight from the definition. The thing to take away here is that the shift caused by the step function is accounted for when you change the limits of integration.