Start with the Laplace transform definition:
\begin{equation*}
\lap{f(t) u_c(t)} = \int_{0}^{\infty} f(t) u_c(t) e^{-st}\ dt.
\end{equation*}
Since \(u_c(t) = 0\) for \(t \lt c\text{,}\) the integral simplifies to:
\begin{equation*}
\lap{f(t)\cdot u_c(t)} = \int_c^\infty f(t)\ e^{-st}\ dt.
\end{equation*}
This new integral looks a lot like the Laplace transform of
\(f(t)\text{,}\) but it starts at
\(t = c\) instead of
\(t = 0\text{.}\) Luckily, we can shift the graph of
\(f(t)\ e^{-st}\) to the left
\(c\) units to force the integral start back at
\(t = 0\text{.}\)
Note: Shifting a function \(c\) units to the left means adding \(c\) to all of the \(t\) inputs:
\begin{equation*}
\text{before shift:}\quad f(t)\ e^{-st}\qquad\text{after shift:}\quad f(t + c)\ e^{-s(t + c)}
\end{equation*}
Using this idea leads to the desired transform:
\begin{align*}
\int_{c}^{\infty} f(t)\ e^{-st}\ dt
\amp \os{\large\DLBb \text{A}}{=} \int_{0}^{\infty} f(t + c)\ e^{-s(t + c)}\ dt\\
\amp \os{\large\DLBb \text{B}}{=} \int_0^\infty f(t + c) \left(e^{-st} \cdot e^{-sc}\right)\ dt\\
\amp \os{\large\DLBb \text{C}}{=} e^{-cs} \int_0^\infty f(t + c) e^{-st}\ dt\\
\amp = e^{-cs} \lap{f(t + c)}
\end{align*}
\(\DLBb \text{A}, \text{B}, \text{C}\) Details.
\(\DLBb \text{A}\ \): Shifting \(f(t)e^{-st}\) to the left \(c\) units makes it
\begin{equation*}
f(t + c)\ e^{-s(t + c)}
\end{equation*}
and starts the integral at \(t=0\text{.}\)
\(\DLBb \text{B}\ \): Exponential Property
\(e^{-s(t + c)} = e^{-st} e^{-sc}\)
\(\DLBb \text{C}\ \): Factor out the constant,
\(e^{-sc}\text{,}\) as we are integrating in
\(t\text{.}\)
Therefore,
\begin{equation*}
\lap{f(t) \, u_c(t)} = e^{-cs} \lap{f(t + c)}.
\end{equation*}
