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Subsection 6.5.2 Second-Shifting Property

If
\begin{equation*} \mathscr{L}[f(t)] F(s)\text{and}\quad g(t) = \begin{cases} f(t-a), & \text{where} \quad t\geq a\\ 0, & \text{where} \quad t \lt a \end{cases} \end{equation*}
then
\begin{equation*} \mathscr{L}[g(t)] = e^{-as}F(s). \end{equation*}

Proof.

\begin{equation*} \mathscr{L}[g(t)] = \int\limits_{0}^{\infty} e^{-st}g(t)\,dt \end{equation*}
\begin{equation*} = \int\limits_{0}^{a} e^{-st}g(t)\,dt +\int\limits_{a}^{\infty} e^{-st}g(t)\,dt \end{equation*}
\begin{equation*} = \int\limits_{0}^{a} e^{-st}.0\,dt +\int\limits_{a}^{\infty} e^{-st}f(t-a)\,dt \end{equation*}
\begin{equation*} = \int\limits_{0}^{\infty} e^{-s(h+a)}.f(h)\,dh = e^{-as} \int\limits_{0}^{\infty}f(h)\,dh \end{equation*}
\begin{equation*} = e^{-as}F(s) \end{equation*}
[put \(t-a =h\)]