The following proof is given for the special case when \(s\) is a real number. The general case is covered in advanced texts. Using the dummy variables \(\sigma\) and \(\tau\) and the integrals defining the transforms, we can express their product as
\begin{equation*}
F(s)G(s) = \left[\int_0^{\infty}f(\sigma) e^{-s\sigma}\,d\sigma\right]\left[\int_0^{\infty}g(\tau)e^{-s\tau}\,d\tau\right]\text{.}
\end{equation*}
The product of integrals in this equation can be written as an iterated integral:
\begin{equation*}
F(s)G(s) = \int_0^{\infty}\left[\int_0^{\infty}f(\sigma)e^{-s(\sigma+\tau)}\,d\sigma\right]g(\tau)\,d\tau\text{.}
\end{equation*}
Hold \(\tau\) fixed, and use the change of variables \(t=\sigma +\tau\) and \(dt=d\sigma\text{.}\) Then the inner integral in the equation can be rewritten to obtain
\begin{align*}
F(s)G(s) \amp = \int_0^{\infty}\left[\int_{\tau }^{\infty}f(t-\tau)e^{-st}\,dt\right]g(\tau)\,d\tau\\
\amp = \int_0^{\infty}\left[\int_{\tau }^{\infty}f(t-\tau)g(\tau)e^{-st}dt\right]d\tau\text{.}
\end{align*}
The region of integration for this last iterated integral is the wedge-shaped region in the
\((t,\tau)\) plane shown in
Figure 11.10.2. The order of integration in the integral can be reversed to get:
\begin{equation*}
F(s)G(s) = \int_0^{\infty}\left[\int_0^tf(t-\tau)g(\tau)e^{-st}\,d\tau\right]dt\text{.}
\end{equation*}
This equation can be written as
\begin{align*}
F(s)G(s) \amp = \int_0^{\infty}\left[\int_0^tf(t-\tau)g(\tau)\,d\tau\right]e^{-st}\,dt\\
\amp = \mathcal{L}^{-1}\left(\int_0^tf(t-\tau)g(\tau)\,d\tau\right)\text{,}
\end{align*}
which establishes Equation
(11.10.2). Since we can interchange the role of the functions
\(f(t)\) and
\(g(t)\text{,}\) Equation
(11.10.1) follows immediately.