Subsection 11.5.1 From Fourier Transforms to Laplace Transforms
We have seen that certain real-valued functions \(f(t)\) have a Fourier transform and that the integral
\begin{equation*}
g(\omega) = \int_{-\infty}^{\infty}f(t)e^{-i\omega t},dt
\end{equation*}
defines the complex function \(g(\omega)\) of the real variable \(\omega\text{.}\) If we multiply the integrand \(f(t) e^{-i\omega t}\) by \(e^{-\sigma t}\text{,}\) we create a complex function \(G(\sigma +i\omega)\) of the complex variable \(\sigma +i\omega\text{:}\)
\begin{equation*}
G(\sigma +i\omega) =\int\nolimits_{-\infty}^{\infty}f( t) e^{-\sigma t}e^{-i\omega t}dt=\int\nolimits_{-\infty}^{\infty }f(t) e^{-(\sigma +i\omega) t}dt\text{.}
\end{equation*}
The function \(G(\sigma +i\omega)\) is called the two-sided Laplace transform of \(f(t)\text{,}\) and it exists when the Fourier transform of the function \(f(t) e^{-\sigma t}\) exists. From the Fourier transform theory, we can state that a sufficient condition for \(G(\sigma +iw)\) to exist is that
\begin{equation*}
\int_{-\infty}^{\infty}|f(t)|e^{-\sigma t}\,dt \lt \infty
\end{equation*}
exists. For a given function \(f(t)\text{,}\) this integral is finite for values of \(\sigma\) that lie in some interval \(a\lt \sigma b\text{.}\)
The two-sided Laplace transform uses the lower limit of integration, that is, \(t=-\infty\text{,}\) and hence requires a knowledge of the past history of the function \(f(t)\text{,}\) i.e., \(t\lt 0\text{.}\) For most physical applications, one is interested in the behavior of a system only for \(t \ge 0\text{.}\) Mathematically speaking, the initial conditions \(f(0), \, f\,'(0), \, f\,''(0), \ldots\text{,}\) are a consequence of the past history of the system and are often all that is necessary to know. For this reason, it is useful to define the one-sided Laplace transform of \(f(t)\text{,}\) which is commonly referred to simply as the Laplace transform of \(f(t)\text{,}\) which is also defined as an integral:
\begin{equation}
\mathcal{L}\big(f(t)\big)=F(s)=\int_0^{\infty}f(t) e^{-st}\,dt, \text{ where } s=\sigma +i\omega\text{.}\tag{11.5.1}
\end{equation}
If the defining
integral (11.5.1) for the Laplace transform exists for
\(s_0=\sigma_0+i\omega\text{,}\) then values of
\(\sigma\) with
\(\sigma > \sigma_0\) imply that
\(e^{-\sigma t} \lt e^{-\sigma_0t}\text{,}\) and thus
\begin{equation*}
\int_0^{\infty}\left| \,f(t) \right| e^{-\sigma t}\,dt \lt \int_0^{\infty}\left| \,f(t) \right| e^{-\sigma_0 t}\,dt\lt \infty\text{,}
\end{equation*}
and it follows that \(F(s)\) exists for \(s=\sigma +i\omega\text{.}\) Therefore, the Laplace transform \(\mathcal{L}(f(t))\) is defined for all points \(s\) in the right half-plane \(\mathrm{Re}(s) > \sigma_0\text{.}\)
Another way to view the relationship between the Fourier transform and the Laplace transform is to consider the function \(U(t)\) given by
\begin{equation*}
U(t) = \begin{cases}f(t) \amp \text{ for } t \ge 0,\\ \;\; 0 \amp \text{ for } t \lt 0. \end{cases}
\end{equation*}
Then the Fourier transform theory shows us that
\begin{equation*}
U(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}\left[\int_{-\infty}^{\infty}U(t)e^{-i\omega t}\,dt\right]e^{i\omega t}\,d\omega\text{,}
\end{equation*}
and since the integrand \(U(t)\) is zero for \(t\lt 0\text{,}\) this equation can be written as
\begin{equation*}
f(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}\left[\int_0^{\infty}f(t) e^{-i\omega t}\,dt\right]e^{i\omega t}\,d\omega\text{.}
\end{equation*}
Use the change of variable \(s=\sigma +i\omega\) and \(d\omega=\frac{ds}{i}\text{,}\) where \(\sigma>\sigma_0\) is held fixed, then the new limits of integration are from \(s=\sigma -i\omega\) to \(s=\sigma +i\omega\text{.}\) The resulting equation is
\begin{equation*}
f(t)=\frac{1}{2\pi}\int_{\sigma -i\infty}^{\sigma +i\infty}\left[\int_0^{\infty}f(t) e^{-st}\,dt\right]e^{st}\,ds\text{.}
\end{equation*}
Therefore, the Laplace transform is
\begin{equation*}
\mathcal{L}\big(f(t)\big) = F(s) = \int_0^{\infty}f(t) e^{-st}\,dt, \text{ where } s=\sigma +i\omega\text{,}
\end{equation*}
and the inverse Laplace transform is
\begin{equation}
\mathcal{L}^{-1}\big(F(s)\big)=f(t =\frac{1}{2\pi}\int_{\sigma-i\infty}^{\sigma +i\infty}F(s) e^{st}\,ds\text{.}\tag{11.5.2}
\end{equation}
