18.3: A.3- Derivation of the Laplace Transform of a Definite Integral
- Page ID
- 7742
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Suppose that a function \(f(t)\) has Laplace transform \(F(s)=L[f(t)]\), and that we need the transform of the definite integral \(\int_{\tau=-\infty}^{\tau=t \geq 0} f(\tau) d \tau\). Note the lower limit of \(\tau=-\infty\); we usually consider \(f(t)\) only for \(t \geq 0\), but occasionally the integral of \(f(t)\) over previous time, \(t<0\), is also needed.
\[L\left[\begin{array}{l}
\tau=t \geq 0 \\
\iint_{-\infty} f(\tau) d \tau
\end{array}\right]=\int_{t=0}^{t=\infty} \overbrace{ \left[\int_{\tau = -\infty}^{\tau=t \geq 0} f(\tau) d \tau \right] }^u\overbrace{e^{-s t} d t}^{d v}\label{eqn:A.8} \]
Integrating by parts gives
\[L\left[\int_{\tau=-\infty}^{\tau=t \geq 0} f(\tau) d \tau\right]=\left.\left\{\left[\int_{\tau=-\infty}^{\tau=t \geq 0} f(\tau) d \tau\right]\left(\frac{e^{-s t}}{-s}\right)\right\}\right|_{t=0} ^{t=\infty}-\left(\frac{1}{-s}\right) \int_{t=0}^{t=\infty} e^{-s t} \frac{d}{d t}\left[\int_{\tau=-\infty}^{\tau=t \geq 0} f(\tau) d \tau\right] d t\label{eqn:A.9} \]
The derivative of the definite integral in the second right-hand-side term of Equation \(\ref{eqn:A.9}\) is a special case of Leibnitz’s rule (Hildebrand, 1962, p. 360):
\[\frac{d}{d t}\left[\int_{\tau=-\infty}^{\tau=t \geq 0} f(\tau) d \tau\right]=f(t)\label{eqn:A.10} \]
With the simple result Equation \(\ref{eqn:A.10}\), and with evaluation of the limits of the first right-hand-side term, Equation \(\ref{eqn:A.9}\) becomes
\[L\left[\int_{\tau=-\infty}^{\tau=t \geq 0} f(\tau) d \tau\right]=\frac{1}{s} \int_{\tau=-\infty}^{\tau=0} f(\tau) d \tau+\frac{1}{s} \int_{t=0}^{t=\infty} e^{-s t} f(t) d t\label{eqn:A.11} \]
Thus, the final form of the required general transform is
\[L\left[\int_{\tau=-\infty}^{\tau=t \geq 0} f(\tau) d \tau\right]=\frac{1}{s} F(s)+\frac{1}{s} \int_{\tau=-\infty}^{\tau=0} f(\tau) d \tau\label{eqn:A.12} \]
For most applications, we have \(f(t)=0\) for \(t<0\), for which the simpler transform is:
\[L\left[\int_{\tau=0}^{\tau=t \geq 0} f(\tau) d \tau\right]=\frac{1}{s} F(s)\label{eqn:A.13} \]
If we regard the integral of \(f(t)\) as being the first “negative” derivative (antiderivative), then we see that transform Equation \(\ref{eqn:A.12}\) is logically consistent with transform Equation 2.2.9 for a “positive” derivative, with respect to both power of \(s\) and the initial value term.