18.3: A.3- Derivation of the Laplace Transform of a Definite Integral
- Page ID
- 7742
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)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.