18.4: A.4- Applications of the Laplace Transform of a Definite Integral
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A relatively painless method for deriving certain inverse transforms is based upon the inverse transform of Equation 18.3.6,
\[L^{-1}\left[\frac{1}{s} F(s)\right]=\int_{\tau=0}^{\tau=t \geq 0} f(\tau) d \tau\label{eqn:A.14} \]
It is necessary to apply carefully the limits of the definite integral, as is illustrated in the following three examples.
\[L^{-1}\left[\frac{1}{s} \frac{1}{s-p}\right]=\int_{\tau=0}^{\tau=t \geq 0} e^{p \tau} d \tau=\frac{1}{p} \int_{\tau=0}^{\tau=t \geq 0} d\left(e^{p \tau}\right)=\frac{1}{p}\left(e^{p t}-1\right), t \geq 0\label{eqn:A.15} \]
\[L^{-1}\left[\frac{1}{s\left(s^{2}+\omega^{2}\right)}\right]=\frac{1}{\omega} \int_{\tau=0}^{\tau=t} \sin \omega \tau d \tau=-\frac{1}{\omega^{2}} \int_{\tau=0}^{\tau=t} d(\cos \omega \tau)=\frac{1}{\omega^{2}}(1-\cos \omega t), t \geq 0\label{eqn:A.16} \]
\[L^{-1}\left[\frac{1}{s} \frac{1}{s}\right]=\int_{\tau=0}^{\tau=t \geq 0} H(\tau) d \tau=\int_{t=0}^{\tau=t \geq 0} d \tau=t, t \geq 0\label{eqn:A.17} \]