6.3 Laplace Transform of the Integral of a Function
If \(f(t)\) is piecewise continuous and satisfies an inequality of the form \(\big |f(t)\big |\leq Me^{\alpha t}\), then \[\mathcal {L}\Big \{\int ^t_0f(\tau ) d\tau \Big \} = \frac {1}{s} \mathcal {L}\{f(t)\},\hspace {0.4cm} s>0,\hspace {0.4cm} s>\alpha \]
From this, we have: \[\mathcal {L}^{-1}\Big \{\frac {1}{s}\hspace {0.1cm} \mathcal {L}\{f(t)\}\Big \} = \int ^t_0f(\tau ) d\tau \]
Let \(\displaystyle {\mathcal {L}\{f(t)\} =\frac {1}{s(s^2 + \omega ^2)}}\). Find \(f(t)\).
Solution.
Note that \(\dfrac {1}{s(s^2+\omega ^2)} = \dfrac {1}{s}\hspace {0.1cm}\mathcal {L}\{g(t)\}\),where \(\displaystyle {\mathcal {L}\{g(t)\} = \frac {1}{s^2 + \omega ^2}}\). From this we get \[g(t) = \mathcal {L}^{-1}\Bigg \{\frac {1}{s^2 + \omega ^2}\Bigg \} = \frac {1}{\omega }\sin \omega t\]
From the theorem, \(\displaystyle {\mathcal {L}^{-1}\Bigg \{\frac {1}{s}\mathcal {L}\{f(t)\}\Bigg \}=\int ^t_0f d\tau }\) so, \begin {align*} \mathcal {L}^{-1}\Bigg \{\frac {1}{s}\cdot \frac {1}{s^2 + \omega ^2}\Bigg \} & = \int ^t_0 \frac {1}{\omega }\sin \omega \tau d\tau \\\\ & = -\frac {1}{\omega ^2}\cos \omega \tau \Bigg |^t_0\\\\ & = \frac {-1}{\omega ^2}[\cos \omega t - 1] = \frac {1}{\omega ^2}(1-\cos \omega t) \end {align*}
\[\therefore \hspace {0.5cm} \mathcal {L}^{-1}\Bigg \{\frac {1}{s(s^2+\omega ^2)}\Bigg \} = \frac {1}{\omega ^2}\hspace {0.1cm}(1 - \cos \omega t)\]
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