Proof. Let \(g(t) = \int _0^t f(\tau )\,d\tau \). Then \(g\) is continuous, \(g(0) = 0\), \(g' = f\) except at the jumps of \(f\), and \(g\) is of exponential order. By Theorem 6.2.1, \(\mathcal {L}\{f\} = \mathcal {L}\{g'\} = s\mathcal {L}\{g\} - g(0) = s\mathcal {L}\{g\}\). □

Solution. Since \(\mathcal {L}^{-1}\left \{\frac {1}{s^2 + \omega ^2}\right \} = \frac {1}{\omega } \sin \omega t\), Theorem 6.3.1 gives \[f(t) = \int ^t_0 \frac {1}{\omega }\sin \omega \tau \, d\tau = \left [-\frac {\cos \omega \tau }{\omega ^2}\right ]^t_0 = \frac {1}{\omega ^2}(1-\cos \omega t) .\]

Solution. Apply Theorem 6.3.1 once more to the result of Example 6.3.2: \[\mathcal {L}^{-1}\left \{\frac {1}{s}\cdot \frac {1}{s(s^2 + \omega ^2)}\right \} = \int _0^t\frac {1 - \cos \omega \tau }{\omega ^2}\,d\tau = \frac {1}{\omega ^2}\left (t - \frac {\sin \omega t}{\omega }\right ) .\]