Proof. Differentiating under the integral sign, which is justified for \(s>\alpha \) by the exponential bound, \[F'(s) = \frac {d}{ds}\int _0^\infty e^{-st}f(t)\,dt = \int _0^\infty (-t)e^{-st}f(t)\,dt = -\mathcal {L}\{tf(t)\} .\] Applying this to \(g(t) = tf(t)\), with \(G = -F'\), gives \(\mathcal {L}\{t^2f(t)\} = -G'(s) = F''(s)\), and by induction the general formula. □

Solution. With \(F(s) = \mathcal {L}\{\sin \omega t\} = \frac {\omega }{s^2 + \omega ^2}\), \[\mathcal {L}\{t \sin \omega t\} = -\frac {d}{ds}\left (\frac {\omega }{s^2 + \omega ^2}\right ) = \frac {2s\omega }{(s^2 + \omega ^2)^2} .\] With \(F(s) = \frac {1}{s + 1}\), \(\mathcal {L}\{t^2e^{-t}\} = F''(s) = \frac {2}{(s+1)^3}\), which agrees with the first shifting theorem.

Solution. Let \(F(s) = \ln (s + 1) - \ln (s - 1)\). Then \(F'(s) = \frac {1}{s + 1} - \frac {1}{s - 1}\), so by Theorem 6.6.1 \[-tf(t) = e^{-t} - e^t,\qquad f(t) = \frac {e^t - e^{-t}}{t} = \frac {2\sinh t}{t} .\]