Proof. Let \(g(t) = \frac {f(t)}{t}\), so \(f = tg\) and, by Theorem 6.6.1, \(F(s) = -G'(s)\). Integrating from \(s\) to \(\infty \), and using \(G(u)\to 0\) as \(u\to \infty \), gives \(\int _s^\infty F(u)\,du = G(s)\). □

Solution. Here it is easier to integrate \(F\) than to invert it. Let \(f(t) = t\,g(t)\) with \(\mathcal {L}\{g\} = \int _s^\infty F(u)\,du\): \[\int ^{\infty }_s \frac {2u}{(u^2 + 1)^2}\,du = \left [-\frac {1}{u^2 + 1}\right ]^{\infty }_s = \frac {1}{s^2 + 1} ,\] so \(g(t) = \sin t\) and \(f(t) = t \sin t\), in agreement with Example 6.6.2.

Solution. By Theorem 6.9.1, \(\mathcal {L}\left \{\frac {\sin t}{t}\right \} = \int _s^\infty \frac {du}{u^2 + 1} = \frac {\pi }{2} - \tan ^{-1}s\). Letting \(s\to 0^+\) (which can be justified) gives \(\int _0^\infty \frac {\sin t}{t}\,dt = \frac {\pi }{2}\).