Proof. Writing \(F(s)G(s)\) as a double integral and substituting \(t = \tau + \sigma \) for the second variable, \[F(s)G(s) = \int _0^\infty e^{-s\tau }f(\tau )\int _0^\infty e^{-s\sigma }g(\sigma )\, d\sigma \,d\tau = \int _0^\infty f(\tau )\int _\tau ^\infty e^{-st}g(t - \tau )\,dt\, d\tau .\] The region of integration is \(0\leq \tau \leq t<\infty \), so reversing the order, as in Fubini’s theorem, \[F(s)G(s) = \int _0^\infty e^{-st}\int _0^t f(\tau )g(t - \tau )\,d\tau \,dt = \mathcal {L}\{(f*g)(t)\} .\] □

Solution. Using \(\sin A\cos B = \frac 12[\sin (A + B) + \sin (A - B)]\), \[\sin t * \cos t = \int ^t_0 \sin \tau \cos (t - \tau )\, d\tau = \frac 12\int _0^t\big [\sin t + \sin (2\tau - t)\big ]d\tau = \frac {t\sin t}{2} ,\] the second term integrating to zero. By the convolution theorem its transform is \(\frac {1}{s^2 + 1}\cdot \frac {s}{s^2 + 1} = \frac {s}{(s^2 + 1)^2}\), which is \(\frac 12\mathcal {L}\{t\sin t\}\) by Example 6.6.2 with \(\omega = 1\).

Solution.

1.

\(H(s) = F(s)G(s)\) with \(F = G = \frac {1}{s^2 + 1}\), so \(f = g = \sin t\) and, using \(\sin A\sin B = \frac 12[\cos (A - B) - \cos (A + B)]\), \begin {align*} h(t) & = \int ^t_0\sin \tau \sin (t-\tau )\, d\tau = \frac {1}{2}\int ^t_0 \big [\cos (2\tau - t) - \cos t\big ]d\tau \\ & = \frac 12\left [\frac {\sin (2\tau - t)}{2}\right ]_0^t - \frac {t\cos t}{2} = \frac {\sin t}{2} - \frac {t\cos t}{2} = \frac {1}{2}(\sin t - t\cos t) . \end {align*}

2.

\(H(s) = \frac {1}{s^2}\cdot \frac {1}{s-a}\), with inverse transforms \(t\) and \(e^{at}\), so \[h(t) = \int ^t_0 \tau e^{a(t-\tau )}\, d\tau = e^{at} \int ^t_0\tau e^{-a\tau }\,d \tau = \frac {1}{a^2}\left (e^{at} -at -1\right ),\] integrating by parts.