6.6 Differentiation of Transforms
\(\displaystyle {\mathcal {L}\{t f(t)\}=F'(s)}\), where \(\displaystyle {F(s) = \mathcal {L}\{f\}}\) \[\text {i.e}\hspace {0.5cm}\mathcal {L}^{-1}\{F'(s)\} = -t f(t)\]
\begin {align*} \mathcal {L}\{t^2 f(t)\} & = \mathcal {L}\{t(t f(t))\}\\ & = \mathcal {L}\{t g(t)\} = -G'(S) \end {align*}
where \(G(s) = \mathcal {L}\{g(t)\}\) and \(g(t) = t f(t)\) \[G(s) = \mathcal {L}\{g(t)\} = \mathcal {L}\{t f(t)\} = -F'(s)\] So that \(\displaystyle {-G'(s) = -\big [- F'(S)\big ]' = F''(s)}\)
\[\therefore \hspace {0.5cm} \mathcal {L}\{t^2 f(t)\} = F''(s)\]
Similarly, \(\displaystyle {\mathcal {L}\{t^3 f(t)\}} = - F''(s)\)
\[\therefore \hspace {0.5cm} \mathcal {L}\{t^n f(t)\} = (-1)^n F^n(s) = (-1)^n \frac {d^n F}{d s^n}\]
Solution.
Here \(f(t) = \sin \omega t\) and \(\displaystyle {F(s) = \mathcal {L}\{\sin \omega t\}= \dfrac {\omega }{s^2 + \omega ^2}}\)
\begin {align*} \mathcal {L}\{t \sin \omega t\} & = - F'(s)\hspace {0.3cm} \text {so}\\ \mathcal {L}\{t \sin \omega t\} & = \frac {-d}{ds}\Big (\frac {\omega }{s^2 + \omega ^2}\Big )\\ & = \frac {2s\omega }{\big (s^2 + \omega ^2\big )^2}\\\\ \end {align*}
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