6.9 Integration of Transforms
\(\displaystyle {\mathcal {L}\Bigg \{\frac {f(t)}{t}\Bigg \}=\int ^{\infty }_s F(u)du}\), where \(F(s) = \mathcal {L}\{f(t)\}\).
Hence, \(\displaystyle {\mathcal {L}^{-1}\Bigg \{\int ^{\infty }_s F(u)du\Bigg \} = \frac {f(t)}{t}}\).
Solution. \(\displaystyle {\frac {f(t)}{t} = \mathcal {L}^{-1}\Bigg \{\int ^{\infty }_s \frac {2u}{(u^2 + 1)^2}du}\)
\begin {align*} \int ^{\infty }_s 2u\big (u^2 + 1\big )^{-2}du & = \frac {-1}{u^2 + 1}\Bigg |^{\infty }_s\\\\ & = \frac {1}{s^2 + 1} \end {align*}
So \(\hspace {0.3cm}\displaystyle {\frac {f(t)}{t}=\mathcal {L}^{-1}\Bigg \{\frac {1}{s^2 + 1}\Bigg \}=\sin t}\)
\[\therefore \hspace {0.5cm} f(t) = t \sin t\]
Questions on this section
Stuck on something here? Ask below and it stays attached to this topic.