I'm trying to find the Laplace transform of


What I've tried: $\int_0^\infty e^{-st} \frac{\sin(t)}{t} dt$ = $\int_0^\infty e^{-st} \sin(t) dt \int_0^\infty e^{-st} ds$ but I don't think this is correct as I am getting undefined results. Thanks in advance for any help.


$L(\sin t) = \frac{1}{1+s^2}$. Hence $$L\left(\frac{\sin t}{t}\right) = \int_s^\infty \frac{du}{u^2+1} = \frac{\pi}{2} - \tan^{-1}{s} $$

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  • $\begingroup$ Hey thank you for the response. I don't understand the justification though. $\endgroup$ – Mathew Sep 6 '16 at 11:40
  • $\begingroup$ This is based on the following: If $L(f(t)) = F(s)$, then $L(\frac{f(t)}{t}) = \int_s^\infty F(u) du$. $\endgroup$ – user348749 Sep 6 '16 at 11:54

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