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Could anyone enlighten me on how to find the Laplace Transform of

$$\frac{1-\cos (t)}{t}$$

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  • $\begingroup$ so which part of the Laplace Transform is most troubling? $\endgroup$
    – Chinny84
    May 18 '15 at 15:16
  • $\begingroup$ Well I tried breaking it up but could not find a form for which I have the formulas for. I'm kind of stumped... $\endgroup$ May 18 '15 at 15:17
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Hint:

Set $$f(t):=\frac{1-\cos t}{t}.$$ We have $$\mathcal{L}\{1-\cos t \}(s)=\mathcal{L} \{tf(t) \}(s)=-F'(s), $$ where $F(s)$ is the Laplace transform of $f(t)$.

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  • $\begingroup$ Might be stupid of me to ask, but why is it equal to $-F'(s)$? $\endgroup$ May 18 '15 at 15:24
  • $\begingroup$ @KevinZ by the General frequency domain differentiation. $\endgroup$
    – user1337
    May 18 '15 at 15:25
  • $\begingroup$ Oh! Is that the one where it is equal to $(-1)^1 \times \frac{dF}{ds}$ ? $\endgroup$ May 18 '15 at 15:27
  • $\begingroup$ @KevinZ indeed. $\endgroup$
    – user1337
    May 18 '15 at 15:28
  • $\begingroup$ Is this correct? $$-F'(s) = \frac{1}{s} - \frac{s}{s^2+1}$$ Integrating both sides: $$-F(s) = \ln(s) - \frac{1}{2}\ln(s^2+1)$$ $$-F(s) = \ln( \frac{s}{{\sqrt{s^2+1}}}) + c $$ Since $\lim_{s\to\infty} F(s) = 0$ $\implies$ $0 = \ln(1) + c$ $\implies c = 0$ Thus $F(s) = -\ln( \frac{s}{{\sqrt{s^2+1}}})$ $\endgroup$ May 18 '15 at 15:37
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$$ f(t)=\frac{g(t)}{t}=\frac{1-\cos t}{t} $$ thus the Laplace transform is $$ F(s)=\int_s^\infty G(z)\,\mathrm dz=\int_s^\infty \left[\frac{1}{z}-\frac{z}{z^2+1}\right]\,\mathrm dz=\left[\log{z}-\log\left(\sqrt{z^2+1}\right)\right]_s^\infty=\log\left(\tfrac{\sqrt{s^2+1}}{s}\right) $$

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