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According to the property of the Laplace transform if $f(t)=\mathcal{{L}}^{-1}_s\left(F(s)\right)$, then $f(t)=-\frac{1}{t}\mathcal{{L}}^{-1}_s\left(\frac{d}{ds}\left(F(s)\right)\right)$.

This works for $F(s)=\ln\left(\frac{s+1}{s+2}\right)=\ln(s+1)-\ln(s+2)$.

So $f(t)=-\frac{1}{t}\mathcal{{L}}^{-1}_s\left(\frac{1}{s+1}-\frac{1}{s+2}\right)=-\frac{1}{t}{e^{-t}-e^{-2t}}$.

Why doesn't it work for $F(s)=\ln(s)$? According to the above method $f(t)=-\frac{1}{t}$.

Will it work for $F(s)=\ln\left(s+\frac{1}{s}\right)$?

Thank you.

eMathHelp
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1 Answers1

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Short answer: $\ln(s)$ (and $\ln(s+\frac{1}{s})$ for that matter) is not a Laplace transform of anything, so the employed property will yield bogus. Long answer and in depth explanation:

Inverse laplace transform of $\ln$

Max
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