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To find Inverse Laplace of $$F(s)= \log\frac{s+1}{(s+2)(s+3)}.$$

I have tried to use shifting theorems, but of no use. Should I apply series for log and take inverse laplace of individual terms, if yes under which conditions series is valid. I their any other way to do this?

Thanks

Vlad
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Taylor Ted
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2 Answers2

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First of all, $\log\dfrac{ab}c=\log a+\log b-\log c$

$$L^{-1}\log(s-a)=-\dfrac1t L^{-1}\left(\dfrac{d\{\log(s-a)\}}{ds}\right)=-\dfrac1tL^{-1}\left(\dfrac1{s-a}\right)$$

$$L^{-1}\left(\dfrac1{s-a}\right)=e^{at}$$

lab bhattacharjee
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  • @TaylorTed, See http://planetmath.org/inverselaplacetransformofderivatives – lab bhattacharjee Aug 23 '15 at 10:29
  • Note that the inverse transform of $F(s)$ doesn't exist in the sense of ordinary functions. In terms of distributions, $\mathcal L^{-1}[F] = t_+^{-1} (e^{-3t} + e^{-2t} - e^{-t}) + \gamma \delta(t)$, see here. – Maxim Jul 21 '18 at 12:51
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F(s)=log[(s+1)/(s+2)(s+3)]

f(s)=log(s+1)-log(s+2)-log(s+3)

differentiating with respect to s

f'(s)= (1/s+1)-(1/s+2)-(1/s+3)

taking inverse laplace

-t*f(t)= (e^-1)-(e^-2)-(e^-3)

f(t)= -[(e^-1)-(e^-2)-(e^-3)]/t

Mrunal
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