I would like to calculate the ILT of the function $\log\left(s\right)$. I don't know if my calculations are right. Since $$F(s)=\log\left(s\right),\,\textrm{Re}(s)>0$$ then $$F^{\prime}\left(s\right)=\frac{1}{s}$$ so if we put $$f\left(t\right)=L^{-1}\left(\log\left(s\right)\right)\left(t\right)$$ we have, using the properties of the Laplace transform, that $$L\left(tf\left(t\right)\right)\left(s\right)=L\left(u\left(t\right)\right)\left(s\right)$$ where $u(t)$ is the unit step function. So $$f\left(t\right)=\frac{u\left(t\right)}{t}.$$ Are my calculations correct? Thank you.
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1$\begingroup$ No way to find ILT of logarithm math.stackexchange.com/questions/2038252/… $\endgroup$– RaffaeleCommented Sep 2, 2017 at 12:07
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$\begingroup$ and what is $\mathcal{L}\left\{\frac{u(t)}{t}\right\}$?? $\endgroup$– alexjoCommented Sep 2, 2017 at 20:31
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$\begingroup$ As you see the Laplace transform of $\frac{u(t)}{t}$ doesn't converge. Look instead at the Laplace transform of $\frac{u(t-1)}{t}$ which converges and has a Laplace transform close to $\log s$ (see also the exponential integral function) $\endgroup$– reunsCommented Sep 2, 2017 at 23:24
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$\begingroup$ Just a comment, I am an aficionado and when I need to know if a calculation is feasible (I add this comment as companion of previous, but those have more merit since were mathematical reasonings), for instance your example, then I search in Google the following words: inverse Laplace transform, Wolfram Alpha Language, and it address to me to Wolfram Language Documentation and syntax about such function. Then I ask in this Wolfram Alpha online calculator your problem with this code InverseLaplaceTransform[log(s),s,t], and one can see what was the output. $\endgroup$– user243301Commented Sep 14, 2017 at 18:38
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1 Answer
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The integral of $e^{-s t}/t$ from zero to infinity doesn't exist, one has to consider some regularization of it. In terms of distributions this is done by defining the functional $t_+^{-1}$ as $$(t_+^{-1}, \phi) = \int_0^1 \frac {\phi(t) - \phi(0)} t dt + \int_1^\infty \frac {\phi(t)} t dt,$$ then the Laplace transform can be computed as $$\mathcal L[t_+^{-1}] = (t_+^{-1}, e^{-s t}) = -\ln s - \gamma,$$ where $\gamma$ is Euler's constant, and $$\mathcal L^{-1}[\ln s] = -t_+^{-1} - \gamma \delta(t).$$
