1
$\begingroup$

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.

$\endgroup$
4
  • 1
    $\begingroup$ No way to find ILT of logarithm math.stackexchange.com/questions/2038252/… $\endgroup$
    – Raffaele
    Sep 2 '17 at 12:07
  • $\begingroup$ and what is $\mathcal{L}\left\{\frac{u(t)}{t}\right\}$?? $\endgroup$
    – alexjo
    Sep 2 '17 at 20:31
  • $\begingroup$ 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) $\endgroup$
    – reuns
    Sep 2 '17 at 23:24
  • $\begingroup$ 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. $\endgroup$
    – user243301
    Sep 14 '17 at 18:38
3
$\begingroup$

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 $$\mathcal L^{-1}[\ln s] = -t_+^{-1} - \gamma \delta(t).$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.