0
$\begingroup$

One can look up in this table that the inverse Laplace transform of $\exp(-c s)/s$ with $c\in\mathbb{R}$ is given by:

$$\frac{1}{2\pi i}\lim_{T\to \infty}\int_{\gamma-i T}^{\gamma+iT}ds\frac{e^{s(t-c)}}{s}=\theta(t-c)$$

where $\gamma=const.>0$ and $\theta(x)$ is the Heaviside step function. Looking at the integral, one might think that one way to arrive at the result could be to close the integration contour in a half-circle to the left in the complex plane, and collect a residue at $s=0$. However, that does not seem to give a step function. Or maybe I am missing some subtlety? What is a correct way to explicitly calculate this inverse Laplace transform?

$\endgroup$
  • 1
    $\begingroup$ Why not work backwards and calculate the LT of $\theta(t-c)?$ $\endgroup$ – Adrian Keister Feb 4 '19 at 22:08
  • $\begingroup$ @AdrianKeister That would work nicely, but I am kind of more curious about closing the circle through inverse transform and learning the subtleties involved to get it right... $\endgroup$ – Kagaratsch Feb 4 '19 at 22:14
1
$\begingroup$

You close the contour to the left if $t>c$ and to the right if $t<c$, because you should always use the half-plane where the exponential decays. In the second case, there are no poles inside the contour...

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Wait, but I thought $\gamma>0$ is always true, since it has to be to the right of all the poles in the integrand. Which residue do we pick up if we close the contour to the right? $\endgroup$ – Kagaratsch Feb 4 '19 at 22:23
  • 1
    $\begingroup$ There is no residue, and hence the integral is zero in this case. $\endgroup$ – Vladimir Feb 4 '19 at 22:24
  • $\begingroup$ Aaaah! Perfect, thank you! :D $\endgroup$ – Kagaratsch Feb 4 '19 at 22:24

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.