0
$\begingroup$

So i've been reading these two and got no clue about how to determine the argument when we are dealing with complex logarithm in contour integration.

Integral with two branch cuts

Inverse Laplace Transform of logarithm

So, if i have a complex logarithm as a function and the contour is given by this:

enter image description here

Then, what is the argument of complex logarithm on the upper and the lower path of contour? The first link said we choose $\pi$ as an argument above the branch cut and $-\pi$ as an argument below the branch cut. But the second link with a bit different contour picked $-\pi$ for the above and $\pi$ for the below.

Then, what's the difference? How do we exactly choose an argument for the given contour? I've seen a video about someone doing contour integration and use the argument to be $\arg(z)\in[0,\,2\pi)$ instead and it makes me more confuse.

Hope you can help me. I need to understand this one. Thanks in advance.

$\endgroup$
1
  • 1
    $\begingroup$ The argument is $0$ on $(0,\infty)$, then it is changing continuously on the upper half-disk, thus giving $arg = \pi-\epsilon$ for the points slightly above the negative axis, same for the lower half-disk, giving $-\pi+\epsilon$ for the points slightly below. As a rule of thumb any continuous branch of $\log$ is analytic. If you put the branch cut on the positive axis then you'll get $\epsilon+2\pi k$ above and $2\pi-\epsilon+2\pi k$ below $\endgroup$
    – reuns
    Jan 26 at 15:41

Your Answer

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

Browse other questions tagged or ask your own question.