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I am particularly interested in how Ron Gordon came up with the parametrization in his anser to this question: Inverse Laplace transform $\mathcal{L}^{-1}\left \{ \ln \left ( 1+\frac{w^{2}}{s^{2}}\right ) \right \}$

EDIT: Do we have to incorporate four branch cuts, if for example we wanted to integrate the function $$f\left ( z \right )=\frac{\sqrt{z}}{\ln\left ( z^{4}+1 \right )}$$ Is there a generalization for choosing the parametrization for multiple branch cuts?

EDIT #2: My understanding on how to approach integration of a complex function with one branch cut is clear, and I state this very boldly, I feel like I understand it intuitively(keeping the Riemann surfaces in mind). I am also very comfortable with recognizing when a function has multiple branch cuts, realizing why they are there, and that we must avoid them in order for the domain to stay simply connected. What I don't understand, is how to integrate along a branch cut which isn't necessarily along the real axis(e.g. along the line $\phi=\frac{\pi}{3}$. In the question linked, this paragraph confuses me :

The reader should note that there is a fourfold split here: when $z$ is to the left or right of the imaginary axis, and when $z$ is above or below the real axis. In each of these cases, the log has a negative argument, but the log of the negative argument takes on a different value along each section of the dog-bone. When $z$ is to the left and right of the imaginary axis we respectively parametrize $z=+iy$and $z=-iy$. However, we subtract $2i\pi$ from the log term upon crossing the imaginary axis and add $2i\pi$ upon crossing the real axis.

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  • $\begingroup$ What parametrization? $\endgroup$
    – davidlowryduda
    Dec 10 '15 at 0:02
  • $\begingroup$ @The one along the imaginary axis. For example, why did he use $z=iy$ when $z$ is to the left of the imaginary axis, and $z=-iy$ when to the right? $\endgroup$ Dec 10 '15 at 8:22
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I think you mean how they chose their curve, yes? Like, which path we should integrate around. Their choice is pretty standard. Whenever you have points to dodge along the x or y axis you should almost always handle it the way they did, putting small circles around them and then joining them up.

The choice of $z=\pm iy$ is slightly arbitrary in the sense that there are other choices that would work too. The point is to avoid the multivalued aspect of the $\log(z)$ "function." It's fine to use log as a function while you stay within a small region, but because of it's periodic multivaluedness, if you let your curve be too large $\log$ will lose it's functionness and then things start falling apart.

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  • $\begingroup$ That is clear to me. However, I don't see why he chose $z=-y$ along the y axis on the left side, and $z=iy$ to the right of thr imaginary axis $\endgroup$
    – Shemafied
    Dec 11 '15 at 17:57
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Not quite sure what you're asking, but I wanted to post a picture to possibly clear up the confusion.

In the question you've linked there was an explicit integral to compute... To do this, Ron used the residue theorem in conjunction with a contour avoiding singularities in the path and restricted the path to "one sheet" of the complex plane. See the picture below and try to visualize the cut along the negative $x$-axis.

With the cut Ron choose, the domain is still simply connected. Thus the Residue Theorem applied and along with a few limits he was able to compute the integral.

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  • $\begingroup$ But how do we visualize the other branch cut, connecting the points $z=i\omega$ and $z=-i\omega$? Do you have any suggestions on how to make the question more clear. I just wanted to be able to solve complex integrals with multiple branch cuts, and hoped someone would give help here. $\endgroup$ Dec 9 '15 at 20:24
  • $\begingroup$ oh, have a look at robjohns answer from this post math.stackexchange.com/questions/472166/… The main reason why we need the cut is to make the integral have 1 value... and log is multivalued at the origin. We take the almost full circles since $z= \pm iw$ is like $\ln z $ at $z =0$, so we can't loop around them. $\endgroup$
    – Jeb
    Dec 9 '15 at 20:39

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