Suppose $a$ is real and nonnegative. Say we wanted to compute the above function (for whatever reason, be it to solve an improper real integral, or something else) along the curve $C$, as on the picture. I have chosen the contour as to avoid the branch cut connecting the three branch points. Supposing $arg\left ( z \right ) \in \left [ 0, 2\pi \right )$ I also made parametrisations for each part of the contour. However, I wasn't able to do so for the parts $C_{i}$, $i=1,2,3$.enter image description here

In several integrals like this and this one Ron talks about assigning a phase to the segments. To me it seems like he is assigning the phase as if the branch point was now the origin of the plane, and the phase he added was relative to that point, am I right on this one? With that being said I would say that $$C2: z=iye^{i\pi}$$ $$C3: z=iye^{-i\pi}$$ $$C1: z=iye^{i0}$$ However this doesn't look right as the argument wasn't defined for $\left [ -\pi,\pi \right )$. How do we deal with these branch cuts? And how to know what phase to add? Note that I have asked a similar question here for a different function, but I didn't receive satisfactory answers(due to my poor wording, I guess).

  • $\begingroup$ I still have the feeling that you might have problems expressing what exactly you are asking. Here is an insight: for a complex function, only the branch points are fixed in the sense that they have a absolute existence. The branch cuts can be chosen arbitrary (up to your liking). They should be however kept fixed during the calculation. For your function, you have two branch points at $z=\pm i a$. You can convince yourself that connecting them with with a branch cut in between, you obtain a perfectly well behaved holomorphic function. If this is not clear, I can expand a bit on this. $\endgroup$
    – Fabian
    Dec 22 '15 at 5:55

Let $\Omega=\mathbb{C}\setminus[-\pi i,\pi i]$. If $\gamma$ is any simple closed path in $\Omega$, then $$ \int_{\gamma}\frac{w}{w^2+a^2}dw $$ has value equal to $0,\pm 2\pi i,\pm 4\pi i,\pm 6\pi,\cdots$. For any $z \in \Omega$, define $\gamma_{z}$ to be a path from $0$, to the right of $0$, with a termination point $z\in\Omega$, and define $$ G(z) = a\exp\left\{\int_{\gamma_z}\frac{w}{w^2+a^2}dw\right\}. $$ The function $G$ does not depend on the specific such path $\gamma_z$, and $G(0)=a$. Also, $$ G'(z)=G(z)\frac{z}{z^2+a^2},\;\;\; z\in\Omega. $$ Therefore $G(z)^2$ satisfies the following for $z\in\Omega$: $$ \frac{\frac{d}{dz}G(z)^2}{G(z)^2}=\frac{2z}{z^2+a^2}=\frac{\frac{d}{dz}(z^2+a^2)}{z^2+a^2}, \\ \frac{d}{dz}\frac{z^2+a^2}{G(z)^2}=0,\\ G(z)^2=C(z^2+a^2). $$ Evaluating at $z=0+$ gives $C=1$. Hence, $G(z)^2=z^2+a^2$, or $G(z)$ is a square root of $z^2+a^2$.

The argument of $G$ is $0$ on the positive real axis because $\gamma$ may be chosen to a straight line segment from $0+$ to $z$ on the positive real axis. Then the argument along $C_1$ is found by integrating $$ \int_{0}^{r}\frac{\epsilon+is}{(\epsilon+is)^2+a^2}ds $$

  • $\begingroup$ Where does this reasoning come from? Can we apply the same procedure for the other part of the contour. $\endgroup$ Dec 19 '15 at 17:20
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    $\begingroup$ Yes, you may continue to follow the contour because of how everything is defined in terms of an integral. This sort of procedure for dealing with complex logarithms is very powerful, quite sound, and the most general because it helps you design the branch cuts so that all the closed path integrals give you values that are, inevitably, multiples of $2\pi i$. You can do this with cube roots, square roots, any number of factors. The branch cuts can be finite, infinite, etc.. I think I got this from Conway. $\endgroup$ Dec 19 '15 at 17:43
  • $\begingroup$ I am so sorry, but I have done the integral and the result is some featuring the natural logarithm, and nothing like an $n \pi i$ term. I know this is against the site's policy, but could you just show the work, how you would evaluate the integral? $\endgroup$ Dec 21 '15 at 21:10
  • $\begingroup$ @EmirŠemšić : I have given you a way to define $\sqrt{z^2+a^2}$ so that it has one branch cut between $-ia$ and $ia$, and is holomorphic everywhere else. So you have an integral of a holomorphic function over a closed path whose interior includes the region where the function is holomorphic. What will you get for such an integral: $\oint_{C}\sqrt{z^2+a^2}dz$? $\endgroup$ Dec 21 '15 at 22:37

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