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How meaningful is it to try to integrate around the branch-cut of a function?

For example lets say I have the function $\log(z^2+a^2)$ for $a>0$ and I choose my branch-cuts to be starting at $\pm ia$ and moving up and down the $y-$axis respectively. Now I am trying to integrate around a small circle around such a branch-point by avoiding crossing the branch-cut.

  • Like can I say,

$\lim_{\epsilon \rightarrow 0} [ [\int_{\phi = \pi/2} ^{\phi = -\pi } + \int_{\phi = \pi} ^{\phi = \pi/2} ] z\text{ }\tanh(\pi z)\log(z^2+a^2) dz ] = 0 $

(where $z = ia + \epsilon e^{i\phi}$)

?

  • Or at least can one argue (by some symmetry or parity argument?) the sum of such integrals around the two branch-points $\pm ia$ to be zero where one goes clockwise around both? (..and the angle integrals around each is split and ordered so as to avoid the respective branch-cuts..)

Like can I say,

$\lim_{\epsilon \rightarrow 0} \left [ [\int_{\phi_1 = \pi/2} ^{\phi_1 = -\pi } + \int_{\phi_1 = \pi} ^{\phi_1 = \pi/2} ] z_1\text{ }\tanh(\pi z_1)\log(z_1^2+a^2) dz_1 + [\int_{\phi_2 = \pi} ^{\phi_2 = -\pi/2 } + \int_{\phi_2 = -\pi/2} ^{\phi_2 = -\pi} ] z_2\text{ }\tanh(\pi z_2)\log(z_2^2+a^2) dz_2 \right ] = 0 $

(where $z_1 = ia + \epsilon e^{i\phi_1}$ and $z_2 = -ia + \epsilon e^{i\phi_2}$ )

?


I understand that this query might not be really research level but its something about which asking around I am not getting a clear answer. I had this question also lying around unanswered for days on Mathematics Stackexchange.

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    $\begingroup$ Lightning fast comment: think of doing the integral on the Riemann surface of the multivalued function. Then, the integral is no longer over a closed loop. The only thing you can really say about it (without invoking special information about the integrand) is that, as usual, its value doesn't change if you deform the integration path while keeping its endpoints fixed and not passing through any singularities. $\endgroup$ Commented Oct 16, 2015 at 8:55
  • $\begingroup$ ^That seems to mean thats its okay to interprete my limits as evaluating to $0$ as I was saying? $\endgroup$
    – user6818
    Commented Oct 16, 2015 at 20:54

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