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The problem: I am integrating complex logarithms over an angle $\phi$ over $[0,2\pi]$. It is quite complex (pun not intended) and I called Mathematica in to aid me. I am calculating an energy of a system (which must be real-valued) and getting a complex result.

The integration (done by Mathematica, which formally chooses the logarithm's branch cut as $]-\infty,0]$) yields

$2\pi (i \pi + 3 \ln{(-r)})$

With that branch cut, $\ln{(-r)} = \ln{|r|} + i\pi, (r>0)$, which results in a remaining imaginary term. The question I have is: is it allowed to "re-choose" the logarithms branch cut when evaluating the seemingly independent problem of evaluating the logarithm of a negative number, just so, that instead of a imaginary $i\pi$, I get $-i\pi/3$, nicely cancelling with the other term in the expression? In other words: is the integration independent of the choice of the branch cut in the sense that the branch cut can be changed afterwards?

Alternatively, how would I go about calculating the integral with another branch cut (the one I want), so I can see if the "better" branch cut gives a real result?

Thanks!

UPDATE: Here's the integrand itself:

$$\int_\xi^R \rho d\rho \int_0^{2\pi} d\phi \frac{1}{\rho^2} \frac{4\rho^2+r^2-4r\rho \cos{\phi}}{\rho^2 +r^2-2r\rho \cos{\phi}}$$

Splitting in partial fractions and calculating the integral over $\rho$ gives you

$$\int_0^{2\pi} d\phi \left[ \ln{\left|\frac{R}{\xi}\right|} + \frac{2e^{2i\phi}-1}{e^{2i\phi}-1} \left( \ln{(R-r e^{i\phi})}-\ln{(\xi-r e^{i\phi})} \right) + \frac{e^{2i\phi}-2}{e^{2i\phi}-1} \left( \ln{(R-r e^{-i\phi})}-\ln{(\xi-re^{-i\phi})} \right) \right]$$

This last one evaluates to the above complex expression after a call to Integrate[...,{$\phi$,0,2$\pi$}] in Mathematica, if I make the physical assumption that $\xi << r << R$.

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    I'm not sure what "is it allowed" means here. Unless you say which branch of the logarithm you mean in the integrand, you don't have a well-defined problem. Once you say that, you have a well-defined problem, a well-defined integrand and a well-defined integral, and it's not clear how you could then be "allowed" to choose among various possibilities in the result, since the well-defined integral only has one well-defined value. It would be much easier to say something more specific about your problem if you also told us the integral you're dealing with.2011-04-11
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    Also, there's no branch of the logarithm on which $\ln(-r)=\ln|r|-i\pi/3$ (presuming you meant $-i\pi/3$); the branches of the logarithm yield $\ln(-r)=\ln|r|+(2n+1)i\pi$ with $n\in\mathbb{Z}$.2011-04-11
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    @joriki: yes I meant $-i\pi/3$, thanks for that. I thought a branch cut could be chosen arbitrarily, as long as it starts from a branch point (the origin in this case)?2011-04-11
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    You can choose the branch cut arbitrarily, but that doesn't mean you can choose the function values arbitrarily. It just means you can choose where the discontinuity (in this case a jump by $2\pi i$) goes, but the fibre of possible function values for each argument is fixed, and you get it by adding multiples of $2\pi$ to the "canonical" function values.2011-04-11
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    By the way, choosing a branch cut doesn't determine the function values; you can still add multiples of $2\pi i$ to all function values at once without moving the discontinuity. So the function values are not determined by specifying that Mathematica chooses the branch cut as $]-\infty,0]$, you also need to specify one function value, e.g. $\ln 1=0$. (Of course usually logarithms are taken such that they coincide with the real logarithm on the positive real axis.)2011-04-11
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    I've never seen partial fractions done like this. It might work, I'm not sure, but you can avoid all these complications from going complex by doing it the usual way, namely using quadratic denominators if they don't factor over the reals.2011-04-11
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    @ruben: I'm not on a computer with *Mathematica* right now for me to check your work, so one question: have you tried using `Assuming[]`?2011-04-12

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