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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. –  joriki Apr 11 '11 at 11:46
    
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 $n\in\mathbb{Z}$. –  joriki Apr 11 '11 at 11:53
    
@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)? –  rubenvb Apr 11 '11 at 12:10
2  
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. –  joriki Apr 11 '11 at 12:17
    
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.) –  joriki Apr 11 '11 at 12:27
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