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Integral of 1/x in complex variables

  1. Oct 23, 2007 #1


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    Does [tex]\int 1/x dx = ln|x| + c[/tex] in complex variable theory? Or can we relax the absolute value restraint of ln|x|? (as in, can it be ln(x) + c?)
  2. jcsd
  3. Oct 24, 2007 #2

    Gib Z

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    I'm not sure integrals work exactly the same in the complex plane, I highly doubt it but I haven't quite reached there in my textbook. However, choosing the principal branch of the log in the complex plane, we know from Euler that [tex]e^{i\pi}= -1[/tex] So the logs of negative numbers are quite easy to extend from the reals: [tex]\log (-b) = \log b + \log (-1) = \log b + i\pi[/tex]
  4. Oct 24, 2007 #3
    not if you are integrating over a closed path containing the origin. Although this is a correct answer it doesn't give much insight to your question..complex integration is significantly different from it's real analogue.

    for example: lnx as a function is senseless unless you've defined lnx as the complex logarithm, which is actually pretty complicated as GibZ's msg above suggests.
  5. Oct 24, 2007 #4
    It is true that if [tex]\log z[/tex] is any logarithm along some branch B. Then [tex](\log z)' = 1//z[/tex] for all values not on B. No matter how you choose to define the complex logarithm there will be a branch where it will not be holomorphic. But it does exist at least partially.
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