Integral of 1/x in complex variables

[Simfish]

Does \int 1/x dx = ln|x| + c in complex variable theory? Or can we relax the absolute value restraint of ln|x|? (as in, can it be ln(x) + c?)

 

[Gib Z]

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 e^{i\pi}= -1 So the logs of negative numbers are quite easy to extend from the reals: \log (-b) = \log b + \log (-1) = \log b + i\pi

 

[SiddharthM]

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.

 

[Kummer]

It is true that if \log z is any logarithm along some branch B. Then (\log z)' = 1//z 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.