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Complex Countour Integral

  1. Oct 8, 2009 #1
    1. The problem statement, all variables and given/known data

    Let C be the contou starting at z=-1 going around the circle lzl=1 and ending back at z=-1 (counterclockwise) determine. Int. (Log(z)/z) dz.


    2. Relevant equations

    z=e^(Log(z))


    3. The attempt at a solution

    So I've been stuck on this problem for quite some time.

    Im not sure if the integral is suppose to be broken up into:

    Int f(z(t))*z'(t) dt.

    If so, I'm unsure what to call my f(z) and my z(t).
    I tried calculating:

    z(t)=e^it
    z'(t)=ie^it
    Log(z(t))=Log(e^it)=it

    Then

    (0,pi)Int((it/e^it)*ie^it) dt
    = (0,pi)Int (-t) dt
    = -pi.

    I'm sure this isn't correct.
    Any help would be appreciated.
     
  2. jcsd
  3. Oct 8, 2009 #2

    Dick

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    The integral of t from 0 to pi isn't pi. Is it?
     
  4. Oct 8, 2009 #3
    Ha. I got ahead of myself and didn't even compute the integral.

    (0,pi)Int(-t) dt
    = -t^2/2 evaluated at (0,pi)
    = -pi^2/2

    Is this on the right track? Should the bounds be changed? Or is my integral entirely incorrect?
     
  5. Oct 8, 2009 #4

    Dick

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    It looks right to me. Any reason to doubt it?
     
  6. Oct 9, 2009 #5
    why doesn't this work using the Cauchy Integral formula?

    i set f(z)=log z and a=0 but then f(a)=f(0)=-infinity.

    that's how it doesn't work but what's the reason for this?
    is it because f isn't holomorphic on the unit disc? (and it's not holomorphic because it "blows up" at z=0)
     
  7. Oct 9, 2009 #6

    Dick

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    I don't know what that's supposed to be about. You want to integrate Log(z)/z, not Log(z) and the limits are 1 to -1, not 0 to 'a' whatever that is. If you are asking why you can't do it as an antiderivative, you can. As long as you're careful about not crossing branch cuts. Evaluate (Log(z))^2/2 between 1 and -1.
     
  8. Oct 9, 2009 #7
    i havent done anything with the Cauchy integral formula in a while....can you talk me through the procedure here:

    we have [itex]f(a)=\oint \frac{f(z)}{z-a}dz[/itex]

    so why can't we set f(z)=log z and a=0 then the RHS is just [itex]\oint \frac{\log{z}}{z} dz[/itex] and it will be equal to the LHS [itex]f(a)=f(0)=\log{0}=-\infty[/itex]
    i think the reason this doesn't work is because the f i have used in this case isn't holomorphic on the region we are integrating over i.e. at z=0 it has a pole (is that correct???)

    so what should i use for my f(z)?

    thanks.
     
  9. Oct 9, 2009 #8

    Dick

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    Well, yeah. Log isn't holomorphic in the disk. It's worse than a pole, it has a branch cut. The given contour isn't even closed. You can't really integrate around anything. You have to find another way to solve it. There are two alternative methods in this thread.
     
  10. Oct 9, 2009 #9
    ok. so you can't use the formula because log isn't holomorphic. but the contour is closed surely? its a circle of radius 1 from z=-1 to z=-1 isn't it?
     
  11. Oct 9, 2009 #10

    Office_Shredder

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    Yes, but there's a branch cut, so somewhere on that circle log won't even be defined, or at least is discontinuous
     
  12. Oct 9, 2009 #11

    Dick

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    Hmm. I was paying more attention to the limits of integration in the attempt. If the first z=-1 isn't a typo then ryanj123 had better rework the limits.
     
  13. Oct 9, 2009 #12
    would this be possible using a semicircular contour from z=1 to z=-1 then along the real axis to z=1 again but with a small semicircular indent to avoid the origin and then use the Cauchy Residue Theorem?
     
  14. Oct 9, 2009 #13

    Office_Shredder

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    You don't know that the integral around that little indent around the origin goes to zero as it becomes small
     
  15. Oct 9, 2009 #14
    why not? because log goes to infinity at 0?
     
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