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Complex analysis

  1. Mar 3, 2009 #1
    1. The problem statement, all variables and given/known data

    Hey guys.
    So, I need to calculate this integral, I uploaded what I tried to do in the pic.
    But according to them, this is not the right answer, according to them, the right answer is the one I marked in red at the bottom.
    Any idea where this Sin came from?


    2. Relevant equations

    3. The attempt at a solution

    Attached Files:

  2. jcsd
  3. Mar 3, 2009 #2


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    Homework Helper

    One way sin may appear is from combining exponetials

    Try this
    integrate around a pie slice
    one edge
    z=t t in (0,R)
    one edge
    z=t*exp(pi*i/n) t in (0,R)
    the rounded edge
    z=R*exp(pi*i*t/n) t in (0,1)
    let R become large
    express the integral of the pie slice in terms of I then as 2pi*i*Res(f,exp(pi*i/(2n))
  4. Mar 4, 2009 #3
    can you explain to me how you got that term to drop out to 0, when you integrated round the contour from R to -R? did you find a bound for the integrand?
  5. Mar 4, 2009 #4
    When R goes to infinity, the integral of the semi circle goes to zero.
  6. Mar 4, 2009 #5
    yes. but don't you need to find a bound, say M, such that the integrand is less than or equal to M at all points on the semi circle and then say as R goes to infinity the bound goes to 0.
    my question was, what did you use to bound the intergal?
  7. Mar 4, 2009 #6
    Oh, I thought it's right for any semi circle integral so I didn't try to prove it.
    Now I'm not sure.
  8. Mar 4, 2009 #7


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    So for the curved edge
    recall n>=3
    |integral(curved edge)|<|R^4/(R^(2n)+1)|(2pi/n)*R<|R^5/R^(2n)|C<=C/R->0
    The same bound holds for a semicircle (times n)

    ***It is easier to do a wedge than a semi circle as one avoids the sum of n residues only one appears.***
    Last edited: Mar 4, 2009
  9. Mar 4, 2009 #8
    ok. how can you assume [itex]n \geq 3[/itex]
  10. Mar 4, 2009 #9


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    It is given in the problem statement.
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