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Evaluate real integral

  1. Apr 25, 2015 #1
    1. The problem statement, all variables and given/known data

    Use the first example in Lecture 38 to evaluate [itex] \int_{0}^{2\pi} cost\: sint \:dt [/itex]

    http://s8.postimg.org/ccycky76d/Screen_Shot_2015_04_25_at_9_47_41_PM.png

    2. Relevant equations

    Residue theorem, etc.

    3. The attempt at a solution

    I mostly understand his work in the example. On a side note, it seems that his notes are a bit terse, but I digress.

    I understand the application of the Residue Theorem around the closed path. The upper half circle includes the removable singularity z = i. He integrated around the two half circle paths using z = Re, dz =Riedz and z = re, dz = riedz substitutions. I assume that limits of integration for f(z) = f(re) are π to 0 to complete the path circumscribing the difference between the two upper half circles. For the segment integral -R to -r, why is πi in the numerator? And he equated them to segment integral r to R because of symmetry?
     
  2. jcsd
  3. Apr 25, 2015 #2

    Svein

    User Avatar
    Science Advisor

    You are talking about two completely different things here. The problem you specify can be solved easily without referring to "Lecture 38", just remembering the formula for sin(2t).
     
  4. Apr 26, 2015 #3

    Mark44

    Staff: Mentor

    Or even more simply, by using a very simple substitution, u = sin(t).
     
  5. Apr 26, 2015 #4
    Yes, I know that, but the problem says to use the first example in Lecture 38. That's why I'm first trying to fully understand the example and then apply it to the problem.
     
  6. Apr 26, 2015 #5
    Yes, of course, I know how to integrate this with real methods. I'm simply trying to do what the problem instructs.
     
  7. Apr 26, 2015 #6

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    You can transform this integral to one in the complex plane, I guess, but the most naive transformation does not lead to an analytic function.
    Strange problem.
     
  8. Apr 26, 2015 #7
    The best I can figure is to substitute cos θ = (1/2)[z - (1/z)], sin θ = (1/2i)[z - (1/z)], and dz = zidθ and integrate.
     
    Last edited: Apr 26, 2015
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