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Some Complex Integrals

  1. Mar 25, 2010 #1
    Hi

    I find my notes for how to calculate complex integrals woefully inadequate, and I'm hoping someone can explain to me how to do them.

    One that the notes particularly fail for is:

    'Integrate z/(a-exp(-iz)) along the rectangle with vertices at pi, -pi, pi+iR, -pi+iR

    Hence integrate xsinx/(1-2acos(x) +a^2) dx from 0 to infinity, for 0<a<1'


    1)Epic fail because it doesn't tell me what a is in the first one. If |a|=/=1, then surely the integral of the first one would be zero because there are no poles?

    2) You can sort of get the second integral from the first by taking the imaginary part of it - assuming that |z|=1. Which makes it clear that my first point is wrong.

    So, how do I do the first integral? Why that particular contour? What do I take a to be? Why? How does this relate at all to the second integral? How do I do any of this?

    Sorry about the lack of LATEX knowhow, and thanks for any help.
     
  2. jcsd
  3. Mar 27, 2010 #2
    What do you get if you multiply numerator and denominator by

    a - exp(i z)

    ?
     
  4. Mar 27, 2010 #3
    That gives you a denominator somewhat resembling what is required, but you still have a |z| in there that I can't get rid of. (I got the denominator to be a^2+1-2|z|*a*cos(theta) ).

    And even forgetting the issue with that |z| it still leaves the question over what to do with the numerator, ie how to get xsin(x) out of it, and how on earth to make that ridiculous contour any use at all.
     
  5. Mar 27, 2010 #4
    What is [a - exp(iz)] [a - exp(-iz)] ?
     
  6. Mar 27, 2010 #5
    About the numerator, you do get exp(iz) in here...
     
  7. Mar 27, 2010 #6
    My bad, I worked the denominator out wrong.

    After multiplying by that, we have the integrand as:

    z(a-cos(z)-isin(z))/(a^2+1-2acos(z))

    Which is nearly of the required form. But how is the numerator sorted out? We can't just take imaginary parts, because z has both real and imaginary parts and it'd mess up. And also, it looks like we'll have to consider the imaginary axis to get the integral from 0 to infinity that's required - but this isn't part of the contour it tells us to look at...
     
  8. Mar 27, 2010 #7
    I think the integral of xsinx/(1-2acos(x) +a^2) dx is from minus to plus pi.

    The sum of the two integrals parallel to the imaginary axis simplifies to something you can exactly evaluate.
     
    Last edited: Mar 27, 2010
  9. Mar 27, 2010 #8
    I don't understand your first sentence. Are you saying you think there's a typo in the question I'm doing?

    And (I haven't tried it yet, but) I'm fairly certain that integrating along the lines parallel to the imaginary axis won't yield the integral -along- the imaginary axis. So I still see no relevance at all between the integral and the contour.
     
  10. Mar 27, 2010 #9
    I think there is a typo, the integral from zero to infinty of xsinx/(1-2acos(x) +a^2) dx should be the integral from zero to pi. Note that the integral from zero to infinity doesn't converge.

    You extract this integral from the part of the contour on the real axis from minus pi to plus pi. The two parts parallel to the imaginary axis can be taken together (due to the periodicity of the exp(-i z). The z in the numerator makes that the sum of the two integrals doen't completely cancel. What is left can be computed exactly.
     
  11. Mar 27, 2010 #10
    I believe I have it completely sorted now. Thanks very much, you've been a great help. Sorry if I came across as dim :P
     
  12. Mar 27, 2010 #11
    You should also do the integral of

    cos(px)/[(1-2acos(x) +a^2]

    using the same method. I think you can also do this using the standard exp(ix) = z substitution, but now you have to deal with a branch point singularity of p is not an integer.
     
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