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How might one evaluate a contour integral like

  1. Jul 26, 2006 #1
    How might one evaluate an integral equation like the following:

    I = lim k-> 0+ {ClosedContourIntegral around y [1/(z^2 + k^2)]}, where the contour y is a simple, closed, and positively oriented curve that encloses the simple pole at z = i*K?

    Is it possible to evaluate integrals of this form?


  2. jcsd
  3. Jul 28, 2006 #2


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    Why don't you just evaluate the integral inside the limit using residues?
  4. Jul 28, 2006 #3


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    I thought that at first, but since the contour y encloses the simple pole at z=ik and presummably not the other simple pole at z=-ik, what is going to happen when the pole becomes order 2 at k=0? But we're not caring what happens at k=0, just what happens near k=0, so take the curve y as, say, [tex]y: \left| z-ik\right|=k[/tex] so that the other pole at z=-ik is excluded for all k, then by the residue Thm. we have

    [tex]\oint_y \frac{dz}{z^2 + k^2} = 2\pi i \mbox{Res}_{z=ik} \left( \frac{1}{z^2 + k^2}\right) [/tex]

    but we have

    [tex]\frac{1}{z^2 + k^2} = \frac{i}{2k(z + ik)}-\frac{i}{2k(z -i k)}[/tex]

    so it turns out that

    [tex]\oint_y \frac{dz}{z^2 + k^2} = 2\pi i \left( \frac{i}{2k}\right) = -\frac{\pi}{k} [/tex]
    Last edited: Jul 28, 2006
  5. Jul 30, 2006 #4
    So the following would be true?

    [tex]{k} \oint_y \frac{dz}{z^2 + k^2} = -pi[/tex]

    If you took the limit as k approaches 0 from the right of the expression above, would it still equal -pi?
    Last edited: Jul 30, 2006
  6. Jul 31, 2006 #5
    the limit would still be [itex]-\pi[/itex].
    Last edited: Jul 31, 2006
  7. Jul 31, 2006 #6
    Perfect! Thanks for the information everyone!

    Best Regards,

  8. Aug 21, 2006 #7


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    Hello benorin. Looks to me this should be:




    The answer to the original question then would be:

    \lim_{k\to 0}\int_{\Lambda(k)}\frac{1}{z^2+k^2}dz=\infty[/tex]

    where [itex]\Lambda(k)[/itex] is the contour defined above.

    Think so anyway. I'm taking Complex Analysis this semester so . . . may need some help.:confused:
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