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Contour integral with poles on contour

  1. Nov 29, 2013 #1


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    In the process of calculating the integral [itex] \int_0^{2\pi}\frac{\sin{x} \cos{x}}{\sin{x}+\cos{x}}dx [/itex] by contour integration,I got the following:

    -\frac{1}{2}[ \LARGE{\oint} \large{\frac{z^2}{(1-i)z^2+i+1}}dz-\LARGE{\oint}\large{\frac{z^{-2}}{(1-i)z^2+i+1}}dz] [/itex]

    Where the contour of integration for both integrals is the unit circle centered at the origin. The poles are at [itex]z=\pm i \sqrt{i}=\pm \frac{\sqrt{2}}{2}(1-i) [/itex]. As you can see, [itex] |z|=1 [/itex] and so they're on the contour.

    My question is,how should I treat such poles?
    Should I exclude them and calculate the integrals as [itex] \pi i \sum_i r_i [/itex] or should include them and use [itex] 2 \pi i \sum_i r_i [/itex] ?
    How should I decide with what sign each of the residues should appear in the calculation of integrals?

  2. jcsd
  3. Nov 29, 2013 #2


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    The initial integral doesn't look well-defined, it has a pole inside.
  4. Nov 29, 2013 #3
    A pole on the contour is << equivalent to 1/2 pole >>
    That is : Enclose the pole inside the contour with a small half-circle. The corresponding angle is pi instead of 2*pi.
    This is also true for a pole located on an angular point, with the corresponding angle.
  5. Nov 29, 2013 #4


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    Yeah...it'll diverge...I wasn't careful about that.
    But how does this divergence show up in contour integration?

    I did that...but it didn't agree with the result that maple gave me(which is finite!!!).Although wolframalpha seems to be aware of its divergence.
  6. Dec 1, 2013 #5
    Whenever you make such a transformation on an integral and you create simple poles on the contour, then the original integral must be interpreted in the Cauchy principal-valued sense. That is, we should consider:

    [tex]\text{PV}\int_0^{2\pi} \frac{\sin(x) \cos(x)}{\sin(x)+\cos(x)}dx[/tex]

    with poles at ##3\pi/4## and ##7\pi /4##. Now, how about a nice plot showing little indentations around those poles of your particular orientation around them. Either way won't matter as long as we do the calculations correctly in fact, why not do one one way and the other the other way. That'll still work won't it? And while you're at it, review that theorem in your text book about how to compute the integral over an indentation around a simple pole as the radius goes to zero being equal to ##\theta i r## where ##\theta## is the radial angle extended by the indentation.

    Oh yeah, color-code them so we know which one's which.
    Last edited: Dec 1, 2013
  7. Dec 1, 2013 #6


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    If I include the poles in the contour and do it with half circles,I'll get zero which seems to be wrong!
  8. Dec 1, 2013 #7
    It is zero in the same way that:

    $$\text{PV}\int_{-1}^1 \frac{1}{x}dx=0$$

    Ok, compute the antiderivative of:

    $$\int \frac{\sin(x)\cos(x)}{\sin(x)+\cos(s)}$$

    and using that antiderivative, prove that the principle value of that integral is zero by taking the appropriate limiting cases.
  9. Dec 2, 2013 #8


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    Ok, let's do it as follows:
    First,I choose a contour which avoids including the poles.So the integrals divides to five integrals on five different parts of the contour but because the poles are not included,We have [itex] I_1+I_2+I_3+I_4+I_5=0 [/itex]. The PV is [itex] I_1+I_3+I_5=-I_2-I_4 [/itex] so we only have to calculate the integrals around the poles as the part of the contour around them goes to zero.But the boundaries of the integrals around the poles become equal as the radius goes to zero and so the integrals go to zero and we will have zero for the PV.
    Was it good enough?
  10. Dec 2, 2013 #9
    Absolutely not. But that's ok. I'm not your teacher. Lemme's see, what would I do if I was and you pulled that stunt on me? Hummm . . . well, I'd tell you that in the real world, outside of class rooms and nicely-worded problems in textbooks, the world is messy, unkind, brutal, and unforgiving and what you said up there is not good enough to fight them off. What you need to be to survive in that world, is to be direct, explicit, exact, and comprehensive in your dealings with life. Now, this problem won't help much in that endeavor, but if you attacked it as such, it would give you practice for one day when it really counts.

    So not. Rather plots, color-coded, integrals, residues, explanations, and rigor to show unequivocally that it is such.
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