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Simple contour integral

  1. May 10, 2012 #1
    Trying to teach myself contour integration, but I'm not so good at it. I want help with evaluating the closed integral:

    ∫sinθ/(a-sinθ)dθ from -[itex]\pi[/itex] to [itex]\pi[/itex]

    So I substitute z= e, and sinθ = -i/2(z-z-1) and dθ = -ie-iθdz

    So our integral becomes:

    ∫-i/2(z-z-1)/(a+i/2(z-z-1) dz

    Is this correct so far? If so, what do I do from this point? I suppose I want to find the order of the pole for this function?
     
  2. jcsd
  3. May 10, 2012 #2
    Where is the -i/z from dθ=-i/z dz?

    So next you should convince yourself that the integration region is bounded by a simple curve, and then find all the poles inside this curve.
     
  4. May 10, 2012 #3
    oops, I forgot that on here :)

    Okay, so I thought that's what I should do, but how do you determine the order of the pole for such a messy expression? i.e. how can you in general be sure that the pole you get in the denominator is not cancelled off by something in the numerator? :)
     
  5. May 10, 2012 #4
    You simply plug the number into the numerator (remember that numerator here is everything besides the ##(z-a)^m## term where a is the pole) and if the result is not 0 then m is your order.

    Your integral is correct, but in the case of finding poles, it is best to clean it up a bit.

    ##\displaystyle \int_{\partial \mathbb{D}} \frac{-\frac{i}{2} (z-z^{-1})}{a+ \frac{i}{2} (z-z^{-1})} dz##

    (I rewrote it so the integral is easier to view by people)

    You converted it correctly, since you substituted for ##e^{it}##, you are integrating around the unit disk and since ##t \in [-\pi,\pi]## this becomes contour integral now. The easiest way to find the poles would be to multiply through by (2z)/(2z) to get rid of all the fractions and then consider what make the integrand undefined. Like I mentioned above, simply write the integrand as ##\frac{g(z)}{(z-a)^m}## and verify ##g(a) \neq 0##. Then ##m## is the order of the pole ##a##. Then use the Residue Theorem.

    The residues eventually come out to clean numbers, so don't be deterred by the messy algebra.

    Edit: Some of your residues might be outside of ##\mathbb{D}## depending on your choice of ##a##. Remember to take that into account.
     
    Last edited: May 10, 2012
  6. May 11, 2012 #5
    There are three cases to consider here (assuming a is real): a = 0, 0<a≤1 and a>1. Only the latter one is of any interest, since a=0 is trivial, and the integral does not converge if 0<a≤1.
     
  7. May 11, 2012 #6

    HallsofIvy

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    You say you are doing this as a contour integral in the complex plane but you haven't said what contour!
     
  8. May 11, 2012 #7
    a<-1 is also a valid case.
     
  9. May 15, 2012 #8
    Has been a while, but going back into solving this integral.

    So I end up with the integral:

    ∫ (z-1-z)/(2az + iz2-i) dz

    Now the roots of the equations in the denominator are clearly poles. But how do I know, that you can't get higher order poles by reducing the fraction further? And secondly: The z-1 in the numerator is undefined for z=0 - will that make a problem?
     
  10. May 15, 2012 #9
    Clearly z=0 is a pole. In this case it's easy to find all the poles. Just write the integrand as [itex] \frac{P(z)}{Q(z)} [/itex] where P and Q are polynomials. Then find all the roots of Q, and those are the poles.
     
  11. May 15, 2012 #10
    So my expression is not correct for evaluating the integral? Please show me how you get that ratio of polynomials you want? :(
     
  12. May 15, 2012 #11
    You just multiplied the denominator by ##2z## and then multiplied the top by 2/i for some reason? The whole reason for multiplying by ##\frac{2z}{2z}## is to get rid of the zeroes in the numerator (and "transfer" them down to the denominator, so to speak).

    So, go back and multiply ##\displaystyle \int_{\partial \mathbb{D}} \frac{-\frac{i}{2} (z-z^{-1})}{a+ \frac{i}{2} (z-z^{-1})} dz## by 2z/2z and you will hopefully see what to do from there!
     
  13. May 15, 2012 #12
    Well, in your version, where did the 1/iz term go?
    The original integral is:

    ∫(1/2i(z-1-z)/(a-1/2i(z-z-1) dz/iz

    How do you come to your version from that? Seems to me like you left out the iz term under dz, which I then put back in? Maybe I'm just a bit tired, but I should think something is wrong.
     
  14. May 18, 2012 #13
    Hi all, I think I managed to find the poles inside the unit circle, which as ai - i√(a2-1) in agreement with my solutions notes. However, I also found that z=0 is a pole, which my book does agree on. Who is right, and why? :)
     
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