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Contour integral using residue theorem

  1. Dec 4, 2016 #1
    1. The problem statement, all variables and given/known data
    Find the solution of the following integral upload_2016-12-4_1-19-56.png

    2. Relevant equations
    upload_2016-12-4_1-22-35.png

    3. The attempt at a solution
    I applied the above relations getting that
    upload_2016-12-4_1-23-28.png
    Then I was able to factor the function inside the integral getting that
    upload_2016-12-4_1-25-22.png
    From here I should be able to get a solution by simply finding the poles of the function that are inside the bounds of this integral (in this case the unit circle), using that poles to find a residue, and then summing the residue's and multiplying them by 2πi.

    In this case, I see four poles that are "inside" the unit circle: z= -1, z = -1/3 , z = 0, and z = 1. z= - 3 is outside of the unit circle.

    However, when I do this, I get a solution of -8π/9, when I know that the solution should be 2π/9. In fact, the residue of 0 comes back as undefined, and -1 , 1 as zero.

    Clearly I have done something wrong, though I am not sure what exactly.
     
  2. jcsd
  3. Dec 4, 2016 #2

    Delta²

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    Gold Member

    I am rusty in complex analysis and contour integration but what do you mean when you say that the residue of 0 comes back undefined? Maybe you do a mistake on calculating the residue of 0? Can you show us your work on how you calculate all the residues?
     
  4. Dec 4, 2016 #3
    Are you sure there are 4 poles inside the unit circle? What's a pole? Also when you make the substitution ## z=e^{it} ## the integral becomes a closed contour integral (with a circle through it). The substitutions in these are very messy and when you suspect something is wrong, first thing you should do is double, triple check those substitutions.
     
  5. Dec 4, 2016 #4

    lurflurf

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    Homework Helper

    z= -1, z = 1
    are not poles they have residue zero
    z=0 is a pole of order 2
    $$\operatorname{Res}=\lim_{z\rightarrow0}\dfrac{\operatorname{d}}{\operatorname{dx}}z^2\operatorname{f}(z)$$
     
  6. Dec 4, 2016 #5
    Thanks everyone! I forgot about that sneaky aspect of poles being at a higher order.
     
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