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How to use the residue theorem when I have an exponential as the pole?

  1. May 22, 2007 #1
    1. The problem statement, all variables and given/known data

    Q. Use residues and the contour shown (where R > 1) to establish the integration formula

    [tex] \int^{\infty}_{0} \frac{dx}{x^3+1} = \frac{2\pi}{3 \sqrt{3}} [/tex]

    The given contour is a segment of an arc which goes from R (on the x-axis) to Rexp(i*2*pi/3)

    2. Relevant equations




    3. The attempt at a solution

    For the function [itex] f(x) = \frac{dx}{x^3+1} [/tex]
    I have worked out that there is only only one pole inside the contour (R >1) which is [tex] x=e^{\frac{i\pi}{3}} [/tex]

    So the residue would be

    2 * pi * i * Res_([tex] x=e^{\frac{i\pi}{3}} [/tex]) f(x)

    But I'm not sure how to do that residue as I've never calculated one with an e in it before... any suggestions? :) Thanks
    1. The problem statement, all variables and given/known data



    p.s sorry I couldn't get the LaTex to show up, someone told me how to do it, but that is obviously not it...
     
    Last edited: May 23, 2007
  2. jcsd
  3. May 22, 2007 #2
    The opening tags shouldn't have a slashes.
     
  4. May 22, 2007 #3

    Dick

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    To deal with the part of the contour along the complex line substitute t*exp(i*2*pi/3) into the integral. You should be able to convince yourself that this part of the contour has been cleverly picked to have a simple relation to the integral along the x axis. If we call the integral along the x-axis I, then the integral along this line going outward is I*exp(i*2*pi/3). Now convince yourself you can ignore the contribution from the arc in the limit as R->infinity. So the total integral along the contour is I-I*exp(i*2*pi/3). (The minus sign coming because we are integrating counterclockwise around the region). Now equate this to what you get from the residue theorem (it's a simple pole, just like in the 1/(x^2+1) case) and solve for I.
     
  5. May 22, 2007 #4

    Hurkyl

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    Er, it's just a number, like any other. What's giving you problems?


    Incidentally, you have the wrong tags for LaTeX. You want
    [tex]YourLatexHere[/tex]
     
  6. May 22, 2007 #5

    Dick

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    The e doesn't make any difference. It's still lim(z->z0) (z-z0)*f(z). Factor the cubic as (z-z0)*(z-z1)*(z-z2) where the zi's are the three cube roots of -1.
     
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