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Contour integration

  1. Apr 28, 2010 #1
    1. The problem statement, all variables and given/known data

    By an appropriate choice of contour, validate the following relationship by evaluation,

    2. Relevant equations


    3. The attempt at a solution
    Any help on this one would be much appreciated.

    Attached Files:

    Last edited: Apr 28, 2010
  2. jcsd
  3. Apr 28, 2010 #2


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    I'm afraid copy pasting the relevant equations don't constitute as a attempt at a solution. What type of contour are you thinking of and do you know how to find the poles?

    Edit: I believe the answer should be [itex]\pi/2[/itex]
    Last edited: Apr 28, 2010
  4. Apr 28, 2010 #3


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    I'd start by finding all the singularities of the integrand...where are they? What type of singularities are they?
  5. Apr 29, 2010 #4
    Put in z in place of the x's. The singularities are when the bottom line is zero, so theres a singularity at z = i and z = -i. The contour im thinking would be a semi circle on the positive side of the axis so -i isnt needed so z = i is one simple pole. I dont know how to handle the z^4 though.
  6. Apr 29, 2010 #5


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    That is correct so far. You handle the z^4 term in the same manner, [itex](z^2+1)(z^4+1)=0[/itex] gives you [itex](z^2+1)=0[/itex] and [itex](z^4+1)=0[/itex]. Solve for z.
  7. Apr 29, 2010 #6
    Ok, so i split ([tex]z^{2}[/tex] + 1) into (z+i)(z-i) to get the first pole. So am i right in spliting ([tex]z^{4}[/tex] + 1) into ([tex]z^{2}[/tex] + i)([tex]z^{2}[/tex] - i)? Do i then split each of those again?
  8. Apr 29, 2010 #7


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    Don't need to split them further, just solve for z now.
  9. Apr 29, 2010 #8
    Is z = [tex]\sqrt{i}[/tex] the other simple pole? I tried to do it out using that but i didnt get out [tex]\frac{\pi}{6}[/tex] as the final answer
  10. Apr 29, 2010 #9


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    You should have a total of three poles within the contour, [itex]\sqrt{i}[/itex] is one of them.

    As for not getting pi/6, read post #2.
  11. Apr 29, 2010 #10
    Got it! Thanks a lot for all your help.
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