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Complex analysis

  1. Mar 8, 2009 #1
    1. The problem statement, all variables and given/known data

    Hey guys.
    I have this integral, I tried to use trigo, tried to use the complex expression but nothing worked, can I please have some help?

    Thanks a lot.


    2. Relevant equations



    3. The attempt at a solution
     

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  3. Mar 8, 2009 #2

    gabbagabbahey

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    The complex Residues method should work fine; why don't you show us what you tried for that method...
     
  4. Mar 9, 2009 #3
    Ok, so I tried to use the residues theorem.
    I did the substitute but ended up with that really ugly expression, should I use the residues theorem on it?
     

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  5. Mar 9, 2009 #4

    gabbagabbahey

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    I won't be able to read your attempt until admin approves your attachment. You can save time by uploading your file to imageshack.us and posting a link to it instead.
     
  6. Mar 9, 2009 #5
    http://img8.imageshack.us/img8/1059/scan0015p.jpg [Broken]

    Thanks a lot.
     
    Last edited by a moderator: May 4, 2017
  7. Mar 9, 2009 #6

    gabbagabbahey

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    Okay, so far so good....now find the 4 roots of the denominator in order to find the poles....which of those poles lies within your contour?
     
  8. Mar 9, 2009 #7

    gabbagabbahey

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    To make your calculations easier, you should note that if [itex]|z^2|>1[/itex] then so is [itex]|z|[/itex]; and [itex]3+2\sqrt{2}>1[/itex].

    Also note that [tex]\sqrt{3-2\sqrt{2}}=\sqrt{2}-1[/tex]
     
  9. Mar 9, 2009 #8
    Oh, I've only now noticed your calculations tips :confused:

    Anyway, this is what I did:

    http://img27.imageshack.us/img27/4732/scan0016y.jpg [Broken]

    I've to warn you, The numbers are awful.

    Does it seems right?

    Thanks a lot.
     
    Last edited by a moderator: May 4, 2017
  10. Mar 9, 2009 #9

    gabbagabbahey

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    They sure are! :biggrin:

    Luckily, they're also correct:approve:....you can simplify them though...after a little algebra you should find that [itex]A=B=\frac{-1}{\sqrt{2}}[/itex] and so your final result becomes [itex]2\pi(\sqrt{2}-1)[/itex]
     
  11. Mar 9, 2009 #10
    Thanks a lot :smile:
     
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