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Using the Residue Theorem for Real Integrals

  1. Sep 14, 2011 #1
    1. The problem statement, all variables and given/known data
    [tex]I=\int_{-\infty}^{\infty} { dx \over {5x^2+6x+5}}[/tex]


    2. Relevant equations
    The residue theorem.


    3. The attempt at a solution
    I can't use the residue theorem since the denominator has real zeros. How should I solve this?
     
  2. jcsd
  3. Sep 14, 2011 #2

    Hootenanny

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    Yes you can use the residue theorem, you just need to "go around" the poles, i.e. deform your usual semi-circular contour around the poles. Have a look http://www.nhn.ou.edu/~milton/p5013/chap7.pdf" [Broken] and in particular at second 7.6.
     
    Last edited by a moderator: May 5, 2017
  4. Sep 14, 2011 #3
    Maybe you should double-check that business about the real zeros.
     
  5. Sep 14, 2011 #4

    Hootenanny

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    Good catch, I didn't even think to look!
     
  6. Sep 15, 2011 #5
    oops. From now on, I'll never try to guess root. Quadratic equation, we meet again.

    Thanks guys.
     
  7. Sep 15, 2011 #6

    dynamicsolo

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    There is also the quaint old "complete the squares" technique:


    [tex]\int_{-\infty}^{\infty} \frac{dx}{5x^2+6x+5} = \frac{1}{5} \int_{-\infty}^{\infty} \frac{dx}{ (x^2+\frac{6}{5}x+\frac{9}{25}) + (1 - \frac{9}{25} ) }[/tex]

    [tex]= \frac{1}{5} \int_{-\infty}^{\infty} \frac{dx}{ (x+\frac{3}{5})^{2} + (\frac{4}{5})^{2} } . [/tex]

    I believe quadratic polynomials in the denominator never require complex-analytic techniques, though you certainly aren't forbidden to use them...
     
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