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Cauchy residue formula

  1. Jul 23, 2010 #1
    would it be valid (in the sense of residue theorem ) the following evaluation of the divergent integral ?

    [tex] \int_{-\infty}^{\infty} \frac{dx}{x^{2}-a^{2}}= \frac{ \pi i}{a} [/tex]

    also could we differentiate with respect to [tex] a^{2} [/tex] inside the integral above to calculate

    [tex] \int_{-\infty}^{\infty} \frac{dx}{(x^{2}-a^{2})^{2}} [/tex]
  2. jcsd
  3. Jul 23, 2010 #2
    I believe you should consider two cases: the poles are on or off the real axis. If they are on the real-axis, then you can take the Cauchy Principal Value of the integral by considering a half-disc contour with indentations around the poles. The principal value in this case can be shown to be zero. If the poles are complex, then the integral over the same contour is equal to 2 pi i times the residue of the (single) enclosed pole with the integral over the half-circle arc as it's radius goes to infinity, going to zero and therefore, the real integral is 2 pi i times the residue of the enclosed pole.
  4. Jul 23, 2010 #3
    Yes it should be justifiable but then you would essentially be using real methods. I've seen a differentiation under the integral sign theorem in complex analysis but it seems useless since there exist much more general situations in which you can interchange differentiation and integration.
  5. Jul 23, 2010 #4
    You need to employ the notion of a principal value, otherwise the integral doesn't make sense - it depends on the way in which you take limits either side of the singularities. The best way to rigorously justify differentiating such things would be via the theory of distributions.
  6. Jul 23, 2010 #5
    Eh distribution theory is kind of overkill? I mean it's the most general way I've seen used to justify differentiating under the integral but in a lot of cases I've seen much more "low-tech" methods work as well.
  7. Jul 24, 2010 #6
    Well, you want to differentiate a distribution. You will be interchanging several limiting processes (the integral, the definition of the principal value and the differentiation itself) so you can't just differentiate under the integral sign and "expect" it to work.

    However, using distributions the result is pretty much immediate.
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