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Contour integration & the residue theorem

  1. Sep 1, 2015 #1
    When one uses a contour integral to evaluate an integral on the real line, for example [tex]\int_{-\infty}^{\infty}\frac{dz}{(1+x)^{3}}[/tex] is it correct to say that one analytically continues the integrand onto the complex plane and integrate it over a closed contour ##C## (over a semi-circle of radius ##R## closed along the real line between ##-R## and ##R##)? In this case [tex]\int_{C}\frac{dz}{(1+z)^{3}}[/tex] which reduces to the original integral on the real line in the limit as the radius tends to infinity.
     
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  3. Sep 1, 2015 #2

    Geofleur

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    Yes, but you have to show, using Jordan's lemma or otherwise, that the integral over the upper part of the contour does indeed vanish as R goes to infinity.
     
  4. Sep 1, 2015 #3
    OK, but it is analytic continuation then, it just happens to be a trivial case where the complex part vanishes in the limit, right?
     
  5. Sep 1, 2015 #4

    Geofleur

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    I think so, but your integrand blows up at x = -1, and analytic continuation and using a semi-circular contour is not going to help that.
     
  6. Sep 1, 2015 #5
    Yes, sorry. I made up the integral as I was writing the post and didn't think to check that it worked!
     
  7. Sep 1, 2015 #6

    Geofleur

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    Note that to analytically extend a function, it has to be analytic in the first place. Extending just any old real function into the complex plane is not the same as analytically continuing it. Analytic continuation occurs when you have a Laurent series that only converges in some finite region of the complex plane, and then you use the fact that the function is analytic to extend the domain of applicability of the series beyond its original region of convergence.
     
  8. Sep 1, 2015 #7

    Geofleur

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    Originally, I thought you were talking about specifically ## 1/(1+z)^3 ##, so I had in mind the Laurent series for that, which converges in a circular region of radius 1 centered at the origin. You would analytically continue this function to get outside of that region.
     
  9. Sep 1, 2015 #8
    OK, thanks for the info. So, for example, would it be correct to take ##e^{-x}##, which is analytic for all ##x\in\mathbb{R}##, and analytically continue this to ##e^{-z}## which is in turn analytic for all ##z\in\mathbb{C}##?
     
  10. Sep 1, 2015 #9

    Geofleur

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    I think it's OK to call that analytic continuation, because it is extending the domain of an analytic function. Usually the phrase "analytic continuation" is used in the context of extending a function outside the radius of convergence of its Laurent series. Since ## e^{-x} ## has no singularities, its Laurent series has an infinite radius of convergence. At any rate, as long as it's clear what's going on, what we call it is not as important I suppose.
     
  11. Sep 1, 2015 #10
    Ah Ok, I'll bear that in mind. Thanks for your help!
     
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