Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Complex Integration-Residue Theory

  1. Dec 22, 2009 #1
    I'm doing this question, and I have the right answer, but I'm not sure is the method is right. Also, I have a couple of general questions about this method of integration.

    Compute [tex] \int_{-\infty}^{+\infty} \frac{x^2+3}{(x^2+1)(x^2+4)}

    I factorized the bottom line to find the singularities, which are at [tex] \pm i[/tex] and [tex]\pm 2i [/tex].

    First question: in my book, there are examples where the contour is taken as a semicircle in the upper or lower halves of the complex plane. Is it ok to take a contour in the right or left hand side of the complex plane instead? I ask this because all of the singualrities in this question are along the imaginary axis, but say they were all along the real axis? My understanding is that the coutour must enclose the singularities, but a semicircle in the upper or lower halves of the complex plane would run through these (hypothetical) singularities. If it is right to take the the semicircle in the left or right planes, how does this change the way to answer the question, if at all? Or if this is not the right way to take the contour, how should it be done?

    Back to the question:
    Taking the contour as a semicircle in the upper half of the complex plane excludes the singularities at -i and -2i. So computing the residues at each of the enclosed singularities gives a residue of [tex]\frac{1}{3i}[/tex] at i and [tex] \frac{1}{12i}[/tex] at 2i.

    Second question: why is a semicircle the usual choice? How is it any different to a contour that encloses maybe only 1, or 3 of the singularities in this question instead of 2? Each of these give diffenent final answers but I'm wondering what's so special about this particular contour, and it's answer?

    Anyway, I filled in the two residues into the formula stating that the integral is equal to [tex]2\pi i[/tex] times the sum of the residues, and got the answer to the intgral as [tex]\frac{5\pi}{6}[/tex] which is what the answer in the book is.(Also, I don't trust this book as I've found a lot of mistakes, so this could easily be wrong.)

    So basically I think I'm right with the answer, I'm just a little shaky on the method. Any tips are much appreciated.
  2. jcsd
  3. Dec 23, 2009 #2
    You can choose the contour in any way you like. You can apply the residue theorem only if the function is meromorphic inside the contour (i.e. analytic everywhere, except at some poles or essential singularities). Then you get a result of the form:

    Integral over desired interval + integral over additional paths = 2 pi i sum of residues

    Then, the integral over the additional paths can tend to zero, as is usually the case for the semi-circle.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook