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Jordan's Lemma technicalities - where does sin(mz) go

  1. Apr 4, 2012 #1
    1. The problem statement, all variables and given/known data

    I don't understand why it's so great m>0 here

    http://gyazo.com/cfa21b39653275a38ec80722b7faa58b
    From these lectures notes (I think they are generally pretty good)

    Where has the sin(mz) gone?

    When going from 4.67 to 4.68 has

    e^imz = cos(mz) + isin(mz) been used? I'm guessing that was the trick, but I don't see where sin(mz) has pootled off to?
     
  2. jcsd
  3. Apr 4, 2012 #2

    Mark44

    Staff: Mentor

    Yes.
    Instead of looking at ##\int \frac{cos(mx)dx}{a^2 + x^2}##, the writer of the notes is looking at ##\int \frac{e^{mz}dz}{a^2 + z^2}## and getting a value. He also notes that the complex part, which contains the sin(mx) numerator, turns out to be zero, due to the fact that the integrand is an odd function. The integral of an odd function over (-∞, ∞) is zero.
     
  4. Apr 5, 2012 #3

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    At the bottom he says that the sin(mx) integral vanishes, leaving only the cos(mx) part. It says that right in the document!

    RGV
     
  5. Apr 5, 2012 #4
    I think I'm happy with the sin(mx) disappearing. The m variable is causing me grief now:

    So for the variable m, what decides it. If I'm given a functions like

    1/(x+1)^4

    is the variable m in the e^imz equal to the overall degree of the fucntion

    ie would m=2 if

    x^2/(x+1)^4 was my function?

    Thanks
    Thomas
     
  6. Apr 5, 2012 #5

    Mark44

    Staff: Mentor

    Neither m nor a is a variable - both are constants as far as the integration is concerned. The only variables are x and z.
     
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