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From sum to integral

  1. Oct 1, 2012 #1
  2. jcsd
  3. Oct 1, 2012 #2
    For some reason, there was no page 266 for me. But anyway, it's because of the limit definition of an integral, which works for double integrals just as it does a single integral. A nice overview of this can be found here.
     
  4. Oct 1, 2012 #3
    Why integrals are from zero to ##2\pi##. Why not
    ##\rightarrow \frac{1}{16\pi^2}\int^{4\pi}_0\int^{4\pi}_0...##
     
  5. Oct 1, 2012 #4
    I believe both of those should be equal; if you take the integral at 4[itex]\pi[/itex] and then divide by 16[itex]\pi^2[/itex], it should be equal to taking the integral at 2[itex]\pi[/itex] and dividing by 4[itex]\pi^2[/itex], correct?
     
  6. Oct 2, 2012 #5
    So you said that I can go from sum to integral in the way

    [tex]\rightarrow \frac{1}{a^2}\int^{a}_0\int^{a}_0...[/tex]?
     
  7. Oct 2, 2012 #6
    Well look for example ##lim_{N\to \infty}\frac{1}{N}\sum_{q_1,q_2)ln(q_1-2\pi)##. Is it the same? How I go from that sum to integral?
     
    Last edited: Oct 2, 2012
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