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Help with a limit

  1. Jan 10, 2015 #1
    We have the following limit:
    [tex]\lim _{N\rightarrow \infty}N\log\left(1+\frac{(s\log N)^{2}}{4\pi^{2}} \right )-\sum_{n=1}^{N}\log\left(1+\frac{(s\log n)^{2}}{4\pi^{2}} \right )-N\left(\frac{2\log N}{(\log N)^{2}+\frac{4\pi^{2}}{s^{2}}} \right ) [/tex]

    Where png.png is a complex parameter.

    any thoughts are appreciated
     
  2. jcsd
  3. Jan 10, 2015 #2

    mfb

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    Staff: Mentor

    Is this homework?
    Is the third large term inside the sum or outside?
    Did you use ##N = \sum_{n=1}^N 1## and combine the differences of logs to a log of fractions?
     
  4. Jan 10, 2015 #3
    this is not a homework
    the third term is outside the sum
    i tried your suggestion, but wasn't helpful

    thanks for the remarks though
     
    Last edited: Jan 10, 2015
  5. Jan 13, 2015 #4
    the limit is better stated this way, i guess.
    [tex]2\lim_{N\rightarrow \infty}\sum_{n=1}^{N} \left[\int_{\log n}^{\log N}\frac{x}{x^{2}+\frac{4\pi^{2}}{s^{2}}}dx-\left(\frac{\log N}{(\log N)^{2}+\frac{4\pi^{2}}{s^{2}}} \right ) \right ][/tex]
     
  6. Jan 18, 2015 #5

    Svein

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    If we split up the expression, the last part is not dependent of n and can be moved outside the sum. In the integral, if you put u=sx/2π, you get x=2πu/s and therefore dx=2π/s*du. This will make the integral easier to solve. Just remember to change the integration limits (x = log(n) transforms into u=s*log(n)/2π).
     
  7. Jan 18, 2015 #6

    mfb

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    The integral is easy to solve (no substitution necessary, the numerator is 1/2 the derivative of the denominator), but then we are back at the expression in post 1.
     
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