Help with a limit

  • Thread starter mmzaj
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  • #1
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Main Question or Discussion Point

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
 

Answers and Replies

  • #2
34,038
9,879
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?
 
  • #3
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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:
  • #4
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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]
 
  • #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).
 
  • #6
34,038
9,879
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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