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Strange relation

  • #1
[tex]\sum\ (m^2+n^2) = 2\pi\int^{\infty}_{1} (\frac{1}{r^2} r dr) [/tex]

The sum goes over [tex]\mathbb{N}\times \mathbb{N}[/tex]
 
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Answers and Replies

  • #2
Wait, that is what I mean.

[tex]\sum\ (m^2+n^2)^{-1} \sim 2 \pi\int^{\infty}_{1} (\frac{1}{r^2} r dr) [/tex]

The sum goes over [tex]\mathbb{N}\times \mathbb{N}[/tex]

I can make no sense of it.

Edit: Now everything fixed.
 
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  • #3
Dick
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The sum (m^2+n^2) clearly diverges. You must mean 1/(m^2+n^2). Then it's just an integral estimate. And it still diverges.
 
  • #4
HallsofIvy
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And the right side makes no sense. In [itex]2\pi\int^{\infty}_{1} (\frac{1}{r^2} r dr) [/itex], "r" is clearly intended to be a vector (otherwise it reduces to [itex]\int 1/r dr= ln(r)[/itex]) in which case integrating from 1 to infinity doesn't make sense. My guess is that the right side is supposed to be integrated over all of R2.
 
  • #5
The sum (m^2+n^2) clearly diverges. You must mean 1/(m^2+n^2). Then it's just an integral estimate. And it still diverges.
Right, that's what I meant. I'm still fighting with latex.

What you mean by 'just an integral estimate'. Why are both expressions [tex]\sim[/tex]?

And, yes my book says too, it diverges.
 
  • #6
tiny-tim
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[tex]\sum\ (m^2+n^2)^{-1} \sim 2 \pi\int^{\infty}_{1} (\frac{1}{r^2} r dr) [/tex]

The sum goes over [tex]\mathbb{N}\times \mathbb{N}[/tex]

I can make no sense of it.
Hi Icosahedron! :smile:

Looks like you're supposed to integrate something over x and y from 1 to ∞ which has poles "of integral 1/r²" whenever x and y are both whole numbers, and then convert to polar coordinates. :smile:
 
  • #7
Dick
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The right side is the integral of 1/(x^2+y^2) over the plane (excluding some stuff near zero) in polar coordinates. It should be a reasonable estimate for the integer sum (also excluding (0,0), I hope). Just like the integral test for infinite series.
 
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  • #8
Thanks!
 

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