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[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]

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

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- Thread starter Icosahedron
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The sum goes over [tex]\mathbb{N}\times \mathbb{N}[/tex]

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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.**

[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.

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Dick

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HallsofIvy

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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.

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tiny-tim

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Hi Icosahedron![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.

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.

- #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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Thanks!

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