Strange Relation: $\sum$ (m$^2$+n$^2$) = 2$\pi$ Integral

[Icosahedron]

$$\sum\ (m^2+n^2) = 2\pi\int^{\infty}_{1} (\frac{1}{r^2} r dr)$$

The sum goes over $$\mathbb{N}\times \mathbb{N}$$

 

[Icosahedron]

Wait, that is what I mean.

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

The sum goes over $$\mathbb{N}\times \mathbb{N}$$

I can make no sense of it.

Edit: Now everything fixed.

 

[Dick]

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.

 

[HallsofIvy]

And the right side makes no sense. In $2\pi\int^{\infty}_{1} (\frac{1}{r^2} r dr)$, "r" is clearly intended to be a vector (otherwise it reduces to $\int 1/r dr= ln(r)$) 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.

 

[Icosahedron]

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 $$\sim$$?

And, yes my book says too, it diverges.

 

[tiny-tim]

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

The sum goes over $$\mathbb{N}\times \mathbb{N}$$

I can make no sense of it.

Hi Icosahedron!

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.

 

[Dick]

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.

 

[Icosahedron]

Thanks!