# From sum to integral

1. Oct 1, 2012

2. Oct 1, 2012

### danielu13

For some reason, there was no page 266 for me. But anyway, it's because of the limit definition of an integral, which works for double integrals just as it does a single integral. A nice overview of this can be found here.

3. Oct 1, 2012

### LagrangeEuler

Why integrals are from zero to $2\pi$. Why not
$\rightarrow \frac{1}{16\pi^2}\int^{4\pi}_0\int^{4\pi}_0...$

4. Oct 1, 2012

### danielu13

I believe both of those should be equal; if you take the integral at 4$\pi$ and then divide by 16$\pi^2$, it should be equal to taking the integral at 2$\pi$ and dividing by 4$\pi^2$, correct?

5. Oct 2, 2012

### LagrangeEuler

So you said that I can go from sum to integral in the way

$$\rightarrow \frac{1}{a^2}\int^{a}_0\int^{a}_0...$$?

6. Oct 2, 2012

### LagrangeEuler

Well look for example $lim_{N\to \infty}\frac{1}{N}\sum_{q_1,q_2)ln(q_1-2\pi)$. Is it the same? How I go from that sum to integral?

Last edited: Oct 2, 2012