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This question is based on some Physics motivation. Define a distance function $f(\mathbf{r})=\int_{\Omega }d^2k\int_{\Omega }d^2q \cos[(\mathbf{k}-\mathbf{q})\cdot\mathbf{r}]$, where $\mathbf{r},\mathbf{k},\mathbf{q}\in \mathbb{R}^2$ and $\Omega \subset\mathbb{R}^2$ is some finite region.

The question is: At large distance $\left | \mathbf{r} \right |\rightarrow \infty $, what's the asymptotic behavior of this function $f(\mathbf{r})$? Does it decay like $e^{-\left | \mathbf{r} \right |/L}$ or $\left | \mathbf{r} \right |^{-a}$(with some characteristic length $L$ and some positive number $a$)? Or does the large distance behavior depend on the geometry of the region $\Omega$?

Thank you very much.

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closed as off-topic by Gerry Myerson, Daniel Moskovich, David White, Ryan Budney, Todd Trimble Oct 15 '13 at 13:03

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question does not appear to be about research level mathematics within the scope defined in the help center." – Daniel Moskovich, Ryan Budney
If this question can be reworded to fit the rules in the help center, please edit the question.

Simul-posted to without notice to either site:… – Gerry Myerson Oct 15 '13 at 10:15
Simul-posting without notice. – Gerry Myerson Oct 15 '13 at 10:16
This is not research level - can be done analytically. Write the cos as a sum of exponentials and separate. – Carl Oct 15 '13 at 10:17
@ Carl, I see,thanks. Or directly integrate cos, right? – Kai Li Oct 15 '13 at 12:15
Yes, you could use the cos addition formula and then integrate. – Carl Oct 15 '13 at 14:03
up vote 3 down vote accepted

As OP seems to be stuck, I give here the few lines needed to solve this problem as sugested in the comments. Given $$ f(\mathbf{r})=\int_{\Omega }d^2k\int_{\Omega }d^2q \cos[(\mathbf{k}-\mathbf{q})\cdot\mathbf{r}]= \frac{1}{2}\int_{\Omega }d^2ke^{i\mathbf{k}\cdot\mathbf{r}}\int_{\Omega }d^2qe^{-i\mathbf{q}\cdot\mathbf{r}} +\frac{1}{2}\int_{\Omega }d^2ke^{-i\mathbf{k}\cdot\mathbf{r}}\int_{\Omega }d^2qe^{i\mathbf{q}\cdot\mathbf{r}}. $$ These integrals are easily done to give $$ f(\mathbf{r})=4\frac{(1-\cos(\pi x))(1-\cos(\pi y))}{x^2y^2} $$ from which the asymptotic behavior can be readily gotten. This makes clear that the question is not appropriate here.

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@ Jon Thanks. Does the numerator in your answer should be $(1-\cos(\pi x))(1-\cos(\pi y))$? – Kai Li Oct 15 '13 at 14:14
Thanks, just fixed. – Jon Oct 15 '13 at 14:54

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