For $n\in\mathbb N$ put $\displaystyle a_n=\int_0^{2\pi}\int_0^{2\pi}\frac{\cos2n(x-y)}{\sqrt{|x-y|}}dxdy$ and $b_n=\displaystyle \int_0^{2\pi}\int_0^{2\pi}\frac{\sin2n(x-y)}{\sqrt{|x-y|}}dxdy$. Can one give asymptotics for $a_n$ and $b_n$? In particular does there exist $0<p<1$ such that the series $$\sum_{n=1}^\infty(a_n^2+b_n^2)^{p}$$ is convergent?
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$\begingroup$ The double integral involving cosine doesn't seem to exist. The integrand certainly blows up all along the line $x=y$. $\endgroup$ – Greg Martin May 11 '15 at 23:19
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$\begingroup$ @GregMartin no, it won't around $x=y$, $|x-y|^{-1/2}$ has growth as $r^{-1/2}$, integral of that is finite, if $n$ is finite. $\endgroup$ – Yimin May 11 '15 at 23:23
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$\begingroup$ @Yimin good point! $\endgroup$ – Greg Martin May 12 '15 at 7:03
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For $a_n$,
$\displaystyle\int\int \frac{\cos(2n(x-y))}{\sqrt{|x-y|}} dxdy = \int_0^{2\pi}\left(\int_{r=-y}^{2\pi-y} \frac{\cos(2nr)}{\sqrt{|r|}}dr\right) dy \\ =\displaystyle \int_0^{2\pi}\left(\int_{r=0}^{y} + \int_{r=0}^{2\pi -y} \frac{\cos(2nr)}{\sqrt{r}}dr\right) dy$
inside integral,
$\int_0^x\frac{\cos(2nr)}{r^{1/2}}dr = \frac{C}{\sqrt{n}} \int_0^{2nx^2} \cos(t^2)dt$
thus the original integral is less than
$\frac{C}{\sqrt{n}}\sup_{0\le x\le 2\pi} \int_0^{2nx^2}\cos(t^2)dt$
it is about Fresnel integral. The integral is always finite. Thus
your integral is with decay rate as $\frac{1}{\sqrt{n}}$.
For the other part is the same way, $|b_n|\le C\frac{1}{\sqrt{n}}$, then
$\displaystyle\sum (a_n^2 + b_n^2)^p \sim C \sum \frac{1}{n^{p}}$
when $p<1$, it is not convergent.
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$\begingroup$ actually, you can bound $|a_n|,|b_n|$ from below also, with rate as $1/\sqrt{n}$. $\endgroup$ – Yimin May 11 '15 at 23:50
