1
$\begingroup$

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?

$\endgroup$
3
  • $\begingroup$ 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
  • $\begingroup$ @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
  • $\begingroup$ @Yimin good point! $\endgroup$ – Greg Martin May 12 '15 at 7:03
1
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.