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Given $f(x,y) = \sin(x^2+y^2)$ I must give a detailed explanation of the convergence of the integral of $f$. The problem suggests to analyze the following limit $$ \displaystyle \lim_{R\to \infty} \int_0^R \cos^2(x)\mathrm{d}x $$

as well as analyzing the existance of the extended integral: $$ \int_\mathbb{R} \cos^2(x) \mathrm{d}x$$

It also suggests to give two exhaustions of a set to determine the existence of the integral first integral such that both limits are different.

It is easy to prove both the limit and the extended integral do not exist. What I'm having trouble with is, on how to use this three sugestions to determine the convergence of the integral of $f(x,y)$.

I tried using the fact that $\sin(x^2+y^2) = \sin(x^2)\cos(y^2)+\sin(y^2)\cos(x^2)$ and find a relationship between $\cos(x^2)$ and $\cos^2(x)$ to bound the function. However since the extended integral of $\cos^2(x)$ over $\mathbb{R}$ does not exist, I cannot yet see how does the suggestions help out with the convergence.

Edit: Change of variable is not allowed

sputnik
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  • The integral of $f$ over what domain? All of $\mathbb R^2$? Also are you sure the problem suggested $\cos^2(x)$ and not $\cos(x^2)$? – eyeballfrog Apr 20 '22 at 19:52
  • @eyeballfrog Yes, it would be the integral over $\mathbb{R}^2$ and the problem suggests $\cos^2(x)$, not $\cos(x^2)$. – sputnik Apr 20 '22 at 21:33
  • OK, this is odd. $\lim_{R\rightarrow\infty}\int_{-R}^R\int_{-R}^R\sin(x^2+y^2)dxdy = \pi$ but $\lim_{R\rightarrow\infty} \int_0^{2\pi}\int_0^R \sin(r^2)rdrd\theta$ doesn't exist. – eyeballfrog Apr 20 '22 at 21:52
  • @eyeballfrog I believe that's becuase the change of variable suggested is not biyective over the region of integration – sputnik Apr 20 '22 at 22:39

1 Answers1

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You can try a polar transformation: $$r = \sqrt{x^2+y^2} \\ \theta = \text{arctan}(\frac{y}{x})$$ which has Jacobian determinant $r$. Your integral becomes $\int\limits^{2\pi}_{0}\int\limits^{\infty}_{0}r\text{sin}(r^2)dr d\theta$. Can you take it from here?

Kyler S
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  • Thnaks! Nonetheless, I forgot to mention that change of variable is not allowed in this problem. – sputnik Apr 20 '22 at 19:00
  • You are right, based on @eyeballfrog's comment, I don't think we can use change of variable. I don't see why the problem asks for $cos^2(x)$ other than using the bound $cos(x^2) \leq cos^2(x)$ and $sin(x^2) \leq cos^2(x)$ – Kyler S Apr 21 '22 at 16:20
  • Note that the bound you are suggesting is not true for all $x \in \mathbb{R}$. It must be split into the adecuate intervals for it to hold. – sputnik Apr 21 '22 at 20:15
  • Yep, that bound is incorrect in general. I don't see a way to generalize all regions over which the inequality holds. Is it possible that $cos^2(x)$ is an error in the problem? – Kyler S Apr 22 '22 at 02:48
  • Possibly useful: https://math.stackexchange.com/questions/2937980/examining-convergence-of-improper-integral?rq=1 – Kyler S Apr 22 '22 at 02:51