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The open set to check for integrability is $U = \{(x,y) \in \mathbb{R}^2 | 0<x^2+y^2<4\}$. I've considered a succesion of sets $U_n = \{(x,y) \in \mathbb{R}^2 | \frac{1}{n^2}<x^2+y^2<4\}$, that comply with $U_n \subset U_{n+1}$ and $U = \bigcup_{n=1}^{\infty}U_n$.

Now, from that, I can obtain that $\lim_{n\to \infty}\int -\ln{n}<\lim_{n \to \infty}\int\ln{\sqrt{x^2+y^2}}<\lim_{n \to \infty}\int \ln{2} \implies -\lim_{n \to \infty}\frac{1}{n}<\lim_{n\to \infty}\int\ln{\sqrt{x^2+y^2}}<\lim_{n\to \infty}n \ln{2} \implies0<\lim_{n\to \infty}\int\ln{\sqrt{x^2+y^2}}<\infty$.

Does this prove something about whether or not the function is integrable? How else can I use exhaustion to prove its integrability?

Thanks a lot!

Eduardo V. Kuri
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  • It seems to me that you simply assume the limit in question exists, i.e. you assume what you actually have to prove. Note that if we have $a_n\leq b_n\leq c_n$ and $\lim_{n\to\infty}a_n$, $\lim_{n\to\infty}c_n$ exist, then the existence of $\lim_{n\to\infty}b_n$ doesn't follow (e.g. take $a_n=-1,c_n=1,b_n=(-1)^n$). Also, please include the differentials so it's clear what variable we're integrating with respect to. – bjorn93 Apr 19 '22 at 22:18
  • Sorry, where did I assume the limit existed? I just bounded the function between two other functions, and never actually concluded the function of the limit existed or not. – Eduardo V. Kuri Apr 19 '22 at 22:24
  • Show that the integrals over $U_n$ are monotonically decreasing and bounded below. The existence of the limit follows. – eyeballfrog Apr 20 '22 at 18:01

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Switch to polar coordinates. $\int_U ln(\sqrt{x^2+y^2})dxdy=2\pi\int\limits_0^2 ln(r) rdr$$=8\pi(\frac{ln(2)}{2}-\frac{1}{4})$.

To avoid integration, elementary calculus may be used to get bounds on the integrand. Maximum is at $r=2$ and minimum is at $r=e^{-1}$. Bounded continuous function over finite interval is integrable.

herb steinberg
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