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In the book of Analysis on Manifolds by Munkres, at page 151, question 3, it is asked that

Let $U$ be the open set in $\mathbb{R}^2$ consisting of all $x$ with $\|x\|<1$. Let $$ f(x,y) = \frac{1}{x^2 + y^2} $$ for $(x,y) \ne 0$. Determine whether $f$ is integrable over $U\setminus 0$ and over $\mathbb{R}^2\setminus \overline{U}$; if so, evaluate.

I have given the following answers, but, as you can see below, in both cases I have found that the integral does not exists, so is everything correct in my answers, because the author says "if so evaluate", which means in general at least one of the integral exists and we are going to evaluate it :).

My answers:

Let $g(r,\theta) = (r \cos\theta, r \sin \theta),$ where $g: (0,1)\times (0,2\pi) \to U - \{(x,y) \in \mathbb{R}^x | y = 0, x \geq 0\}.$ Note that the excluded set $\{(x,y) \in \mathbb{R}^x | y = 0, x \geq 0\}$ is of measure zero, hence it does not contribute the integral that we are going to evaluate.

For $U - \vec 0,$ let $C_n = [1/n, 1-1/n] \times [1/n, 2\pi - 1/n]$ be a rectifiable set in $(0,1)\times (0,2\pi)$ s.t $C_n \subset Int (C_{n+1}) $ and their union is $(0,1)\times (0,2\pi)$. Since for each $n \geq 2$, this set is a rectifiable set s.t $n+1$-th set contains $n$-th one and the union of these sets form a cover for $(0,1)\times (0,2\pi)$, we can evaluate the improper integral of any function over $(0,1)\times (0,2\pi)$ by first integrating it on $C_n$, and then taking the limit as $n \to \infty$.

Now, by Fubini's theorem, we have $$\int_{1/n}^{2\pi - 1/n} \int_{1/n}^{1-1/n} \frac{1}{r^2} * |r| = ln(\frac{1 - 1/n}{1/n} ) * (2\pi - 2/n),$$ but this function blows up as $n\to \infty$, hence it is not bounded, hence the improper integral does not exists.

For $\mathbb{R}^n - \bar U$, we form the partition $$[1+ 1/n, n] \times [1/n, 2\pi - 1/n],$$ for $n\geq 2$, and apply the same reasoning, and get $$ln(\frac{n}{1 + 1/n} ) * (2\pi - 2/n),$$ but this also blows up as $n \to \infty$, hence it is not bounded, hence the integral does not exists.


Question:

Is there anything wrong with the answers given above ? Is there any flaw ?

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  • $\begingroup$ @Surb Yes, I know from Calculus courses, but the thing is I cannot find the error in my above argument, either. $\endgroup$ – onurcanbektas Jul 8 '18 at 9:53
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    $\begingroup$ @Did Yes, I know English, and I do know that it does not mean that, but the very fact that that statement is there persuade me to question the validity of my answer, hence my question. I mean as far the question concerns, that sentence is not related to the it, so you can just ignore it. $\endgroup$ – onurcanbektas Jul 9 '18 at 3:21
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    $\begingroup$ "so you can just ignore it." Hmmm... but your whole question revolves around this sentence. $\endgroup$ – Did Jul 9 '18 at 5:21
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    $\begingroup$ @Did If that is the case, if I just copy-paste the whole question, and just delete that sentence and ask a new question like that, will I get a different answers ? I really curious how that sentence changes the math that we are doing in here ? $\endgroup$ – onurcanbektas Jul 9 '18 at 12:01
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    $\begingroup$ Sorry but you seem too much interested in wordplays and rhetorics, rather than addressing the mathematical points raised about your question, for my present mood. No big deal. $\endgroup$ – Did Jul 9 '18 at 17:05
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Since for each $n\ge 2$, this set is a rectifiable set s.t $n+1$-th set contains $n-th$ one and the union of these sets form a cover for $(0,1)\times (0,2\pi)$, we can evaluate the improper integral of any function over $(0,1)\times (0,2\pi)$ by first integrating it on $C_n$, and then taking the limit as $n \to \infty$

This is wrong... e.g. take $f(x) = \frac{1}{x}$ and the domain $C = (-1,1)\setminus\{0\}$ then $f$ is not integrable over $C$ because $$\int_C \, |f(x)| \, dx = \infty$$ but by your "assumption" it would hold with $$C_n = (-1,1)\setminus\left(\frac{-1}{n},\frac{1}{n}\right)$$ that:

$$\int_D\; f(x) \, dx = \lim_{n\to\infty} \int_{C_n} f(x) \, dx = \lim_{n\to\infty} 0 = 0$$ because $$ \int_{C_n} f(x) \, dx = 0$$ for all $n\ge 1$

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  • $\begingroup$ Just for the positive values: you have $[1/n, 1-1/n]$, but $\int_{1/n}^{1-1/n} 1/n = ln(\frac{1-1/n}{1/n } ) = ln(n-1) $, and this blows up as $\to \infty$ ? Your computations are hidden from us, I cannot understand how do you compute that value ? $\endgroup$ – onurcanbektas Jul 9 '18 at 3:33
  • $\begingroup$ No you have $$\left[\frac{1}{n},1\right)$$ and so you get $$\int_\frac{1}{n}^1 \frac{1}{x} dx = -\ln\frac{1}{n} = \ln(n)$$ and yes, this tends to infinity as $n \to \infty$, that's why $$\int_C \, |f(x)| \, dx = \infty$$ holds. But on the other hand you have $$\int_{C_n} f(x) \, dx = \int_{-1}^{-\frac{1}{n}} \frac{1}{x} dx + \int_\frac{1}{n}^1 \frac{1}{x} dx = -\ln(n) + \ln(n) = 0$$ $\endgroup$ – Gono Jul 10 '18 at 8:01
  • $\begingroup$ I'm still confused: $\int_{-1}^{-1/n} 1/x = [ln|x| ]|_{-1}^{-1/n} = ln(1/n) - 0 \to - \infty,$, but we have also $ln(n)$, so $ln(1/n) + ln(n) = l(1) = 0$ ? $\endgroup$ – onurcanbektas Jul 11 '18 at 5:31
  • $\begingroup$ Yeah… that's why $f$ is not integrable on $C$ but on each $C_n$ with integral value $0$. $\endgroup$ – Gono Jul 11 '18 at 5:40
  • $\begingroup$ Ok, let me correct one thing: We are integrating $|f|$, so we should take the integral $\int_{-1}^{-1/n} -1/x$, but this leads $[ln(|1/x|)]_{-1}^{-1/n} = ln(n) - 0 \to \infty$ as $n\to \infty.$ $\endgroup$ – onurcanbektas Jul 11 '18 at 5:53

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