Let $U$ be the open set in $\mathbb{R}^2$ consisting of all $x$ with $\|x\|<1$. Let $f(x,y)=1/(x^2+y^2)$ for $(x,y)\neq 0$. Determine whether $f$ is integrable over $U-0$ and over $\mathbb{R}^2-\bar{U}$; if so, evaluate.
This question has already been posted here integral over $U - 0 $ and $R^2 - \overline{U}$ without any solution.
I have thought to do the following: By change of polar coordinates we know that $r^2=x^2+y^2$ and if $g(r,\theta)=(r\cos\theta, r\sin\theta)$ is the diffeomorphism or change of variable, we have that $\det D_g(r,\theta)=r$, with which we will have that $\int_{U-0}\frac{1}{x^2+y^2}dxdy=\int_{0}^{2\pi}\int_{r\in (0,1]}\frac{1}{r^2}rdrd\theta=\int_{0}^{2\pi}\int_{r\in (0,1]}\frac{1}{r}drd\theta$ , and
$\int_{\mathbb{R}^2-\bar{U}}\frac{1}{x^2+y^2}dxdy=\int_{0}^{2\pi}\int_{1}^{\infty}\frac{1}{r^2}rdrd\theta=\int_{0}^{2\pi}\int_{1}^{\infty}\frac{1}{r}drd\theta=\int_{0}^{2\pi}(\infty-1)d\theta=\infty$.
I do not know what I can conclude from this, on the one hand I do not know how to calculate $\int_{0}^{2\pi}\int_{r\in (0,1]}rdrd\theta$ because $r\in (0,1]$ and on the other hand I do not know if the second integral if it results in infinite? Could someone help me please?
Is this reasoning correct? Thank you.
If you want to have a rigorous reasoning you can compute the integrals : $$\int_{\epsilon \le x^2 + y^2 \le 1} f(x,y) \mathrm d x\mathrm d y$$ and $$\int_{1 \le x^2 + y^2 \le R} f(x,y) \mathrm d x\mathrm d y$$ and $\epsilon \to 0$, $R \to \infty$. To compute the above intergrals you can use the polar coordinates substitution and you have $$\int_{\epsilon \le x^2 + y^2 \le 1} f(x,y) \mathrm d x\mathrm d y = \int_{\sqrt \epsilon}^1 \int_0^{2\pi} \frac1r \mathrm dr \mathrm d \theta = -\pi \ln \epsilon \to_{\epsilon \to 0} \infty$$ and $$\int_{1 \le x^2 + y^2 \le R} f(x,y) \mathrm d x\mathrm d y = \int_1^\sqrt R\int_0^{2\pi} \frac1r \mathrm dr \mathrm d\theta = \pi \ln R \to_{R\to\infty} \infty$$ This proves that $f$ is not integrable on the two sets.