Convergence of $\int_\Bbb{R^2} \frac{\log(x^2+y^2)}{x^2+y^2}$ I tried to figure out if $\int_\Bbb{R^2} \frac{\log(x^2+y^2)}{x^2+y^2}$ converges
I think I should split it into two integrals  $$\lim_{k\to\infty}\int_{B_\frac{1}{k}} \frac{\log(x^2+y^2)}{x^2+y^2} + \lim_{r\to\infty}\int_{B_r} \frac{\log(x^2+y^2)}{x^2+y^2}$$
where     $B_\frac{1}{k}=${$ x \in \Bbb{R^2} |$ $ \frac{1}{k}\leq \|x\|\leq \frac{1}{2}  $}   and  $B_r=${$ x \in \Bbb{R^2} | $ $ \frac{1}{2}\leq \|x\|\leq r $}    and with changing coordinates {$x^2+y^2=r^2, \arctan\frac{y}{x}=\phi$} we get $$\lim_{k \to \infty} \int^{2\pi} _0\int^{\frac{1}{2}} _\frac{1}{k}  \frac{r\log(r^2)}{r^2} drd\phi + \lim_{r \to \infty} \int^{2\pi} _0\int^{r} _\frac{1}{2}  \frac{r\log(r^2)}{r^2} drd\phi$$
 but Im not sure if that correct or if I can use again change of variables for $\log(r^2) = t, \frac{2r}{r^2}dr=dt$
thanks ahead

You can form the following chain of inequalities
$$ \int_4^\infty \frac{\log r^2}{r} dr > \int_4^\infty \frac{1}{r} dr = \lim_{b \to \infty} \log b - \log 4 = \infty.$$
So your integral diverges at $\infty$. In fact, it also diverges at $0$.
