I am interested in understanding the behavior of solutions to $$ -\Delta u(x) = |x|^{-2} \, \text{ for }\, x \in B_1$$ with some boundary condition (for example $u = 0$ on $\partial B_1$).
Let $n \geq 3.$ I would like to show something like $$\int_{B_1} |x-y|^{2-n} |y|^{-2} d y \simeq |\log |x|| \,\text{ as } \, x \to 0.$$ Any ideas on how to see this?
Here's an answer with non-sharp behavior at the origin.
Let's show the upper bound, i.e. $$ \int_{B_1} |x-y|^{2-n}|y|^{-2}\, dy \lesssim_n |\ln|x||, $$ for $x$ near $0$. The lower bound is actually easier to handle.
The idea is simply to consider separately the singularities for both terms in the integrand. For this let $\varepsilon\in (0,1)$ and suppose $\varepsilon/2<|x|<\varepsilon$. Decompose $$ B_1= B_{\varepsilon/4}\cup (B_{4\varepsilon}\setminus B_{\varepsilon/4})\cup (B_1\setminus B_{4\varepsilon}), $$ and break up the integral over $B_1$ into the pieces corresponding to the above decomposition. Call each one of these integrals, in order of appearance $I$, $II$, and $III$. In what follows, all the implicit constants involved depend only on the dimension $n$.
Handling $I$ is easy, since if $y\in B_{\varepsilon/4}$, then $\varepsilon\lesssim |x-y|$ and so $$ I\lesssim \int_{B_{\varepsilon/4}} \varepsilon^{2-n}|y|^{-2} \, dy \lesssim 1. $$
$II$ is similar, since in this case $|y|\approx \varepsilon$ and so $$ II\lesssim \varepsilon^{-2} \int_{B_{4\varepsilon}\setminus B_{\varepsilon/4}} |x-y|^{2-n}\, dy \lesssim \varepsilon^{-2} \int_{B(x, 10\varepsilon)} |x-y|^{2-n}\, dy \lesssim 1. $$
Finally, $III$ we use that for $|y|>4\varepsilon$, we have $|x-y|\approx |y|$ and so $$ III\lesssim \int_{B_1\setminus B_{4\varepsilon}} |y|^{-n}\, dy \lesssim |\ln \varepsilon|. $$
Combining all these estimates we arrive at $$ \int_{B_1} |x-y|^{2-n}|y|^{-2}\, dy \lesssim 1 + |\ln |x|| \lesssim |\ln|x||. $$
To get the lower bound simply note $$ \int_{B_1} |x-y|^{2-n}|y|^{-2}\, dy \geq \int_{B_1\setminus B_{2\varepsilon}} |x-y|^{2-n}|y|^{-2}\, dy, $$ and proceed as in $III$.