Let $f = \chi_{B(0,1)}$. Can anyone help me with the behavior of the following convolution $$f * |\cdot|^{-\alpha}(x) = \int_{B(0,1)}\frac{1}{|x-y|^{\alpha}}dy,$$for the cases $|x| > 1$ and $|x| < 1$?
For instance, in $\mathbb{R}$ I thought about picking $x \in (-1,1)$ and write $$\int_{-1}^1\frac{1}{|x-y|^\lambda}dy = \int_{-1}^x\frac{1}{(x-y)^\lambda}dy + \int_x^1\frac{1}{(y-x)^\lambda}dy = \frac{(1-x)^{1-\lambda}}{1-\lambda}+\frac{(x-1)^{1-\lambda}}{1-\lambda} \leq \frac{C}{1-\lambda},$$where $C$ is a finite number. There's an issue with that which is, when $\lambda \rightarrow 0$ that quantity seems to blowup, and it sounds like an issue. Whereas if $|x| > 1$ then we can't give this bound since $x$ can be arbitrarly big. So, in some way, the integral should behave on the distance from $x$ to the origin but I am not able to put this rigorously.
Thank you!