Find an estimation (using polar coordinates) Consider the function
$$
f(x,y):=\lVert x\rVert^{1-n}\ln(\lVert x\rVert)(\arctan(\lVert x-y\rVert))^{-\alpha},~~0<\alpha<n,~~n>1,~~(x,y)\in\Omega\times\Omega,~~~\Omega\subset\mathbb{R}^n
$$
with $x\neq y$.
I am searching for an estimation
$$
\lvert f(x,y)\rvert\leq\frac{\lvert a(x,y)\rvert}{\lVert x-y\rVert^{\alpha}}
$$
with $a\in L^{\infty}(\Omega\times\Omega)$. The hint is, to use polar coordinates. 
Here (Find a weakly singular kernel function for an estimation of a kernel) the same task was asked for the function $g(x,y):= (\arctan(\lVert x-y\rVert))^{-\alpha}$ and I found, using the main value theorem, that
$$
\lvert g(x,y)\rvert\leq\frac{(\lVert x-y\rVert^2+1)^{\alpha}}{\lVert x-y\rVert^{\alpha}}.
$$
So I think to estimate now f, I have to use this result, getting for now
$$
\lvert f(x,y)\rvert\leq\lvert\lVert x\rVert^{1-n}\ln(\lVert x\rVert)\rvert\frac{(\lVert x-y\rVert^2+1)^{\alpha}}{\lVert x-y\rVert^{\alpha}}
$$
But I do not know how I can continue now, especially using polar coordinates.
Hope you can help me.
Edit: Correction of the task!
There was a mistake in the function f!
It has to be
$$
f(x,y):=\lVert x\rVert\ln(\lVert x\rVert)(\arctan(\lVert x-y\rVert))^{-\alpha}.
$$

Easy Peasy. The new function ||x|| ln ||x|| is continuous (although negative near he orign), so it is bounded on $\Omega$ if $\Omega$ is bounded, so you can just use your old estimate times the bound on xlnx.
