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 $$ f(x,y):=\lVert x\rVert\ln(\lVert x\rVert)(\arctan(\lVert x-y\rVert))^{-\alpha}. $$
One can guess that a missing hypothesis is that $\Omega$ is bounded, say $\|x\|\leqslant c$ for every $x$ in $\Omega$, for some $c\geqslant2$, then $|\|x\|\log\|x\||\leqslant c\log c$ for every $x$ in $\Omega$. Since $\|x-y\|\leqslant2c$ for every $x$ and $y$ in $\Omega$, the task is complete if one shows that $$ (\arctan t)^{-\alpha}\leqslant Ct^{-\alpha}, $$ for some finite $C$, for every $0\lt t\leqslant2c$. Equivalently, one wants $$ \arctan t\geqslant C't, $$ for some positive $C'$, for every $0\lt t\leqslant2c$. Since $\arctan$ is concave, $C'=\arctan(2c)/(2c)$ fits. Finally, since $2c\geqslant4$ and $\arctan4\geqslant1$, for each positive $\alpha$, $a(x,y)=A_\alpha$ works, with $$ A_\alpha=2^\alpha\cdot c^{\alpha+1}\log c. $$