Consider the fundamental solution of Laplace's equation in 2d and 3d: $$\Phi(x,y):= \begin{cases} \frac{1}{2\pi}\ln\frac{1}{|x-y|},\quad x,y\in{\mathbb R}^2\\ \frac{1}{4\pi}\frac{1}{|x-y|},\quad x,y\in{\mathbb R}^3. \end{cases}$$ Let $D\subset{\mathbb R}^m$ $(m=2,3)$ be a bounded domain of class $C^1$.
Question: How can I show the following asymptotic behavior of $\Phi$?
$$\Phi(x,y)= \begin{cases} \frac{1}{2\pi}\ln\frac{1}{|x|}+O(\frac{1}{|x|}),\quad x,y\in{\mathbb R}^2\\ O(\frac{1}{|x|}),\quad x,y\in{\mathbb R}^3 \end{cases}$$ for $|x|\to\infty$, which holds uniformly for all directions $x/|x|$ and all $y\in\partial D$.
Note: The statement above appears without a proof in Rainer Kress's Linear Integral Equations (2nd edition), Chapter 6.
[My thoughts.]
According to the definition of the big O notation, for the case in ${\mathbb R}^2$, it suffices to show that there exist constants $C,M>0$ such that $$ \frac{1}{2\pi}\bigg(\ln\frac{1}{|x-y|}-\ln\frac{1}{|x|}\bigg)\leq C\frac{1}{|x|},\quad \forall |x|>M. $$
This boils down to doing the estimate $$ \ln\frac{|x|}{|x-y|}\leq \tilde{C}\frac{1}{|x|} $$ which I have no idea how to go on.
Similarly for the case in ${\mathbb R}^3$, one needs to show for some $M$ that $$ \frac{1}{|x-y|}\leq C\frac{1}{|x|},\quad \forall |x|>M. $$ This one is much simpler: one can simply rewrite $|x-y|$ and $x$ in terms of Cartesian coordinates.