Problem: Suppose $f$ is integrable with compact support and $n ≥ 3$. Then the Newton potential satisfies $$u(x) = CΦ(x) + O(|x|^{−n+1})$$ as $|x| → ∞$, where $C := \int_{\mathbb{R^n}} f(y)d^ny$. (Hint: The inverse triangle inequality $||x| − |y|| ≤ |x − y|$ might be useful.)
This is how we defined the Newtonian potenial for $n\geq 3$: $$ u(x)=\int_{\mathbb{R^n}} \frac{1}{n(n-2)V_n|x-y|^{n-2}}f(y)d^ny$$
My Approach:
Since $f$ has compact support, I was thinking maybe to proceed like this: let $f$ be supported in an open ball $B_r$.
Outside this ball, the integral equals zero, so we can confine ourselves to this ball. Now I should somehow eliminate $y$ from the denominator, so I proceeded in this way: set $|x| ≥ 2r$ and $|y|≤r$, then the following must hold: $$|x-y|≥|x|-|y|≥1/2 |x| $$
But now the factor $1/2$ becomes $2^{2-n}$ when plugged into the integral so I guess it has to be done in another way.