Let $$v(x) = \int_{B(0,R)} \frac{c}{||x-y||^{n-2}}\mathrm{d}y$$ Show that for $x \in B(0,R)^c$ we have $$v(x) = c_1||x||^{2-n} + c_0$$
Where $B(0,R) \subset \mathbb{R}^n$ is the ball centered at $0$ and of radius $R$ and $c_1$ and $c_2$ are constants. I'm afraid my calculus is a bit rusty, I tried changing to polar coordinates but im getting confused with the term $||x-y||$.
Hint. The change of measures theorem states that if $T:\mathrm{X} \to \mathrm{X}'$ is a homeomorphism and $\mu$ is a measure on $\mathrm{X}$ then, for the image measure of $\mu$ via $T,$ denoted as $T(\mu),$ the following relation holds:
$$\int\limits_{\mathrm{X}'} f \circ T^{-1}\ dT(\mu) = \int\limits_{\mathrm{X}} f\ d\mu$$
Take $\mathrm{X} = B(0; R) \setminus \{0\}$ and $\mathrm{X}' = (0, R) \times \mathbf{S}_{n-1}$ with $T:x \mapsto (\|x\|, \|x\|^{-1} x).$
You may want to observe that if $\sigma_{n-1}$ is surface measure on $\mathbf{S}_{n-1}$ and $\lambda_n$ the Lebuesgue measure on $\mathbf{R}^n$ then $T(\lambda_n)=\lambda_1 \otimes \sigma_{n-1}.$