I know about the integrability of $~\frac{1}{|x|^\alpha}$ in the unit ball (center in $~0~$ and radius $~1~$) when $~\alpha<N~$, but how can I calculate$$\int_C \frac{1}{|x-y|^\alpha}\qquad \text{where $~C~$ is the unit ball and $y$ satisfy $|y|>1$}$$
About $~\alpha~$, I'm thinking in $~\alpha=N+2s~$, where $~N~$ is the space dimension and $~0<s<1~$.
On
I have the same question as you, but my $y$ is in the ball, the link is here, r9m kindly told me a new method which is called Rearrangement inequalityhere, you can following the computation in here to understand how to calculate the rearrangement. I have tried when the $y$ is in the ball, I think for the outside case, it should be the same.
I think this is more intuitive than directly computation.
Take hyperspherical coordinates s.t. $\boldsymbol y$ is on the polar axis. Then $$(\boldsymbol x - \boldsymbol y) \cdot (\boldsymbol x - \boldsymbol y) = r^2 + y^2 - 2 r y \cos \theta.$$ The integral over the remaining $n - 2$ angles will give the surface area of an $(n-2)$-dimensional sphere of radius $1$. Therefore $$\int_{x < 1} |\boldsymbol x - \boldsymbol y|^{-\alpha} d\boldsymbol x = S_{n - 2} \int_0^1 \int_0^\pi (r^2 + y^2 - 2 r y \cos \theta)^{-\alpha/2} \hspace {1px} r^{n - 1} \sin^{n - 2} \theta \, d\theta dr = \\ \frac {2 \pi^{n/2}} {\Gamma {\left( \frac n 2 \right)}} \int_0^1 \frac {r^{n - 1}} {(r + y)^\alpha} \hspace {1px} {_2 \hspace {-1px} F_1} {\left( \frac {n - 1} 2, \frac \alpha 2; n - 1; \frac {4 r y} {(r + y)^2} \right)} dr$$ (in one dimension, the integrand is understood as the limit at $n = 1$).