I'm having a trouble integrating (in $\mathbb{R}^2$) the following formula:
$$\frac{t}{|B(x,t)|}\int_{B(x,t)} \frac{||y||}{(t-||x-y||^2)^{\frac{1}{2}}} dy $$
where $B(x,t)$ is the ball with center $x$ and radius t,
$|B(x,t)|$ is the area of the ball
After a change of variables:
$$\frac{t}{|B(x,t)|}\int_{B(x,t)} \frac{||y||}{(t-||x-y||^2)^{\frac{1}{2}}} dy =\frac{1}{|B(0,t)|}\int_{B(0,1)} \frac{||x+tz||}{(1-||z||^2)^{\frac{1}{2}}} dz$$
In polar coordinates: $$ \frac{1}{|B(0,t)|}\int_{0}^{2\pi}\int_{0}^{1} \frac{||x+t\ r \ e^{i\theta}||}{(1-r^2)^{\frac{1}{2}}} dr d\theta$$
But even in polar coordinates I cannot solve it.
Any advice will be helpfull. Thanks!!