I am trying very hard to to calculate an integral on a 2d disk $B(0,R)$, after I changed it to the spherical coordinate I do not know how to integrate it
$$\int_{\partial B(0,R)}\int_0^R\log \frac{|Re_1-rw|}{|Re_1-\frac{R^2w}{r}|}drdw$$ here the $|\cdot|$ denote the distance of the two vectors in Euclidean norm and $e_1=(1,0)'$.
Any help will be appreciated!
Without loss of generality let $R=1$. Express $w$ in terms of the azimuth $\phi$: $w=(\cos\phi,\sin\phi)$. Then $$\begin{align} \ln\lvert e_1-rw\rvert&=\tfrac{1}{2}\ln(1-2r\cos\phi+r^2)\\ \ln\lvert e_1-w/r\rvert&=\tfrac{1}{2}\ln(1-2r^{-1}\cos\phi+r^{-2})=\ln\lvert e_1-rw\rvert-\ln r\\ \ln\frac{\lvert e_1-rw\rvert}{\lvert e_1-w/r\rvert}&=\ln r \end{align}$$ so $\int_0^1\ln\frac{\lvert e_1-rw\rvert}{\lvert e_1-w/r\rvert}\mathrm{d}r=\int_0^1\ln r\,\mathrm{d}r=\left.r(\ln r-1)\right\rvert_{0}^1=-1\text{.}$
Actually, we also know that
$$-\int_{-\pi}^{\pi}\tfrac{1}{2}\ln(1-2r\cos\phi+r^2)\mathrm{d}\phi=\begin{cases}0&r<1 \\ -\ln r & r>1\end{cases}\text{.}$$ People who study harmonic functions call this the mean value property, and people who study physics call this the shell theorem.