Show $\int_{\partial \mathbb{D}} \log\lvert x-z\rvert dz \to \int_{\partial \mathbb{D}} \log\lvert y-z\rvert dz$

Question

Suppose $y\in \partial\mathbb{D}$ where $\mathbb{D}$ is the unit disk in $\mathbb{R}^2$. Is there a way to prove that $$ \lim_{\substack{x\to y \\ x\in \mathbb{D}}}\int_{\partial \mathbb{D}} \log\lvert x-z\rvert \,\mathrm{d}{z} = \int_{\partial \mathbb{D}} \log\lvert y-z\rvert \,\mathrm{d}{z} $$

Thanks in advance for any help! I tried a direct approach which didn't work. Is there another way to do it?

Answer 1

Note that by symmetry $\int_{\partial \mathbb{D}} \log |x-z|dx$ just depends on $|x|$. Therefore, we just need to show $\int_{\partial \mathbb{D}} \log |r-z|dz \to \int_{\partial \mathbb{D}} \log |1-z|dz$ as $r \to 1$. Wolfram says all of these integrals are $0$.