Show that $\int_{\partial\mathbb{D}} \log\lvert x-y\rvert \,\mathrm{d}y = 0$ whenever $x\in\mathbb{D}$

Question

As the title suggests, I am trying to show that $$ \int_{\partial\mathbb{D}} \log\lvert x-y\rvert \,\mathrm{d}y = 0 $$ whenever $x\in\mathbb{D}$. In this problem, $\mathbb{D}$ is the unit disk in $\mathbb{R}^2$.

I am not sure how to do this. I was thinking of using the mean value property since $f(y) = \log\lvert x-y\rvert$ is harmonic away from the point $x$. Sadly, I wasn't able to get anywhere with this.

Thanks in advance for any help/suggestions!

Answer 1

We can approach this problem using the mean value property for another function. I will give you a little outline and you can fill the details in each step to complete the proof:

1. The function
   $$f(x)= \int_{\partial \mathbb{D}} \log|x-y|dy$$ is well defined and harmonic on $\mathbb{D}$.
2. $f$ is radially symmetric since for any rotation matrix $A$ (orthogonal transformation)

$$f(Ax)= \int_{\partial \mathbb{D}} \log|Ax-y|dy=f(x)= \int_{\partial \mathbb{D}} \log|Ax-Ay|dy=f(x)$$ 3. Apply the mean value property on every sphere with radius $r\in (0,1)$ and use the fact that $f(0)=0$ to get

$$2\pi f(x)=0$$

for any $x$ with norm $r$.