TITLE: Checking whether $\int_{B(0,r)} \int_{B(0,r)} \ln(\|x-y\|)dxdy > 0$?
QUESTION [9 upvotes]: Today I was discussing with a classmate about the sign of the integral
$$\int_{B(0,r)} \int_{B(0,r)} \ln(\|x-y\|)dxdy,$$
where $B(0,r)$ denotes the ball of center $0$ and radius $r$ in $\mathbb{R^2}$. My friend said that this integral is negative for every $r>0$ because the function $\ln |x-y|$ is "very negative" at $x=y$. I don't agree and I told him that I think there exists a critical $r$ from which the integral is positive. However, I don't know how to prove it. I rewrite it by using polar coordinates as
$$\int_0^r\int_0^r\int_0^{2\pi}\int_0^{2\pi} \ln(\sqrt{r_1^2+r_2^2-2r_1r_2\cos(t_1-t_2)})r_1r_2dt_1dt_2dr_1dr_2.$$
I computed this integral with the software Mathematica and I obtained positive values with, for example, $r=2$. However, this proof is not valid for my friend. Does anyone know how to prove it rigorously?

REPLY [5 votes]: Let $A = \int_{B(0,1)} \ln(\|x\|)dx$ and choose $r$ so that $\pi r^2 \gt 4\pi + \frac{|A|}{\ln 2}$.
For any fixed $y \in B(0,r)$, write $$B_1= \{x \in B(0,r):\|x-y\|\le 1 \}$$ and $$B_2= \{x \in B(0,r):\|x-y\|\ge 2 \}$$
Then we have:
$$\int_{B(0,r)} \ln(\|x-y\|)dx \ge \int_{B_1} \ln(\|x-y\|)dx + \int_{B_2} \ln(\|x-y\|)dx
$$
$$\ge \int_{B(0,1)} \ln(\|x|)dx + \int_{B_2}(\ln 2) dx$$
$$\ge A + (\pi r^2 - 4\pi)(\ln 2) \gt 0$$
With $C = A + (\pi r^2 - 4\pi)(\ln 2)$, this implies: $$\int_{B(0,r)} \int_{B(0,r)} \ln(\|x-y\|)dxdy \gt \int_{B(0,r)} C dy = \pi r^2 C \gt 0$$
Now observe that the function $$f(r) = \int_{B(0,r)} \int_{B(0,r)} \ln(\|x-y\|)dxdy$$
is continuous, $f(1) \lt 0$ and, by the above calculations, $f(r) \gt 0$ for large enough $r$, so there is a minimum value $r_0$ where $f(r_0) = 0$ and it's easy to see that $f(r) \gt 0$ for all $r \gt r_0$