Expected reciprocal distance from the centre of a disc

Despite the singularity of $1/r$ at the origin, the expectation is finite — a cancellation of the integrand’s blow-up with the vanishing density near $r = 0$. This is a distinctive feature of 2D: the density $f_R(r) = 2r$ is linear in $r$, precisely offsetting the $1/r$ factor.

In $d$ dimensions, the radial density of a uniform point in the unit ball is $d \, r^{d-1}$, so $$\mathbb{E}[1/R] = \int_0^1 \frac{1}{r} \cdot d \, r^{d-1} \, dr = d \int_0^1 r^{d-2} \, dr = \frac{d}{d-1},$$ provided $d \geq 2$. (For $d = 1$, the integral diverges, because the density is $1$ everywhere and $\int 1/r \, dr = \infty$ near $r = 0$.) In the ball this gives $3/2$; on a $4$-ball $4/3$; and so on. The 2D answer $2$ fits $d/(d-1)$ with $d = 2$.

The same finiteness holds for $\mathbb{E}[1/|P_1 - P_2|]$ with two independent uniform points in the unit disc: near coincidence the relative-displacement area element $\rho\,d\rho\,d\theta$ cancels the $1/\rho$ singularity, exactly as the factor $2r$ did above, so the integrable singularity gives a finite value. In fact $\mathbb{E}[1/|P_1 - P_2|] = \tfrac{16}{3\pi} \approx 1.698$. (It is in one dimension that the reciprocal-distance expectation would diverge, since there the density does not vanish at $0$.)

import numpy as np
r = np.sqrt(np.random.rand(10**7))
print(f"sim: {(1/r).mean():.5f}   exact: 2.0")
# sim: 1.99498   exact: 2.0