Expected distance from a random point in a square to its centre

Solution. With the centre at the origin, the expected distance is $$\mathbb{E}[|P|] = \int_{-1/2}^{1/2} \int_{-1/2}^{1/2} \sqrt{x^2 + y^2} \, dx \, dy.$$

By symmetry, restrict to the first quadrant and multiply by $4$: $$\mathbb{E}[|P|] = 4 \int_0^{1/2} \int_0^{1/2} \sqrt{x^2 + y^2} \, dx \, dy.$$

Use polar coordinates in the first quadrant. The square $[0, 1/2]^2$ breaks into two regions by the diagonal $y = x$. In the lower triangle ($0 \leq y \leq x \leq 1/2$) the angle $\theta$ runs in $[0, \pi/4]$ and the radius goes from $0$ to $1/(2 \cos \theta)$. By symmetry, the upper triangle contributes the same integral. $$\begin{align*} \mathbb{E}[|P|] &= 8 \int_0^{\pi/4} \!\! \int_0^{1/(2\cos\theta)} \!\! r^2 \, dr \, d\theta \\ &= 8 \int_0^{\pi/4} \frac{1}{3} \cdot \frac{1}{8 \cos^3 \theta} \, d\theta = \tfrac{1}{3} \int_0^{\pi/4} \sec^3 \theta \, d\theta. \end{align*}$$

The standard integral is $\int \sec^3 \theta \, d\theta = \tfrac{1}{2}\big[ \sec \theta \tan \theta + \ln|\sec \theta + \tan \theta| \big]$. Evaluating from $0$ to $\pi/4$: $$\int_0^{\pi/4} \sec^3 \theta \, d\theta = \tfrac12 \big[ \sqrt{2} \cdot 1 + \ln(\sqrt{2} + 1) \big] - 0 = \tfrac{\sqrt{2} + \ln(1 + \sqrt{2})}{2}.$$

Hence $$\boxed{\mathbb{E}[|P|] = \frac{\sqrt{2} + \ln(1 + \sqrt{2})}{6} \approx 0.3826.} \qedhere$$

import numpy as np
# Uniform in [-0.5, 0.5]^2; measure distance to centre (origin).
pts = np.random.rand(2, 10**7) - 0.5
d = np.sqrt(pts[0]**2 + pts[1]**2)
exact = (np.sqrt(2) + np.log(1 + np.sqrt(2))) / 6
print(f"sim: {d.mean():.5f}   exact: {exact:.5f}")
# sim: 0.38256   exact: 0.38260