Expected Distance of Closer of Two Random Points to center of Unit Square

Question

Given two random points picked in the unit square, what is the expected distance from the center of the square to the closer of the two points?

The expected distance from the center of a square to a random point is known to be about 0.382597. Precisely, it is given by the universal parabolic constant over 6 (P/6). I believe I have managed to calculate an approximate answer of 0.3015435, but all methods I have tried so far to calculate a nice integral have failed.

Answer 1

Using Mathematica, we can symbolically compute the PDF of $$D_{(1)} = \min (D_1, D_2), \quad D_i \sim \sqrt{X^2 + Y^2}, \\ X, Y \sim \operatorname{Uniform}(-1/2, 1/2)$$ with the command

FullSimplify[PDF[OrderDistribution[{ProbabilityDistribution[
 FullSimplify[PDF[TransformedDistribution[
    Sqrt[x^2 + y^2], {Distributed[x, UniformDistribution[{-1/2, 1/2}], 
     Distributed[y, UniformDistribution[{-1/2, 1/2}]}], d], 
 0 < d < 1/Sqrt[2]], {d, 0, 1/Sqrt[2]}], 2}, 1], d], 
0 < d < 1/Sqrt[2]]

which gives output equivalent to

$$f_{D_{(1)}}(d) = \begin{cases} 4 \pi d (1 - \pi d^2) & 0 \le d\leq \frac{1}{2} \\ 4 d (\pi - 4 \tan ^{-1}t)(1 - \pi d^2 - t + 4 d^2 \tan ^{-1}t)& \frac{1}{2} < d < \frac{1}{\sqrt{2}} \end{cases}$$ where $t = \sqrt{4d^2-1}$. A plot of this function is shown below:

We can also compute an expectation but Mathematica gives the result in terms of polylogarithms and I did not attempt to simplify it further. Numerical approximation gives $$\operatorname{E}[D_{(1)}] \approx 0.30154355840431487492052719049429100813792402431564\ldots.$$