Average Distance from Center of a Unit Square to Its Perimeter

You start at the center of the unit square and then pick a random direction to move in, with all directions being equally likely. You move along this chosen direction until you reach a point on the perimeter of the unit square. On average, how far can you expect to have traveled?

Let's assume that you start at the origin, which is the center of the unit square $B_\infty = \{ (x,y) \in \mathbb{R}^2 \mid \max \{ |x|, |y| \} = \frac{1}{2} \}.$ Further, let's assume that your randomly chosen bearing makes an angle of $\theta$ with the positive $x$-axis, so $\theta \sim U(0,2\pi),$ and let's assume that you move at some unit speed, so your position through time will be $r(t) = (t \cos \theta, t \sin \theta).$

Since you will hit the $B_\infty$ exactly when $t \max \{ |\cos \theta|, |\sin \theta| \} = \frac{1}{2}$, therefore, the distance you would travel from the origin is given by $$d(\theta) = \frac{1}{2 \max \{ |\cos \theta|, |\sin \theta| \} } = \frac{1}{2} \min \{ |\sec \theta|, |\csc \theta| \}.$$ Relying on the periodicity of $d(\theta)$ to break the entire integral into 8 equal parts, we get thatthe expected distance is therefore \begin{align*}\hat{d} &= \frac{1}{2\pi} \int_0^{2\pi} d(\theta) \,d\theta \\ &= \frac{1}{2\pi} \left( 8 \int_0^{\pi/4} \frac{1}{2} \sec \theta \,d\theta \right) \\ &= \frac{2}{\pi} \Biggl.\ln \left| \sec \theta + \tan \theta \right| \Biggr|^{\pi/4}_0 \\ &= \frac{2}{\pi} \ln \left( 1 + \sqrt{2} \right) \approx 0.561099852339\dots.\end{align*}

We can extend this problem to the slightly different problem of starting at the origin, pointing a random direction and traveling at a uniform, unit Euclidean speed until we get to any the perimeter of any half-unit ball, say $B_p = \{ (x,y) \in \mathbb{R}^2 \mid |x|^p + |y|^p = \frac{1}{2^p} \}.$ As you could possibly surmise, the official question was $B_\infty,$ which is the limit as $p \to \infty.$ In this case, we would get $$d_p (\theta) = \frac{1}{2 \sqrt[p]{\cos^p \theta + \sin^p \theta}}$$ and $$\bar{d}_p = \frac{2}{\pi} \int_0^{\pi/4} \frac{d\theta}{\sqrt[p]{\cos^p \theta + \sin^p \theta}} = \frac{4}{\pi} \int_0^{\sqrt{2}-1} \frac{dt}{\sqrt[p]{(1-t^2)^p + 2^p t^p}},$$ where the second integral is obtains through the mystical tangent half-angle substitution. In general, this is not an analytical answer, though in certain other cases it simplifies. Obviously, for the case of $p=2$ everything simplifies to $d_2(\theta) = \frac{1}{2}, \forall \theta \in [0,2\pi)$ and so $\bar{d}_2 = \frac{1}{2}$ as well. For $p=1$, we have $$d_1(p) = \frac{1}{2 (\cos \theta + \sin \theta)}$$ and \begin{align*}\bar{d}_1 &= \frac{4}{\pi} \int_0^{\sqrt{2}-1} \frac{dt}{(1-t^2) + 2t}\\ &= \frac{4}{\pi} \int_0^{\sqrt{2}-1} \frac{dt}{1+2t-t^2}\\ &= \frac{\sqrt{2}}{\pi} \int_0^{\sqrt{2}-1} \frac{1}{1 + \sqrt{2} - t} + \frac{1}{\sqrt{2} - 1 + t} \,dt \\ &= \frac{\sqrt{2}}{\pi} \left. \ln \left| \frac{\sqrt{2} - 1 + t}{\sqrt{2} + 1 - t} \right| \right|_0^{\sqrt{2}-1}\\ & = \frac{\sqrt{2}}{\pi} \left( \ln \left(\frac{2\sqrt{2}-2}{2} \right) - \ln \left( \frac{\sqrt{2}-1}{\sqrt{2}+1} \right) \right)\\ &= \frac{\sqrt{2}}{\pi} \ln ( 1 + \sqrt{2} ) \approx 0.396757510512\dots.\end{align*} For completeness's sake, below is a table for various other values of $p.$

$p$	$\bar{d}_p$
$1$	$0.396757510512\dots$
$1.5$	$0.466430018867\dots$
$2$	$0.500000000000\dots$
$2.5$	$0.518545527769\dots$
$3$	$0.529817160992\dots$
$4$	$0.542203450927\dots$
$5$	$0.548481398011\dots$
$10$	$0.557672594762\dots$
$100$	$0.561063099606\dots$
$1000$	$0.56109948237\dots$
$\infty$	$0.561099852339\dots$