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A planar Brownian motion starting at $1$ exits the disk of radius $r>1$ before leaving a wedge of opening $\alpha \in (0,2\pi ]$, centered on the positive axis, with probability $(4/\pi )\arctan (r^{-\pi /\alpha })$. The power-map reduction for Brownian exit from a wedge, the reflection identity for Brownian exit from a half-disc and the Möbius calculation of circular Brownian exit give the formula. Its large-radius decay exponent is $\pi /\alpha$.

Scale the disk to the unit disc, so that the starting point is $a=1/r\in (0,1)$. Use the Möbius transformationwhich maps the disk onto itself, sends $a$ to $0$, and sends $1$ to $1$. By conformal invariance of planar Brownian motion, the exit image is the circular exit of a Brownian motion starting at $0$. Rotational invariance of planar Brownian motion makes that exit uniform in angle.

The endpoints of the right semicircle satisfyIts image is the arc through $1$ between these points, of angular length $2\phi$. Thus the Möbius calculation of circular Brownian exit givesSince $E_{+}$ and $E_{-}$ partition the exit almost surely, part (b) yieldsThe last equality uses $2\arctan (1/r)\in (0,\pi /2)$, so the double-angle tangent identity uses the stated principal branch. As a check, the boundary angle derivative of ${\psi }_a$ is $(1-a^2)/|e^{i\theta }-a{|}^2$; integrating this circular exit density over the right semicircle gives exactly the integral supplied in the question.