Top-edge exit probability of a Brownian half-strip ($\frac{1}{2}-\frac{1}{\pi }\arctan \frac{1+\cosh (\pi p/a)\cos (\pi q/a)}{\sinh (\pi p/a)\sin (\pi q/a)}$)

For planar Brownian motion started at $p+iq$, $p>0$ and $0<q<a$, in $\{x+iy:x>0,0<y<a\}$, the displayed value is its probability of exiting through $y=a$. The half-strip conformal map by hyperbolic cosine sends the top edge to $(-\infty ,-1)$ and the starting point to $A+iB$, where $A=\cosh (\pi p/a)\cos (\pi q/a)$ and $B=\sinh (\pi p/a)\sin (\pi q/a)$. Integrating the Cauchy exit law from a Brownian half-plane over that ray proves the formula. As $p\to \infty$ it tends to $q/a$.