= Top-edge exit probability of a Brownian half-strip
{c}
{title2=$\frac12-\frac1\pi\arctan\frac{1+\cosh(\pi p/a)\cos(\pi q/a)}{\sinh(\pi p/a)\sin(\pi q/a)}$}
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 $q/a$.
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