 Solution

The printed upper-tail identity cannot be a probability formula: it vanishes at zero and increases in $x$. Its correction isEquivalently the cumulative distribution function is $1/2+{\pi }^{-1}\arctan x$. Differentiating that cumulative function gives the densityThe normalization and sign can also be checked directly by conformal invariance of planar Brownian motion: the Cayley transform $(z-i)/(z+i)$ sends the starting point to the center of the unit disc. Its exit angle is uniform by rotational symmetry. The inverse on the unit circle gives $X=-\cot (\Theta /2)$, $\Theta$ uniform on $(0,2\pi )$, whose change-of-variable density is exactly the one above.

For a start $z_0=x_0+i{y}_0$, $y_0>0$, translation and positive scaling send it to $i$. Thus the exit coordinate has law $x_0+y_{0}C$, where $C$ is standard Cauchy. The corresponding formulas areThis is the upper-half-plane Poisson kernel. The printed density is correct; it is the preceding asserted tail from which it is meant to follow that needs correction.