= Solution
{c}
The printed upper-tail identity cannot be a probability formula: it vanishes at zero and increases in $x$. Its correction is
$$
\boxed{\mathbb P_i(X_{T_{\mathbb H}}>x)=\tfrac12-\frac1\pi\arctan x.}
$$
Equivalently the cumulative distribution function is $1/2+\pi^{-1}\arctan x$. Differentiating that cumulative function gives the density
$$
\frac1{\pi(1+x^2)}.
$$
The 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 $z_0=x_0+iy_0$, $y_0>0$, translation and positive scaling send it to $i$. Thus the exit coordinate has law $x_0+y_0C$, where $C$ is standard Cauchy. The corresponding formulas are
$$
\boxed{\mathbb P_{z_0}(X_{T_{\mathbb H}}\in ds)
=\frac{y_0\,ds}{\pi((s-x_0)^2+y_0^2)},\quad
\mathbb P_{z_0}(X_{T_{\mathbb H}}>a)
=\frac12-\frac1\pi\arctan\frac{a-x_0}{y_0}.}
$$
This 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.
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