= Solution
The <strong law for Brownian motion> says $W_s/s\to0$ almost surely. For sufficiently large $s$, therefore $2W_s-s\leq-s/2$, giving
$$
V_\infty:=\int_0^\infty e^{2W_s-s}\,ds<\infty
\quad\text{almost surely}.
$$
Thus $J_t=\int_0^t e^{W_s-s/2}\,dB_s$ has finite terminal <quadratic variation>. Stopping when its bracket reaches each integer makes it an $L^2$-bounded <martingale>, which converges by the <Martingale convergence theorem>. Patching these limits shows that
$$
\boxed{J_\infty=\int_0^\infty e^{W_s-s/2}\,dB_s}
$$
is a well-defined finite almost sure limit. This construction uses local square integrability; a global $L^2$ bound is not required.
Conditional on the entire $W$ path, the independent <Brownian motion> $B$ still supplies a centered Gaussian <stochastic integral>, with variance $V_\infty\in(0,\infty)$. The <conditionally Gaussian stochastic integral with an independent integrator> therefore has no atoms, even after averaging over $W$.
Also $e^{-W_t+t/2}\to\infty$ almost surely. For each fixed $x$, the relation $X_t=e^{-W_t+t/2}(x-J_t)$ then gives, outside the null event $J_\infty=x$,
$$
X_t\to+\infty\iff J_\infty<x,\qquad
X_t\to-\infty\iff J_\infty>x.
$$
Part (b) identifies the cumulative distribution:
$$
\boxed{\mathbb P(J_\infty\leq x)=\frac12+\frac{\arctan x}{\pi},
\qquad f_{J_\infty}(x)=\frac1{\pi(1+x^2)}.}
$$
Hence $J_\infty$ has the <Standard Cauchy distribution>. This establishes the <Cauchy law of an infinite-horizon Brownian exponential integral>.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2015/iii/paper-30-cauchy-law.png]
{title=Standard Cauchy density and cumulative distribution, with the latter equal to the positive-divergence probability}
{height=360}
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