This is a cool little result, the proof of which uses the machinery of local time. On p. 72, Prob 1 asks to show that $\int\_0^1 dt/x(t)$ exists, where $x(t)$ is a continuous time brownian motion.


In the very first step, they stated,


$\int\_{|x(t)| > \epsilon, t \leq 1} dt/x(t) = 2 \int\_{\epsilon}^{\infty}[\tau(1,b) - \tau(1, -b)] \frac{db}{b}$ 


where $\tau$ is the local time up to time 1 at the two locations. 


Can someone elucidate why this is true.



Neat little proof using local time from Ito McKean
This is a consequence of the occupation time formula satisfied by local time: If $f:\Bbb R\to\Bbb R$ is bounded and Borel measurable, then $\int\_0^u f(x(t))\,dt =\int\_{\Bbb R}\tau(u,b)f(b)\,db$, almost surely. Use this with $u=1$ and $f(b) =b^{-1}1\_{(-\infty,\epsilon)\cup(\epsilon,+\infty)}(b)$.


