 Solution

Put $\tau =T_r\wedge T_R$. First establish almost sure finiteness. The Itô formula gives the martingale $|B_t{|}^2-2t$ for planar Brownian motion. Stop it at $t\wedge T_R$. The stopped path has norm at most $R$, soBy monotone convergence, $\mathbb{E}{T}_R<\infty$, and hence $\tau <\infty$ almost surely.

The function $h(y)=\log |y|$ is a harmonic function away from $0$: in polar coordinates its Laplacian is $h^{\prime \prime }(\rho )+h^{\prime }(\rho )/\rho =-{\rho }^{-2}+{\rho }^{-2}=0$. The Itô formula shows $h(B_{t\wedge \tau })$ is a local martingale. It is bounded between $\log r$ and $\log R$, so it is a true martingale and passage to $\tau$ is justified by dominated convergence. Writing $p={\mathbb{P}}_x(T_r<T_R)$ and using continuity at exit givesSolving yields the planar Brownian annulus hitting probabilityThe stopping argument stays away from the singularity of the logarithm until the calculation is complete.