 Solution

For $X_i=k/U_i$, the indicator is always one, and the importance weight reduces toAveraging these independent contributions givesAs a direct normalization check, its expectation is ${\int }_0^{1}k/[\pi (u^2+k^2)]du={\pi }^{-1}\arctan (1/k)$. For $k>0$ this equals $1/2-{\pi }^{-1}\arctan k$, the exact standard Cauchy distribution tail probability. The substitution $X=k/U$, rather than $k/(1-U)$ with the same symbol left unchanged, explains why the displayed denominator contains $U^2$.