 Matched cutoff estimate for a narrow Lorentzian integral

Split the integral at $ϵ\ll \delta \ll 1$. Replacing the numerator by one below the cutoff gives ${ϵ}^{-1}\arctan (\delta /ϵ)$, with an error bounded by $\frac{1}{2}\log (1+{\delta }^2/{ϵ}^2)$. Above the cutoff, replacing the denominator by $x^2$ has relative error $O({ϵ}^2/{\delta }^2)$, and integration by parts gives $1/\delta +\log \delta +O(1)$. The inner $-1/\delta$ cancels the outer $1/\delta$, and all errors are $o(1/ϵ)$, proving the leading constant independently of the intermediate cutoff.