 Shrinking arc integral at a simple pole

At a simple pole, write $f(z)=c/(z-a)+h(z)$ with $h$ holomorphic near $a$ and $c$ its residue. On the arc $z=a+ϵe^{it}$, the singular term integrates exactly to $ic(\beta -\alpha )$. The remainder has absolute integral at most $ϵ(\beta -\alpha )\sup |h|$, which tends to zero. Reversing orientation reverses the sign. In particular a clockwise upper semicircle contributes $-i\pi c$ in the shrinking limit.

A higher-order pole need not have such a limit: $(z-a{)}^{-2}$ has residue zero but integral $2/ϵ$ on a positively oriented upper semicircle. Thus the simple-pole assumption controls all the potentially divergent terms of the principal part of a Laurent series.