= Shrinking arc integral at a simple pole
{title2=$\int_{a+\epsilon e^{it},\,\alpha\leq t\leq\beta}f(z)\,dz\to i(\beta-\alpha)\operatorname{res}(f;a)$}
At a <simple pole>, write $f(z)=c/(z-a)+h(z)$ with $h$ holomorphic near $a$ and $c$ its <residue>. On the arc $z=a+\epsilon e^{it}$, the singular term integrates exactly to $ic(\beta-\alpha)$. The remainder has absolute integral at most $\epsilon(\beta-\alpha)\sup|h|$, 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 $2/\epsilon$ 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>.
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