= Jordan's lemma
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= Jordan lemma
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Let $a>0$ and let $g$ be continuous on the upper semicircle $\gamma_R:Re^{i\theta}$, $0\leq\theta\leq\pi$. If $M_R=\sup_{\gamma_R}|g|\to0$, then
$$
\left|\int_{\gamma_R}e^{iaz}g(z)\,dz\right|\leq\frac\pi a M_R\longrightarrow0.
$$
Indeed, $|e^{iaz}|=e^{-aR\sin\theta}$. Symmetry and $\sin\theta\geq2\theta/\pi$ on $[0,\pi/2]$ bound the arc integral by $2RM_R\int_0^{\pi/2}e^{-2aR\theta/\pi}d\theta\leq\pi M_R/a$.
A rational factor behaving like $1/z$ therefore suffices, even though the crude length-times-supremum bound would only be of constant order. For $a<0$, use a lower semicircle instead. At $a=0$, the conclusion can fail: the integral of $1/z$ over the upper semicircle is $i\pi$ for every radius. The exponential's sign and the decay assumption are both part of the lemma.
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