Jordan's lemma

Let $a>0$ and let $g$ be continuous on the upper semicircle ${\gamma }_R:R{e}^{i\theta }$, $0\le \theta \le \pi$. If $M_R={\sup }_{{\gamma }_R}|g|\to 0$, thenIndeed, $|e^{iaz}|=e^{-aR\sin \theta }$. Symmetry and $\sin \theta \ge 2\theta /\pi$ on $[0,\pi /2]$ bound the arc integral by $2R{M}_R{\int }_0^{\pi /2}{e}^{-2aR\theta /\pi }d\theta \le \pi M_R/a$.

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.

For real $a\ne 0$ and $|x|<\pi$, integrate $z\sin (xz)/((a^2+z^2)\sin \pi z)$ around a thin positively oriented strip containing the integers. Its integer residues give the series. Closing the two strip sides outward instead gives minus the sum of the residues at $z=\pm ia$, each $\sinh (ax)/(2\sinh (a\pi ))$. The factor $|x|<\pi$ ensures exponential decay on those closures. The bilateral series is interpreted by symmetric partial sums; at $a=0$ its continuous extension is $-x$ with the zero-index summand defined to be zero.