= Hyperbolic-sine infinite product
{title2=$\sinh(\pi z)/(\pi z)=\prod_{r\ge1}(1+z^2/r^2)$}
This <infinite product> converges uniformly on compact subsets of the complex plane. Pair the factors at $iz$ and $-iz$ in the <Weierstrass product for the reciprocal gamma function>, and use the <Gamma reflection formula> together with $\Gamma(1+w)=w\Gamma(w)$. This gives $[\Gamma(1+iz)\Gamma(1-iz)]^{-1}=\sinh(\pi z)/(\pi z)$ and the displayed product. The removable value at $z=0$ is one. Matching zeros alone would not exclude multiplication by a nonvanishing <entire function>; the normalized product fixes that ambiguity.
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