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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.

The frequency-independent measure factor cancels in the ratio:Upon analytic continuation in frequency, this has poles at ${\omega }_1=0,\pm 2\pi ir/\beta$ and zeros at ${\omega }_2=0,\pm 2\pi ir/\beta$, for $1\le r\le N$, apart from cancellations when numerator and denominator vanish together. They match those of $\sinh (\beta {\omega }_2/2)/\sinh (\beta {\omega }_1/2)$. To justify equality, rather than merely matching this divisor, use the hyperbolic-sine infinite productIt follows from pairing the Weierstrass product for the reciprocal gamma function at opposite imaginary arguments and using the Gamma reflection formula. Normal convergence on compact sets follows from $\sum r^{-2}<\infty$. Thus, for positive real frequencies,The cutoff ratio determines the frequency dependence, agreeing with the thermal partition function of a quantum harmonic oscillator.

The printed assertion about an absolute ${\lim }_N{\mathcal{Z}}_N$ needs a normalization qualification: independence of $A_N$ from $\omega$ alone does not ensure existence of a nonzero limit. Set $B_N=A_N{\prod }_{r=1}^N{\nu }_r^{-2}$. Then ${\mathcal{Z}}_N=B_N{\omega }^{-1}{\prod }_r(1+{\omega }^2/{\nu }_r^2{)}^{-1}$, so if $B_N\to B\ne 0$, its limit is $B\beta /[2\sinh (\beta \omega /2)]$. The choice $B_N=1/\beta$ gives the required limit. Conversely, $B_N=(1+(-1{)}^N/2)/\beta$ is a positive frequency-independent normalization with no absolute limit, though every ratio above still converges. Matching zeros and poles alone also permits nonvanishing entire functions; the convergent normalized product is what rules them out here.

Taking reciprocals in the finite product defining the gamma function givesLet $H_n={\sum }_{k=1}^n{k}^{-1}$. Sinceand $H_n-\log n\to \gamma$, where $\gamma$ is the Euler--Mascheroni constant, passage to the limit gives the Weierstrass product for the reciprocal gamma function

The logarithmic derivative of this identity isSince $k^{-1}-(z+k{)}^{-1}=z/[k(z+k)]$, the digamma function therefore satisfies

For real $z>0$, termwise differentiation gives the trigamma functionThus $\psi$ is strictly increasing on the positive real axis. At the two positive integers needed here, telescoping givesThe intermediate value theorem supplies a zero in $(1,2)$, and strict increase makes it unique. This is the positive zero of the digamma function.