<s>[INST] Limit of infinite product $\lim_{x\to\infty} \prod_{n\in\mathbb Z} \frac{(x-1)^2+n^2}{(x+1)^2+n^2}$ What is the limit
$$\lim_{x\to\infty} \prod_{n\in\mathbb Z} \frac{(x-1)^2+n^2}{(x+1)^2+n^2} \:?$$
The infinite product is well-defined pointwise since $\frac{(x-1)^2+n^2}{(x+1)^2+n^2}-1 = O(1/n^2)$ for fixed $x$.
The motivation of the above limit comes from a complex integration. To show that the integration over a semicircle vanishes, I need to find the above limit.
 [/INST] We have $$\prod_{n\in \Bbb Z} \frac{(x-1)^2+n^2}{(x+1)^2+n^2} = \frac{(x-1)^2}{(x+1)^2}\left(\prod_{n\ge 1} \frac{(x-1)^2+n^2}{(x+1)^2+n^2}\right)^2$$
and using the known product
$$\prod_{n\ge 1}\frac{z^2+n^2}{n^2} = \frac{\sinh(\pi z)}{\pi z}$$
we have
$$\prod_{n\ge 1} \frac{(x-1)^2+n^2}{(x+1)^2+n^2} =\prod_{n\ge 1} \frac{\frac{(x-1)^2+n^2}{n^2}}{\frac{(x+1)^2+n^2}{n^2}} = \frac{\prod_{n\ge 1}\frac{(x-1)^2+n^2}{n^2}}{\prod_{n\ge 1}\frac{(x+1)^2+n^2}{n^2}} $$
$$=\frac{\frac{\sinh(\pi(x-1))}{\pi(x-1)}}{\frac{\sinh(\pi(x+1))}{\pi(x+1)}} = \frac{(x+1)\sinh(\pi(x-1))}{(x-1)\sinh(\pi(x+1))}$$
so the original product is $$\prod_{n\in \Bbb Z} \frac{(x-1)^2+n^2}{(x+1)^2+n^2} = \frac{\sinh^2(\pi(x-1))}{\sinh^2(\pi(x+1))}$$
Now you should be able to calculate the limit when $x$ tends to infinity.
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