Limit of infinite product $\lim_{x\to\infty} \prod_{n\in\mathbb Z} \frac{(x-1)^2+n^2}{(x+1)^2+n^2}$

Question

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.

Answer 1

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.

Answer 2

As an option you can also use the following approach to find the limit. $$\ln\prod(x)=\ln\prod_{n\in \Bbb Z} \frac{(x-1)^2+n^2}{(x+1)^2+n^2}=\ln\prod_{n\in \Bbb Z}\bigg(1- \frac{4x}{(x+1)^2+n^2}\bigg)$$ $$=\sum_{n=-\infty}^\infty\ln\Big(1-\frac{4x}{(x+1)^2+n^2}\Big)$$ Decomposing the logarithm into the series $(\frac{4x}{(x+1)^2+n^2}<1$ at $x\to\infty$) $$\ln\prod(x)=-\sum_{n=-\infty}^\infty\frac{4x}{(x+1)^2+n^2}-\frac{1}{2}\sum_{n=-\infty}^\infty\Big(\frac{4x}{(x+1)^2+n^2}\Big)^2-...$$ Given that $\sum_{n=-\infty}^\infty\frac{1}{n^2+a^2}=\frac{\pi}{a}\coth\pi a$ $$-\sum_{n=-\infty}^\infty\frac{4x}{(x+1)^2+n^2}=-\frac{4\pi x}{x+1}\coth\pi(x+1)\to-4\pi \,\,\text{at}\,\,x\to\infty$$ It is not difficult to show that all other terms of the logarithm decomposition have an additional power of $\frac{1}{x}$ and $\to 0$ at $x\to\infty$.

For example, the second term of the decomposition $$\sum_{n=-\infty}^\infty\Big(\frac{x}{(x+1)^2+n^2}\Big)^2=-\frac{x^2}{2(x+1)}\frac{\partial}{\partial x}\sum_{n=-\infty}^\infty\frac{1}{(x+1)^2+n^2}$$ $$=-\frac{x^2}{2(x+1)}\frac{\partial}{\partial x}\frac{\pi}{x+1}\coth\pi(1+x)\to0\,\,\text{at}\,\,x\to \infty$$ Therefore, $$\lim_{x\to\infty}\ln\prod(x)=-4\pi\,\, \Rightarrow\,\,\lim_{x\to\infty}\prod(x)=e^{-4\pi}$$ what, of course, also follows from the closed form of the product obtained by @jjagmath.