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let $x,y$ be real numbers,show that $$\prod_{k=-\infty}^{+\infty}\dfrac{x^2+(4k+1-y)^2}{x^2+(4k+3-y)^2}=\dfrac{1+e^{\pi x}-2e^{\frac{\pi}{2}x}\sin{\dfrac{\pi}{2}y}}{1+e^{\pi x}+2e^{\frac{\pi}{2}x}\sin{\dfrac{\pi}{2}y}}$$

This is a question that a physicist asked me. He said that he could prove the equation in a physical way, but not in a mathematical way. I thought about it for a while and might need to use the following conclusion to solve it, but it didn't work out $$\prod_{k=1}^{+\infty}\left(1+\dfrac{1}{k^2}\right)=\dfrac{\sinh{\pi/2}}{\pi/2}$$？

math110
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Firstly, define the complex number $z=x+yi$. Then, we are asked to prove that $$ \prod_{k\in\mathbb{Z}}\left|\frac{z-(4k+1)i}{z-(4k+3)i}\right|^2=\frac{1+e^{\pi x}-2e^{\frac{\pi}{2}x}\sin{\dfrac{\pi}{2}y}}{1+e^{\pi x}+2e^{\frac{\pi}{2}x}\sin{\dfrac{\pi}{2}y}}. $$ Now, let's rewrite the right-hand side of this equality: $$ \frac{1+e^{\pi x}-2e^{\frac{\pi}{2}x}\sin{\dfrac{\pi}{2}y}}{1+e^{\pi x}+2e^{\frac{\pi}{2}x}\sin{\dfrac{\pi}{2}y}} = \frac{\left(e^{\frac{\pi}{2}x}-\sin\dfrac{\pi}{2}y\right)^2+\cos^2\dfrac{\pi}{2}y}{\left(e^{\frac{\pi}{2}x}+\sin\dfrac{\pi}{2}y\right)^2+\cos^2\dfrac{\pi}{2}y} = \left|\frac{e^{\frac{\pi}{2}x}+i\left(\cos\dfrac{\pi}{2}y+i\sin\dfrac{\pi}{2}y\right)}{e^{\frac{\pi}{2}x}-i\left(\cos\dfrac{\pi}{2}y+i\sin\dfrac{\pi}{2}y\right)}\right|^2= \\ =\left|\frac{e^{\frac{\pi}{2}x}+ie^{i\frac{\pi}{2}y}}{e^{\frac{\pi}{2}x}-ie^{i\frac{\pi}{2}y}}\right|^2=\left|\frac{e^{\frac{\pi}{2}(x-iy)}+i}{e^{\frac{\pi}{2}(x-iy)}-i}\right|^2 =\left|\frac{e^{\frac{\pi}{2}\overline{z}}+i}{e^{\frac{\pi}{2}\overline{z}}-i}\right|^2=\left|\frac{e^{\frac{\pi}{2}z}-i}{e^{\frac{\pi}{2}z}+i}\right|^2. $$ Thus, we need to prove that $$ \prod_{k\in\mathbb{Z}}\left|\frac{z-(4k+1)i}{z-(4k+3)i}\right|=\left|\frac{e^{\frac{\pi}{2}z}-i}{e^{\frac{\pi}{2}z}+i}\right|. $$ Let $w=z-i$, then the last equality can be rewritten as $$ \prod_{k\in\mathbb{Z}}\left|\frac{w-4ki}{w-(4k+2)i}\right|=\left|\frac{e^{\frac{\pi}{2}w}-1}{e^{\frac{\pi}{2}w}+1}\right|. $$

Update. As David mentioned in the comments, in order to compute the product $\prod_{k\in\mathbb{Z}}\frac{t-4k}{t-(4k+2)}$ we can use the following formula: $$ \frac{\sin\pi z}{\pi z}=\prod_{k=1}^{\infty}\left(1-\frac{z^2}{k^2}\right), $$ or $$ \sin\pi z=\pi z\prod_{k\in\mathbb{Z}\backslash\{0\}}\frac{k-z}{k}. $$ For $z=t/4$ and $z=(t-2)/4$ we have $$ \sin\frac{\pi t}{4}=\frac{\pi t}{4}\prod_{k\in\mathbb{Z}\backslash\{0\}}\frac{4k-t}{4k} $$ and $$ \sin\frac{\pi (t-2)}{4}=\frac{\pi (t-2)}{4}\prod_{k\in\mathbb{Z}\backslash\{0\}}\frac{4k+2-t}{4k}, $$ respectively. Hence, $$ \frac{\sin\pi t/4}{\sin\pi (t-2)/4}=\prod_{k\in\mathbb{Z}}\frac{t-4k}{t-(4k+2)}. $$ Therefore, $$ \prod_{k\in\mathbb{Z}}\left|\frac{w-4ki}{w-(4k+2)i}\right|=[t=w/i]=\prod_{k\in\mathbb{Z}}\left|\frac{t-4k}{t-(4k+2)}\right|=\left|\frac{\sin\pi t/4}{\sin\pi (t-2)/4}\right|= \\ =\left|\frac{e^{i\pi t/4}-e^{-i\pi t/4}}{e^{i\pi (t-2)/4}-e^{-i\pi (t-2)/4}}\right|=\left|\frac{e^{i\pi t/4}-e^{-i\pi t/4}}{-ie^{i\pi t/4}-ie^{-i\pi t/4}}\right|=\left|\frac{e^{\pi w/2}-1}{e^{\pi w/2}+1}\right|, $$ so we're done.

richrow
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    You should be able to get your final product by rearranging $\tfrac{\sin (\pi z)}{\pi z} = \prod (1-(z/k)^2)$ – David E Speyer Aug 24 '20 at 13:15
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    Note that $$ \left| {\frac{{e^{\frac{\pi }{2}x} + ie^{i\frac{\pi }{2}y} }}{{e^{\frac{\pi }{2}x} - ie^{i\frac{\pi }{2}y} }}} \right| = \left| {\frac{{e^{\frac{\pi }{2}(x - iy)} + i}}{{e^{\frac{\pi }{2}(x - iy)} - i}}} \right| $$ and not $$ \left| {\frac{{e^{\frac{\pi }{2}(x - iy)} + i}}{{e^{\frac{\pi }{2}(x + iy)} - i}}} \right|. $$ This latter expression is just 1 because is of the form $|\bar t|/|t|$. – Gary Aug 24 '20 at 16:12
  • Indeed. I will edit my answer. – richrow Aug 24 '20 at 16:21
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    Note that you can then take conjugates under the absolute value to have $x+iy=z$ in place of $x-iy$. You should be able to finish the proof afterwards. – Gary Aug 24 '20 at 16:22