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Question Number 142150 by iloveisrael last updated on 27/May/21

$$\underset{n=1}{\prod ^{\mathrm{\infty }}}(1+\frac{1}{n^4})=?$$

Commented by liberty last updated on 27/May/21

$$\underset{n=1}{\prod ^{\mathrm{\infty }}}(1+\frac{1}{n^4})=-\frac{\sin \left(\sqrt[4]{-1}\pi \right)\sin \left(\sqrt[4]{{(-1)}^3}\pi \right)}{{\pi }^2}$$

Answered by Dwaipayan Shikari last updated on 27/May/21

$$\underset{n=1}{\prod ^{\mathrm{\infty }}}(1+\frac{1}{n^4})=\underset{n=1}{\prod ^{\mathrm{\infty }}}(1+{\left(\frac{\sqrt{i}}{n}\right)}^2)(1-{\left(\frac{\sqrt{i}}{n}\right)}^2)$$$$=\frac{sinh\left(\sqrt{i}\pi \right)}{4{\pi }^2}.\frac{sin\left(\sqrt{i}\pi \right)}{i}=-\frac{(e^{\frac{1+i}{\sqrt{2}}\pi }-e^{-\frac{1+i}{\sqrt{2}}\pi })(e^{\frac{i}{\sqrt{2}}-\frac{1}{\sqrt{2}}\pi }-e^{-\frac{i}{\sqrt{2}}+\frac{1}{\sqrt{2}}\pi })}{4{\pi }^2}$$$$=-\frac{e^{\sqrt{2}i\pi }-e^{-\sqrt{2}\pi }-e^{\sqrt{2\pi }}+e^{-\sqrt{2}i\pi }}{4{\pi }^2}=\frac{cosh\left(\sqrt{2}\pi \right)-cos\left(\sqrt{2}\pi \right)}{2{\pi }^2}$$

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