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Question Number 134442 by mnjuly1970 last updated on 03/Mar/21

$$nice.....calculus...$$$$provethat::$$$$\mathit{\varphi }=\underset{n=1}{\prod ^{\mathrm{\infty }}}(1+\frac{1}{n^4})=\frac{cosh\left(2\pi \right)-cos\left(2\pi \right)}{4{\pi }^2}$$$$.......................$$

Answered by Dwaipayan Shikari last updated on 03/Mar/21

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

Commented by mnjuly1970 last updated on 03/Mar/21

$$verynice...thanksalotsir..$$

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