Question Number 134442 by mnjuly1970 last updated on 03/Mar/21
$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{nice}\:…..\:{calculus}… \\ $$$$\:\:\:\:{prove}\:{that}\::: \\ $$$$\:\:\:\:\:\:\:\:\:\boldsymbol{\phi}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\prod}}\left(\mathrm{1}+\frac{\mathrm{1}}{{n}^{\mathrm{4}} }\right)=\frac{{cosh}\left(\mathrm{2}\pi\right)−{cos}\left(\mathrm{2}\pi\right)}{\mathrm{4}\pi^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:………………….. \\ $$
Answered by Dwaipayan Shikari last updated on 03/Mar/21
$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\prod}}\left(\mathrm{1}+\frac{\mathrm{1}}{{n}^{\mathrm{4}} }\right)=\underset{{n}=\mathrm{1}} {\overset{\infty} {\prod}}\left(\mathrm{1}+\frac{{i}}{{n}^{\mathrm{2}} }\right)\left(\mathrm{1}−\frac{{i}}{{n}^{\mathrm{2}} }\right)=\frac{{sinh}\left(\pi\sqrt{{i}}\right){sin}\left(\pi\sqrt{{i}}\right)}{\pi^{\mathrm{2}} {i}} \\ $$$$=\frac{{sinh}\left(\frac{\pi}{\:\sqrt{\mathrm{2}}}\left(\mathrm{1}+{i}\right)\right){sin}\left(\frac{\pi}{\:\sqrt{\mathrm{2}}}\left(\mathrm{1}+{i}\right)\right)}{\pi^{\mathrm{2}} {i}}=−\frac{{e}^{\frac{\pi}{\:\sqrt{\mathrm{2}}}+\frac{\pi}{\:\sqrt{\mathrm{2}}}{i}} −{e}^{−\frac{\pi}{\:\sqrt{\mathrm{2}}}−\frac{\pi}{\:\sqrt{\mathrm{2}}}{i}} }{\mathrm{4}\pi^{\mathrm{2}} }.\left({e}^{\frac{\pi}{\:\sqrt{\mathrm{2}}}{i}−\frac{\pi}{\:\sqrt{\mathrm{2}}}} −{e}^{−\frac{\pi}{\:\sqrt{\mathrm{2}}}{i}+\frac{\pi}{\:\sqrt{\mathrm{2}}}} \right) \\ $$$$=−\frac{{e}^{\sqrt{\mathrm{2}}\pi{i}} −{e}^{\sqrt{\mathrm{2}}\pi} −{e}^{−\sqrt{\mathrm{2}}\pi} +{e}^{−\sqrt{\mathrm{2}}\pi{i}} }{\mathrm{4}\pi^{\mathrm{2}} }=\frac{{e}^{\sqrt{\mathrm{2}}\pi} +{e}^{\sqrt{\mathrm{2}}\pi} −{e}^{\sqrt{\mathrm{2}}\pi{i}} −{e}^{−\sqrt{\mathrm{2}}\pi{i}} }{\mathrm{4}\pi^{\mathrm{2}} } \\ $$$$=\frac{{cosh}\left(\sqrt{\mathrm{2}}\pi\right)−{cos}\left(\sqrt{\mathrm{2}}\pi\right)}{\mathrm{2}\pi^{\mathrm{2}} } \\ $$
Commented by mnjuly1970 last updated on 03/Mar/21
$${very}\:{nice}\:…{thanks}\:{alot}\:{sir}.. \\ $$