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Question Number 148780 by Jonathanwaweh last updated on 31/Jul/21

Answered by Kamel last updated on 31/Jul/21

$$$$$$\mathrm{\Omega }=\underset{n=2}{\prod ^{+\mathrm{\infty }}}e{(1-\frac{1}{n^2})}^{n^2}=e^{\underset{n=2}{\sum ^{+\mathrm{\infty }}}(n^{2}Ln(1-\frac{1}{n^2})+1)}$$$$=e^{\underset{n=2}{\sum ^{+\mathrm{\infty }}}{n}^2(-\underset{p=1}{\sum ^{+\mathrm{\infty }}}\frac{1}{p}\frac{1}{n^{2p}}+\frac{1}{n^2})}=e^{-\underset{p=1}{\sum ^{+\mathrm{\infty }}}\frac{1}{p+1}\underset{n=2}{\sum ^{+\mathrm{\infty }}}\frac{1}{n^{2p}}}=e^{-\underset{p=1}{\sum ^{+\mathrm{\infty }}}\frac{1}{p+1}(\zeta \left(2p\right)-1)}$$$$Wehave:\underset{k=0}{\sum ^{+\mathrm{\infty }}}\zeta \left(2k\right)x^{2k}=-\frac{\pi x}{2}cot\left(\pi x\right)$$$$So\underset{k=0}{\sum ^{+\mathrm{\infty }}}(\zeta \left(2k\right)-1)x^{2k+1}=-\frac{\pi x^2}{2}cot\left(\pi x\right)-\frac{x}{1-x^2}$$$$\therefore \underset{p=0}{\sum ^{+\mathrm{\infty }}}\frac{\zeta \left(2p\right)-1}{p+1}=-{\int }_0^1(\pi x^{2}cot\left(\pi x\right)+\frac{2x}{1-x^2})dx$$$$\int (x^{2}cot\left(\pi x\right)+\frac{2x}{1-x^2})dx=\frac{{ixLi}_2\left(e^{-2\pi ix}\right)}{{\pi }^2}+\frac{{Li}_3\left(e^{-2\pi ix}\right)}{2{\pi }^3}-\frac{{ix}^3}{\pi }+\frac{x^{2}Ln(1-e^{2\pi ix})-\pi Ln(1-x^2)}{\pi }$$$$\therefore {\int }_0^1(\pi x^{2}cot\left(\pi x\right)+\frac{2x}{1-x^2})dx=Ln\left(\pi \right)$$$$So:\underset{p=1}{\sum ^{+\mathrm{\infty }}}\frac{\zeta \left(2p\right)-1}{p+1}=\frac{3}{2}-Ln\left(\pi \right)$$$$Then:\underset{\mathit{n}=2}{\mathbf{\prod }^{+\mathrm{\infty }}e}{(1-\frac{1}{{\mathit{n}}^2})}^{{\mathit{n}}^2}={\mathit{e}}^{\mathit{L}\mathit{n}\left(\mathit{\pi }\right)-\frac{3}{2}}=\frac{\mathit{\pi }}{\mathit{e}\sqrt{\mathit{e}}}$$$$\mathit{K}\mathit{A}\mathit{M}\mathit{E}\mathit{L}\mathit{B}\mathit{E}\mathit{N}\mathit{A}\mathit{I}\mathit{C}\mathit{H}\mathit{A}$$

Commented by Tawa11 last updated on 02/Aug/21

$$\mathrm{Weldone}\mathrm{sir}$$

Answered by Kamel last updated on 31/Jul/21

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