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Question Number 201106 by mnjuly1970 last updated on 29/Nov/23

$$$$$$calculate...$$$$$$$$\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{\zeta \left(2n\right)}{2^n.n}=?$$$$$$

Answered by witcher3 last updated on 30/Nov/23

$$\zeta \left(2\mathrm{n}\right)\mathrm{\Gamma }\left(2\mathrm{n}\right)={\int }_0^{\mathrm{\infty }}\frac{{\mathrm{x}}^{2\mathrm{n}-1}}{{\mathrm{e}}^{\mathrm{x}}-1}\mathrm{dx}$$$$\mathrm{S}=\sum _{\mathrm{n}⩾1}\frac{\zeta \left(2\mathrm{n}\right)}{2^{\mathrm{n}}\mathrm{n}}=\sum _{\mathrm{n}⩾1}\frac{1}{2^{\mathrm{n}}\mathrm{\Gamma }\left(2\mathrm{n}\right).\mathrm{n}}{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{x}}^{2\mathrm{n}-1}}{{\mathrm{e}}^{\mathrm{x}}-1}\mathrm{dx}$$$$={\int }_0^{\mathrm{\infty }}\frac{1}{\mathrm{x}}\sum _{\mathrm{n}⩾1}\left[\frac{{\mathrm{x}}^{2\mathrm{n}}}{\left(2\mathrm{n}\right)!.2^{\mathrm{n}}}\right].\frac{\mathrm{dx}}{{\mathrm{e}}^{\mathrm{x}}-1}$$$$={\int }_0^{\mathrm{\infty }}\frac{\mathrm{ch}\left(\frac{\mathrm{x}}{\sqrt{2}}\right)-1}{\mathrm{x}({\mathrm{e}}^{\mathrm{x}}-1)}\mathrm{dx}=\mathrm{S}$$$$\mathrm{S}\left(\mathrm{a}\right)={\int }_0^{\mathrm{\infty }}\frac{\mathrm{ch}\left(\mathrm{ax}\right)-1}{\mathrm{x}({\mathrm{e}}^{\mathrm{x}}-1)}\mathrm{dx};\mathrm{S}\left(0\right)=0$$$${\mathrm{s}}^{\prime }\left(\mathrm{a}\right)={\int }_0^{\mathrm{\infty }}\frac{\mathrm{sh}\left(\mathrm{ax}\right)}{{\mathrm{e}}^{\mathrm{x}}-1}\mathrm{dx}$$$$={\int }_0^{\mathrm{\infty }}\frac{{\mathrm{e}}^{\mathrm{ax}}-{\mathrm{e}}^{-\mathrm{ax}}}{{\mathrm{e}}^{\mathrm{x}}-1}\mathrm{dx},{\mathrm{e}}^{-\mathrm{x}}=\mathrm{t}$$$$={\int }_0^1\frac{{\mathrm{t}}^{-\mathrm{a}}-{\mathrm{t}}^{\mathrm{a}}}{1-\mathrm{t}}\mathrm{dt}=\mathrm{\Psi }(\mathrm{a}+1)-\mathrm{\Psi }(1-\mathrm{a})$$$$=\frac{1}{\mathrm{a}}-\pi \cot \left(\pi \mathrm{a}\right)$$$$\mathrm{s}\left(\mathrm{a}\right)=\ln \left(\mathrm{a}\right)-\ln (\sin \left(\pi \mathrm{a}\right)+\mathrm{c}$$$$=\ln \left(\frac{\mathrm{a}}{\sin \left(\pi \mathrm{a}\right)}\right)+\mathrm{c}$$$$\underset{\mathrm{a}\to 0}{\lim s}\left(\mathrm{a}\right)=\mathrm{s}\left(0\right)=0=\ln \left(\frac{1}{\pi }\right)+\mathrm{c}\Rightarrow \mathrm{c}=\ln \left(\pi \right)$$$$\mathrm{S}\left(\mathrm{a}\right)=\ln \left(\frac{\pi \mathrm{a}}{\sin \left(\pi \mathrm{a}\right)}\right)$$$$\mathrm{S}=\mathrm{s}\left(\frac{1}{\sqrt{2}}\right)=\ln \left(\frac{\pi }{\sqrt{2}.\sin \left(\frac{\pi }{\sqrt{2}}\right)}\right)$$$$$$

Commented by mnjuly1970 last updated on 01/Dec/23

$$thanksalotmaster\cancel{⋚}$$

Commented by witcher3 last updated on 01/Dec/23

$$\mathrm{withe}\mathrm{pleasur}\mathrm{Barak}\mathrm{alah}\mathrm{fik}$$

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