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Question Number 134662 by mnjuly1970 last updated on 06/Mar/21

$$....nicecalculus...$$$$provethat::$$$$\mathit{\varphi }=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{\zeta \left(2n\right)}{(n+1)(2n+1)}=\frac{1}{2}$$$$...m.n...$$

Answered by Dwaipayan Shikari last updated on 06/Mar/21

$$\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{2\zeta \left(2n\right)}{2n+1}-\frac{\zeta \left(2n\right)}{n+1}$$$$={\int }_0^{1}2\underset{n=1}{\sum ^{\mathrm{\infty }}}\underset{k=1}{\sum ^{\mathrm{\infty }}}\frac{x^{2n}}{k^{2n}}-\underset{n=1}{\sum ^{\mathrm{\infty }}}\underset{k=1}{\sum ^{\mathrm{\infty }}}\frac{x^n}{k^{2n}}$$$$={\int }_0^1\underset{k=1}{\sum ^{\mathrm{\infty }}}\frac{2x^2}{k^2-x^2}-\frac{x}{k^2-x}dx$$$$={\int }_0^{1}x\underset{k=1}{\sum ^{\mathrm{\infty }}}\frac{1}{k-x}-\frac{1}{k+x}-\frac{1}{2}\sqrt{x}\underset{k=1}{\sum ^{\mathrm{\infty }}}\frac{1}{k-\sqrt{x}}-\frac{1}{k+\sqrt{x}}dx$$$$={\int }_0^{1}x\psi (1+x)-x\psi (1-x)-\frac{1}{2}\sqrt{x}\psi (1+\sqrt{x})+\sqrt{x}\psi (1-\sqrt{x})dx$$$$=1-{\int }_0^{1}x\pi cot\left(\pi x\right)dx-\frac{1}{2}+\frac{1}{2}{\int }_0^1\sqrt{x}\pi cot\left(\pi \sqrt{x}\right)dx$$$$=\frac{1}{2}-{\int }_0^{1}x\pi cot\left(\pi x\right)dx+{\int }_0^{1}t\pi cot\left(\pi t\right)dt=\frac{1}{2}$$

Commented by mnjuly1970 last updated on 06/Mar/21

$$tayebballah...sirpayan...$$$$yoursolutionisverynice$$$$andbetterthanthatme..$$$$mercey...$$

Commented by mnjuly1970 last updated on 06/Mar/21

$$\frac{1}{2}{\int }_0^1\sqrt{x}\pi cot\left(\pi \left(\sqrt{x}\right)\right)dx\stackrel{WHY??}{=}{\int }_0^{1}t\pi cot\left(\pi t\right)dt$$$$pleaseexplain....$$

Answered by mnjuly1970 last updated on 07/Mar/21

Answered by mnjuly1970 last updated on 07/Mar/21

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