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Question Number 197819 by mnjuly1970 last updated on 30/Sep/23

$$$$$$findthevalueof:$$$$\mathit{\varphi }=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^{n-1}{H}_{2n}}{n}=?$$$$where,H_n=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}$$

Answered by witcher3 last updated on 04/Oct/23

$${\int }_0^1-{\mathrm{x}}^{\mathrm{n}-1}\ln (1-\mathrm{x})\mathrm{dx}=\frac{{\mathrm{H}}_{\mathrm{n}}}{\mathrm{n}}$$$$\mathrm{\Sigma }{(-1)}^{\mathrm{n}-1}\frac{{\mathrm{H}}_{2\mathrm{n}}}{\mathrm{n}}=2\sum _{\mathrm{n}⩾1}{\int }_0^1{(-1)}^{\mathrm{n}}{\mathrm{x}}^{2\mathrm{n}-2}\ln (1-\mathrm{x})$$$$=2{\int }_0^1-\frac{\ln (1-\mathrm{x})}{(1+{\mathrm{x}}^2)}$$$$=-2{\int }_0^{\frac{\pi }{4}}\ln (\cos \left(\mathrm{t}\right)-\sin \left(\mathrm{t}\right))-\ln \left(\cos \left(\mathrm{t}\right)\right)\mathrm{dt}$$$$=-2{\int }_0^{\frac{\pi }{4}}\ln \left(\sqrt{2}\cos (\mathrm{t}+\frac{\pi }{4})\right)-\ln \left(\cos \left(\mathrm{t}\right)\right)\mathrm{dt}$$$$=-2\ln \left(\sqrt{2}\right).\frac{\pi }{4}-2{\int }_0^{\frac{\pi }{4}}\ln \left(\mathrm{tg}\left(\mathrm{t}\right)\right)\mathrm{dt}$$$$=-\frac{\ln \left(2\right)\pi }{4}+2\mathrm{G}$$$$$$

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