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Question Number 158691 by mnjuly1970 last updated on 07/Nov/21

$$$$$$\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^n{\mathcal{H}}_{2n}}{2n}=?$$

Answered by qaz last updated on 08/Nov/21

$$\underset{\mathrm{n}=1}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^{\mathrm{n}}}{2\mathrm{n}}{\mathrm{H}}_{2\mathrm{n}}$$$$=\underset{\mathrm{n}=1}{\sum ^{\mathrm{\infty }}}{(-1)}^{\mathrm{n}-1}{\int }_0^1{\mathrm{x}}^{2\mathrm{n}-1}\ln (1-\mathrm{x})\mathrm{dx}$$$$={\int }_0^1\frac{\ln (1-\mathrm{x})}{1+{\mathrm{x}}^2}\mathrm{dx}$$$$={\int }_0^{\pi /4}\ln (1-\tan \mathrm{x})\mathrm{dx}$$$$={\int }_0^{\pi /4}\ln (\cos \mathrm{x}-\sin \mathrm{x})-\mathrm{lncos}\mathrm{xdx}$$$$={\int }_0^{\pi /4}\ln \left(\sqrt{2}\sin (\frac{\pi }{4}-\mathrm{x})\right)-\mathrm{lncos}\mathrm{xdx}$$$$=\frac{\pi }{8}\ln 2+{\int }_0^{\pi /4}\mathrm{lntan}\mathrm{xdx}$$$$=\frac{\pi }{8}\ln 2-\mathrm{G}$$

Commented by mnjuly1970 last updated on 08/Nov/21

$$verynicesirqaz$$

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