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Question Number 201329 by MrGHK last updated on 04/Dec/23

Answered by witcher3 last updated on 04/Dec/23

$$=\sum _{\mathrm{n}⩾0}\frac{{(-1)}^{\mathrm{n}}}{2\mathrm{n}+1}\sum _{\mathrm{m}⩾0}{(-1)}^{\mathrm{m}}{\int }_0^1{\mathrm{x}}^{\mathrm{m}+2\mathrm{n}+1}\mathrm{dx}$$$$={\int }_0^1\sum _{\mathrm{n}⩾0}\frac{{(-1)}^{\mathrm{n}}{\mathrm{x}}^{2\mathrm{n}+1}}{2\mathrm{n}+1}.\sum _{\mathrm{n}⩾0}{(-\mathrm{x})}^{\mathrm{m}}\mathrm{dx}$$$$={\int }_0^1\frac{{\tan }^{-1}\left(\mathrm{x}\right)}{1+\mathrm{x}}\mathrm{dx}$$$$={\int }_0^1{\tan }^{-1}\left(\mathrm{x}\right)\ln (\mathrm{x}+1)-{\int }_0^1\frac{\ln (1+\mathrm{x})}{1+{\mathrm{x}}^2}\mathrm{dx}$$$$=\frac{\pi }{4}\ln \left(2\right)-{\int }_0^{\frac{\pi }{4}}\ln (1+\mathrm{tg}\left(\mathrm{t}\right))\mathrm{dt}$$$$=\frac{\pi \ln \left(2\right)}{4}-{\int }_0^{\frac{\pi }{4}}\ln \left(\sqrt{2}\sin (\frac{\pi }{4}+\mathrm{t})\right)-\ln \left(\cos \left(\mathrm{t}\right)\right)\mathrm{dt}$$$$=\frac{\pi }{8}\ln \left(2\right)$$$$\sum _{\mathrm{n}}\frac{{(-1)}^{\mathrm{n}}}{2\mathrm{n}+1}\sum _{\mathrm{m}⩾0}\frac{{(-1)}^{\mathrm{m}}}{\mathrm{m}+2\mathrm{n}+2}=\frac{\pi \ln \left(2\right)}{8}$$

Commented by MrGHK last updated on 05/Dec/23

$$wonderful$$

Commented by witcher3 last updated on 05/Dec/23

$$\mathrm{thank}\mathrm{you}$$

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