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Question Number 201388 by MrGHK last updated on 05/Dec/23

Answered by witcher3 last updated on 05/Dec/23

$$=\sum _{\mathrm{n}⩾1}\frac{{(-1)}^{\mathrm{n}}}{\mathrm{n}}\sum _{\mathrm{m}⩾0}{\int }_0^1{(-1)}^{\mathrm{m}}{\mathrm{x}}^{\mathrm{n}+2\mathrm{m}}\mathrm{dx}$$$$={\int }_0^1\frac{1}{1+{\mathrm{x}}^2}\sum _{\mathrm{n}⩾1}\frac{{(-\mathrm{x})}^{\mathrm{n}}}{\mathrm{n}}$$$$={\int }_0^1\frac{-\ln (1+\mathrm{x})}{1+{\mathrm{x}}^2}\mathrm{dx}$$$$\mathrm{x}=\mathrm{tg}\left(\mathrm{y}\right)\mathrm{done}\mathrm{befor}$$

Commented by MrGHK last updated on 05/Dec/23

$$\mathbf{yeah}-\frac{\mathit{\pi }}{8}ln\left(2\right)$$

Answered by mnjuly1970 last updated on 06/Dec/23

$$\mathrm{\Omega }=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^n}{n}\underset{m=0}{\sum ^{\mathrm{\infty }}}{\int }_0^1{(-1)}^m{x}^{n+2m}dx$$$$={\int }_0^1\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^n}{n}\underset{m=0}{\sum ^{\mathrm{\infty }}}{(-1)}^m{x}^{n+2m}dx$$$$={\int }_0^1\underset{m=0}{\sum ^{\mathrm{\infty }}}{(-1)}^m{x}^{2m}\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^n}{n}{x}^{n}dx$$$$=-{\int }_0^1\frac{ln(1+x)}{1+x{}^2}dx=-\frac{\pi }{8}ln\left(2\right)$$

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