Question Number 201388 by MrGHK last updated on 05/Dec/23
Answered by witcher3 last updated on 05/Dec/23
$$=\underset{\mathrm{n}\geqslant\mathrm{1}} {\sum}\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\mathrm{n}}\underset{\mathrm{m}\geqslant\mathrm{0}} {\sum}\int_{\mathrm{0}} ^{\mathrm{1}} \left(−\mathrm{1}\right)^{\mathrm{m}} \mathrm{x}^{\mathrm{n}+\mathrm{2m}} \mathrm{dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\underset{\mathrm{n}\geqslant\mathrm{1}} {\sum}\frac{\left(−\mathrm{x}\right)^{\mathrm{n}} }{\mathrm{n}} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{−\mathrm{ln}\left(\mathrm{1}+\mathrm{x}\right)}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\mathrm{dx} \\ $$$$\mathrm{x}=\mathrm{tg}\left(\mathrm{y}\right)\:\mathrm{done}\:\mathrm{befor} \\ $$
Commented by MrGHK last updated on 05/Dec/23
$$\boldsymbol{\mathrm{yeah}}\:−\frac{\boldsymbol{\pi}}{\mathrm{8}}{ln}\left(\mathrm{2}\right) \\ $$
Answered by mnjuly1970 last updated on 06/Dec/23
$$\:\:\:\:\Omega=\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}}\:\underset{{m}=\mathrm{0}} {\overset{\infty} {\sum}}\int_{\mathrm{0}} ^{\:\mathrm{1}} \left(−\mathrm{1}\right)^{{m}} {x}^{\:{n}+\mathrm{2}{m}} {dx} \\ $$$$\:\:\:\:\:\:=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}}\:\underset{{m}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{m}} {x}^{{n}+\mathrm{2}{m}} {dx} \\ $$$$\:\:\:=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \:\underset{{m}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{m}} {x}^{\mathrm{2}{m}} \underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}}{x}^{\:{n}} {dx} \\ $$$$\:\:\:=\:−\int_{\mathrm{0}} ^{\:\mathrm{1}} \:\frac{{ln}\left(\mathrm{1}+{x}\right)}{\mathrm{1}+{x}\:^{\mathrm{2}} }{dx}=\:−\frac{\pi}{\mathrm{8}}\:{ln}\left(\mathrm{2}\right)\: \\ $$