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Question-201329

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Question Number 201329 by MrGHK last updated on 04/Dec/23
Answered by witcher3 last updated on 04/Dec/23
=Σ_(n≥0) (((−1)^n )/(2n+1))Σ_(m≥0) (−1)^m ∫_0 ^1 x^(m+2n+1) dx =∫_0 ^1 Σ_(n≥0) (((−1)^n x^(2n+1) )/(2n+1)).Σ_(n≥0) (−x)^m dx =∫_0 ^1 ((tan^(−1) (x))/(1+x))dx =∫_0 ^1 tan^(−1) (x)ln(x+1)−∫_0 ^1 ((ln(1+x))/(1+x^2 ))dx =(π/4)ln(2)−∫_0 ^(π/4) ln(1+tg(t))dt =((πln(2))/4)−∫_0 ^(π/4) ln((√2)sin((π/4)+t))−ln(cos(t))dt =(π/8)ln(2) Σ_n (((−1)^n )/(2n+1))Σ_(m≥0) (((−1)^m )/(m+2n+2))=((πln(2))/8)
$$=\underset{\mathrm{n}\geqslant\mathrm{0}} {\sum}\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\mathrm{2n}+\mathrm{1}}\underset{\mathrm{m}\geqslant\mathrm{0}} {\sum}\left(−\mathrm{1}\right)^{\mathrm{m}} \int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{x}^{\mathrm{m}+\mathrm{2n}+\mathrm{1}} \mathrm{dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \underset{\mathrm{n}\geqslant\mathrm{0}} {\sum}\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} \mathrm{x}^{\mathrm{2n}+\mathrm{1}} }{\mathrm{2n}+\mathrm{1}}.\underset{\mathrm{n}\geqslant\mathrm{0}} {\sum}\left(−\mathrm{x}\right)^{\mathrm{m}} \mathrm{dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{x}\right)}{\mathrm{1}+\mathrm{x}}\mathrm{dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{tan}^{−\mathrm{1}} \left(\mathrm{x}\right)\mathrm{ln}\left(\mathrm{x}+\mathrm{1}\right)−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}\left(\mathrm{1}+\mathrm{x}\right)}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\mathrm{dx} \\ $$$$=\frac{\pi}{\mathrm{4}}\mathrm{ln}\left(\mathrm{2}\right)−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \mathrm{ln}\left(\mathrm{1}+\mathrm{tg}\left(\mathrm{t}\right)\right)\mathrm{dt} \\ $$$$=\frac{\pi\mathrm{ln}\left(\mathrm{2}\right)}{\mathrm{4}}−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \mathrm{ln}\left(\sqrt{\mathrm{2}}\mathrm{sin}\left(\frac{\pi}{\mathrm{4}}+\mathrm{t}\right)\right)−\mathrm{ln}\left(\mathrm{cos}\left(\mathrm{t}\right)\right)\mathrm{dt} \\ $$$$=\frac{\pi}{\mathrm{8}}\mathrm{ln}\left(\mathrm{2}\right) \\ $$$$\underset{\mathrm{n}} {\sum}\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\mathrm{2n}+\mathrm{1}}\underset{\mathrm{m}\geqslant\mathrm{0}} {\sum}\frac{\left(−\mathrm{1}\right)^{\mathrm{m}} }{\mathrm{m}+\mathrm{2n}+\mathrm{2}}=\frac{\pi\mathrm{ln}\left(\mathrm{2}\right)}{\mathrm{8}} \\ $$
Commented by MrGHK last updated on 05/Dec/23
wonderful
$${wonderful} \\ $$
Commented by witcher3 last updated on 05/Dec/23
thank you
$$\mathrm{thank}\:\mathrm{you} \\ $$

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