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Question Number 162297 by mathmax by abdo last updated on 28/Dec/21
find ∫_0 ^1 ln(1−x)ln(1+x)dx
$$\mathrm{find}\:\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{ln}\left(\mathrm{1}−\mathrm{x}\right)\mathrm{ln}\left(\mathrm{1}+\mathrm{x}\right)\mathrm{dx} \\ $$
Answered by Lordose last updated on 28/Dec/21
Ω = ∫_0 ^( 1) log(1−x)log(1+x)dx =^(x=1−u) ∫_0 ^( 1) log(u)log(2−u)du Ω = ∫_0 ^( 1) log(u)log(2(1−(u/2)))du = log(2)∫_0 ^( 1) log(u) + ∫_0 ^( 1) log(u)log(1−(u/2))du Ω = −log(2) − Φ Φ = ∫_0 ^( 1) log(u)log(1−(u/2))du =^(u=2x) 2∫_0 ^( (1/2)) log(2x)log(1−x)dx Φ = 2log(2)∫_0 ^( (1/2)) log(1−x)dx + 2∫_0 ^(1/2) log(x)log(1−x)dx Φ = log^2 (2) + log(2) + 2Δ Δ =^(IBP) ((x−1)log(1−x)−x)log(x)∣_0 ^(1/2) − ∫_0 ^(1/2) (((x−1)log(1−x))/x)dx − ∫_0 ^(1/2) 1dx Δ = (1/2)log(2) − (1/2)log^2 (2) − ∫_0 ^(1/2) log(1−x)dx + ∫_0 ^(1/2) ((log(1−x))/x)dx − (1/2) Δ = (1/2)log(2) − (1/2)log^2 (2) − (1/2)log(2) − (1/2) + Li_2 ((1/2)) + (1/2) Δ = (1/2)log^2 (2) − 1 + (𝛑^2 /(12)) − ((log^2 (2))/2) Φ = log^2 (2) + log(2) +2(−1+(𝛑^2 /(12))) = log^2 (2)+log(2) − 2 + (𝛑^2 /6) Ω = −log(2) − log^2 (2) − log(2) + 2 −(𝛑^2 /6) Ω = 2 − (𝛑^2 /6) − 2log(2) − log^2 (2)
$$ \\ $$$$\Omega\:=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \mathrm{log}\left(\mathrm{1}−\mathrm{x}\right)\mathrm{log}\left(\mathrm{1}+\mathrm{x}\right)\mathrm{dx}\:\overset{\mathrm{x}=\mathrm{1}−\mathrm{u}} {=}\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \mathrm{log}\left(\mathrm{u}\right)\mathrm{log}\left(\mathrm{2}−\mathrm{u}\right)\mathrm{du} \\ $$$$\Omega\:=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \mathrm{log}\left(\mathrm{u}\right)\mathrm{log}\left(\mathrm{2}\left(\mathrm{1}−\frac{\mathrm{u}}{\mathrm{2}}\right)\right)\mathrm{du}\:=\:\mathrm{log}\left(\mathrm{2}\right)\int_{\mathrm{0}} ^{\:\mathrm{1}} \mathrm{log}\left(\mathrm{u}\right)\:+\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \mathrm{log}\left(\mathrm{u}\right)\mathrm{log}\left(\mathrm{1}−\frac{\mathrm{u}}{\mathrm{2}}\right)\mathrm{du} \\ $$$$\Omega\:=\:−\mathrm{log}\left(\mathrm{2}\right)\:−\:\Phi \\ $$$$\Phi\:=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \mathrm{log}\left(\mathrm{u}\right)\mathrm{log}\left(\mathrm{1}−\frac{\mathrm{u}}{\mathrm{2}}\right)\mathrm{du}\:\overset{\mathrm{u}=\mathrm{2x}} {=}\mathrm{2}\int_{\mathrm{0}} ^{\:\frac{\mathrm{1}}{\mathrm{2}}} \mathrm{log}\left(\mathrm{2x}\right)\mathrm{log}\left(\mathrm{1}−\mathrm{x}\right)\mathrm{dx} \\ $$$$\Phi\:=\:\mathrm{2log}\left(\mathrm{2}\right)\int_{\mathrm{0}} ^{\:\frac{\mathrm{1}}{\mathrm{2}}} \mathrm{log}\left(\mathrm{1}−\mathrm{x}\right)\mathrm{dx}\:+\:\mathrm{2}\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \mathrm{log}\left(\mathrm{x}\right)\mathrm{log}\left(\mathrm{1}−\mathrm{x}\right)\mathrm{dx} \\ $$$$\Phi\:=\:\mathrm{log}^{\mathrm{2}} \left(\mathrm{2}\right)\:+\:\mathrm{log}\left(\mathrm{2}\right)\:+\:\mathrm{2}\Delta \\ $$$$\Delta\:\overset{\boldsymbol{\mathrm{IBP}}} {=}\left(\left(\mathrm{x}−\mathrm{1}\right)\mathrm{log}\left(\mathrm{1}−\mathrm{x}\right)−\mathrm{x}\right)\mathrm{log}\left(\mathrm{x}\right)\mid_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \:−\:\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \frac{\left(\mathrm{x}−\mathrm{1}\right)\mathrm{log}\left(\mathrm{1}−\mathrm{x}\right)}{\mathrm{x}}\mathrm{dx}\:−\:\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \mathrm{1dx} \\ $$$$\Delta\:=\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{log}\left(\mathrm{2}\right)\:−\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{log}^{\mathrm{2}} \left(\mathrm{2}\right)\:−\:\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \mathrm{log}\left(\mathrm{1}−\mathrm{x}\right)\mathrm{dx}\:+\:\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \frac{\mathrm{log}\left(\mathrm{1}−\mathrm{x}\right)}{\mathrm{x}}\mathrm{dx}\:−\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Delta\:=\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{log}\left(\mathrm{2}\right)\:−\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{log}^{\mathrm{2}} \left(\mathrm{2}\right)\:−\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{log}\left(\mathrm{2}\right)\:−\:\frac{\mathrm{1}}{\mathrm{2}}\:+\:\boldsymbol{\mathrm{Li}}_{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)\:+\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Delta\:=\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{log}^{\mathrm{2}} \left(\mathrm{2}\right)\:−\:\mathrm{1}\:+\:\frac{\boldsymbol{\pi}^{\mathrm{2}} }{\mathrm{12}}\:−\:\frac{\mathrm{log}^{\mathrm{2}} \left(\mathrm{2}\right)}{\mathrm{2}}\: \\ $$$$\Phi\:=\:\mathrm{log}^{\mathrm{2}} \left(\mathrm{2}\right)\:+\:\mathrm{log}\left(\mathrm{2}\right)\:+\mathrm{2}\left(−\mathrm{1}+\frac{\boldsymbol{\pi}^{\mathrm{2}} }{\mathrm{12}}\right)\:=\:\mathrm{log}^{\mathrm{2}} \left(\mathrm{2}\right)+\mathrm{log}\left(\mathrm{2}\right)\:−\:\mathrm{2}\:+\:\frac{\boldsymbol{\pi}^{\mathrm{2}} }{\mathrm{6}} \\ $$$$\Omega\:=\:−\mathrm{log}\left(\mathrm{2}\right)\:−\:\mathrm{log}^{\mathrm{2}} \left(\mathrm{2}\right)\:−\:\mathrm{log}\left(\mathrm{2}\right)\:+\:\mathrm{2}\:−\frac{\boldsymbol{\pi}^{\mathrm{2}} }{\mathrm{6}} \\ $$$$\Omega\:=\:\mathrm{2}\:−\:\frac{\boldsymbol{\pi}^{\mathrm{2}} }{\mathrm{6}}\:−\:\mathrm{2log}\left(\mathrm{2}\right)\:−\:\mathrm{log}^{\mathrm{2}} \left(\mathrm{2}\right) \\ $$

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