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Question Number 157469 by mnjuly1970 last updated on 23/Oct/21

calculate : Ω:= ∫_(0 ) ^( 1) (( ln(1+x).ln(1−x))/(1+x)) dx =?

$$ \\ $$$$\:\:\:\:{calculate}\:: \\ $$$$\:\Omega:=\:\int_{\mathrm{0}\:} ^{\:\mathrm{1}} \frac{\:{ln}\left(\mathrm{1}+{x}\right).{ln}\left(\mathrm{1}−{x}\right)}{\mathrm{1}+{x}}\:{dx}\:=? \\ $$$$\:\:\:\: \\ $$

Answered by qaz last updated on 23/Oct/21

∫_0 ^1 ((ln(1+x)ln(1−x))/(1+x))dx =∫_1 ^2 ((lnxln(2−x))/x)dx =∫_(1/2) ^1 (((ln2+lnx)(ln2+ln(1−x)))/x)dx =∫_(1/2) ^1 ((ln^2 2+ln2(lnx+ln(1−x))+lnxln(1−x))/x)dx =ln^3 2+ln2∙(−(1/2)ln^2 2+∫_(1/2) ^1 ((ln(1−x))/x)dx)+∫_(1/2) ^1 ((lnxln(1−x))/x)dx =(1/2)ln^3 2+ln2(Li_2 ((1/2))−Li_2 (1))−lnx∙Li_2 (x)∣_(1/2) ^1 +Li_3 (1)−Li_3 ((1/2)) =(1/2)ln^3 2−(π^2 /6)ln2−ln2∙Li_2 ((1/2))−ln2∙Li_2 ((1/2))+ζ(3)−Li_3 ((1/2)) =−(π^2 /4)ln2+(4/3)ln^3 2+(1/8)ζ(3) −−−−−−−−−−−−−−−−− Li_2 ((1/2))=−(1/2)ln^2 2+(π^2 /(12)) Li_3 ((1/2))=−(π^2 /(12))ln2+(1/6)ln^3 2+(7/8)ζ(3)

$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}\left(\mathrm{1}+\mathrm{x}\right)\mathrm{ln}\left(\mathrm{1}−\mathrm{x}\right)}{\mathrm{1}+\mathrm{x}}\mathrm{dx} \\ $$$$=\int_{\mathrm{1}} ^{\mathrm{2}} \frac{\mathrm{lnxln}\left(\mathrm{2}−\mathrm{x}\right)}{\mathrm{x}}\mathrm{dx} \\ $$$$=\int_{\mathrm{1}/\mathrm{2}} ^{\mathrm{1}} \frac{\left(\mathrm{ln2}+\mathrm{lnx}\right)\left(\mathrm{ln2}+\mathrm{ln}\left(\mathrm{1}−\mathrm{x}\right)\right)}{\mathrm{x}}\mathrm{dx} \\ $$$$=\int_{\mathrm{1}/\mathrm{2}} ^{\mathrm{1}} \frac{\mathrm{ln}^{\mathrm{2}} \mathrm{2}+\mathrm{ln2}\left(\mathrm{lnx}+\mathrm{ln}\left(\mathrm{1}−\mathrm{x}\right)\right)+\mathrm{lnxln}\left(\mathrm{1}−\mathrm{x}\right)}{\mathrm{x}}\mathrm{dx} \\ $$$$=\mathrm{ln}^{\mathrm{3}} \mathrm{2}+\mathrm{ln2}\centerdot\left(−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}^{\mathrm{2}} \mathrm{2}+\int_{\mathrm{1}/\mathrm{2}} ^{\mathrm{1}} \frac{\mathrm{ln}\left(\mathrm{1}−\mathrm{x}\right)}{\mathrm{x}}\mathrm{dx}\right)+\int_{\mathrm{1}/\mathrm{2}} ^{\mathrm{1}} \frac{\mathrm{lnxln}\left(\mathrm{1}−\mathrm{x}\right)}{\mathrm{x}}\mathrm{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}^{\mathrm{3}} \mathrm{2}+\mathrm{ln2}\left(\mathrm{Li}_{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)−\mathrm{Li}_{\mathrm{2}} \left(\mathrm{1}\right)\right)−\mathrm{lnx}\centerdot\mathrm{Li}_{\mathrm{2}} \left(\mathrm{x}\right)\mid_{\mathrm{1}/\mathrm{2}} ^{\mathrm{1}} +\mathrm{Li}_{\mathrm{3}} \left(\mathrm{1}\right)−\mathrm{Li}_{\mathrm{3}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}^{\mathrm{3}} \mathrm{2}−\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\mathrm{ln2}−\mathrm{ln2}\centerdot\mathrm{Li}_{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)−\mathrm{ln2}\centerdot\mathrm{Li}_{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)+\zeta\left(\mathrm{3}\right)−\mathrm{Li}_{\mathrm{3}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$=−\frac{\pi^{\mathrm{2}} }{\mathrm{4}}\mathrm{ln2}+\frac{\mathrm{4}}{\mathrm{3}}\mathrm{ln}^{\mathrm{3}} \mathrm{2}+\frac{\mathrm{1}}{\mathrm{8}}\zeta\left(\mathrm{3}\right) \\ $$$$−−−−−−−−−−−−−−−−− \\ $$$$\mathrm{Li}_{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)=−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}^{\mathrm{2}} \mathrm{2}+\frac{\pi^{\mathrm{2}} }{\mathrm{12}} \\ $$$$\mathrm{Li}_{\mathrm{3}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)=−\frac{\pi^{\mathrm{2}} }{\mathrm{12}}\mathrm{ln2}+\frac{\mathrm{1}}{\mathrm{6}}\mathrm{ln}^{\mathrm{3}} \mathrm{2}+\frac{\mathrm{7}}{\mathrm{8}}\zeta\left(\mathrm{3}\right) \\ $$

Commented by mnjuly1970 last updated on 23/Oct/21

thanks alot...

$$\:\:{thanks}\:{alot}... \\ $$

Commented by mnjuly1970 last updated on 23/Oct/21

please recheck the answer

$$\:\:{please}\:{recheck}\:{the}\:{answer} \\ $$

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