Question Number 173909 by mnjuly1970 last updated on 20/Jul/22
![Ω=∫_0 ^( 1) (( ln^( 2) (1−x))/((1+x )^( 2) )) = Li_2 ((1/2) ) −−− Solution −−− Ω =^(i.b.p) {[−(1/(1+x))ln^( 2) (1−x)]_0 ^1 −2∫_0 ^( 1) ((ln(1−x))/((1−x)(1+x)))dx =lim_( x→1^− ) −(1/(1+x)) ln^( 2) (1−x)−∫_0 ^( 1) ((ln(1−x))/(1−x)) +((ln(1−x))/(1+x))dx = lim_( x→1^− ) (1/2) ln^( 2) (1−x)−(1/(1+x))ln^( 2) (1−x)−Φ =lim_( x→1^( −) ) (((x−1)/(2(1+x))))ln^( 2) (1−x)−Φ = −Φ =_(derived) ^(earlier) −(−(π^( 2) /(12)) +(1/2)ln^( 2) (2)) ⇒ Ω = (π^( 2) /(12)) −(1/2) ln^( 2) (2 )=Li_2 ((1/2))](https://www.tinkutara.com/question/Q173909.png)
$$ \\ $$$$\:\:\:\:\Omega=\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\:{ln}^{\:\mathrm{2}} \left(\mathrm{1}−{x}\right)}{\left(\mathrm{1}+{x}\:\right)^{\:\mathrm{2}} }\:=\:{Li}_{\mathrm{2}} \:\left(\frac{\mathrm{1}}{\mathrm{2}}\:\right) \\ $$$$\:\:\:\:−−−\:\:\:{Solution}\:−−− \\ $$$$\:\:\:\:\Omega\:\overset{{i}.{b}.{p}} {=}\left\{\left[−\frac{\mathrm{1}}{\mathrm{1}+{x}}{ln}^{\:\mathrm{2}} \left(\mathrm{1}−{x}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} −\mathrm{2}\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{ln}\left(\mathrm{1}−{x}\right)}{\left(\mathrm{1}−{x}\right)\left(\mathrm{1}+{x}\right)}{dx}\right. \\ $$$$\:\:\:\:\:={lim}_{\:{x}\rightarrow\mathrm{1}^{−} } −\frac{\mathrm{1}}{\mathrm{1}+{x}}\:{ln}^{\:\mathrm{2}} \left(\mathrm{1}−{x}\right)−\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{ln}\left(\mathrm{1}−{x}\right)}{\mathrm{1}−{x}}\:+\frac{{ln}\left(\mathrm{1}−{x}\right)}{\mathrm{1}+{x}}{dx} \\ $$$$\:\:\:\:\:=\:{lim}_{\:{x}\rightarrow\mathrm{1}^{−} } \frac{\mathrm{1}}{\mathrm{2}}\:{ln}^{\:\mathrm{2}} \left(\mathrm{1}−{x}\right)−\frac{\mathrm{1}}{\mathrm{1}+{x}}{ln}^{\:\mathrm{2}} \left(\mathrm{1}−{x}\right)−\Phi \\ $$$$\:\:\:\:={lim}_{\:{x}\rightarrow\mathrm{1}^{\:−} } \:\left(\frac{{x}−\mathrm{1}}{\mathrm{2}\left(\mathrm{1}+{x}\right)}\right){ln}^{\:\mathrm{2}} \left(\mathrm{1}−{x}\right)−\Phi \\ $$$$\:\:\:\:\:=\:−\Phi\:\underset{{derived}} {\overset{{earlier}} {=}}\:−\left(−\frac{\pi^{\:\mathrm{2}} }{\mathrm{12}}\:+\frac{\mathrm{1}}{\mathrm{2}}{ln}^{\:\mathrm{2}} \left(\mathrm{2}\right)\right) \\ $$$$\Rightarrow\:\:\Omega\:=\:\frac{\pi^{\:\mathrm{2}} }{\mathrm{12}}\:−\frac{\mathrm{1}}{\mathrm{2}}\:{ln}^{\:\mathrm{2}} \left(\mathrm{2}\:\right)={Li}_{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$ \\ $$
Commented by mnjuly1970 last updated on 21/Jul/22
$${mercey}\:{sir} \\ $$
Commented by Tawa11 last updated on 21/Jul/22
$$\mathrm{Great}\:\mathrm{sir} \\ $$