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Question Number 163682 by mnjuly1970 last updated on 09/Jan/22

Ω= ∫_0 ^( 1) (Li_( 2) ^ (x ))^( 2) dx = ? ■ m.n −−− −−−

$$ \\ $$$$\:\:\:\:\:\Omega=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \:\left(\mathrm{Li}_{\:\mathrm{2}} ^{\:} \left({x}\:\right)\right)^{\:\mathrm{2}} {dx}\:=\:?\:\:\:\:\:\:\blacksquare\:{m}.{n}\:\:\: \\ $$$$\:\:\:\:\:−−−\:−−− \\ $$$$\:\:\:\: \\ $$

Answered by Kamel last updated on 09/Jan/22

Ω= ∫_0 ^( 1) (Li_( 2) ^ (x ))^( 2) dx = ? ■ m.n −−− −−− Ω=^(IBP) (π^4 /(36))+∫_0 ^1 Ln(1−x)Li_2 (x)dx ∫Ln(1−x)dx=−(1−x)Ln(1−x)−x ∴ Ω=^(IBP) (π^4 /(36))−(π^2 /3)−2∫_0 ^1 (((1−x)Ln^2 (1−x))/x)dx−2∫_0 ^1 Ln(1−x)dx =(π^4 /(36))−(π^2 /3)−4ζ(3)+6

$$ \\ $$$$\:\Omega=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \:\left(\mathrm{Li}_{\:\mathrm{2}} ^{\:} \left({x}\:\right)\right)^{\:\mathrm{2}} {dx}\:=\:?\:\:\:\:\:\:\blacksquare\:{m}.{n}\:\:\: \\ $$$$\:\:\:\:\:−−−\:−−− \\ $$$$\:\:\:\:\:\:\:\Omega\overset{{IBP}} {=}\frac{\pi^{\mathrm{4}} }{\mathrm{36}}+\int_{\mathrm{0}} ^{\mathrm{1}} {Ln}\left(\mathrm{1}−{x}\right){Li}_{\mathrm{2}} \left({x}\right){dx}\: \\ $$$$\int{Ln}\left(\mathrm{1}−{x}\right){dx}=−\left(\mathrm{1}−{x}\right){Ln}\left(\mathrm{1}−{x}\right)−{x} \\ $$$$\:\:\therefore\:\:\Omega\overset{{IBP}} {=}\frac{\pi^{\mathrm{4}} }{\mathrm{36}}−\frac{\pi^{\mathrm{2}} }{\mathrm{3}}−\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\left(\mathrm{1}−{x}\right){Ln}^{\mathrm{2}} \left(\mathrm{1}−{x}\right)}{{x}}{dx}−\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} {Ln}\left(\mathrm{1}−{x}\right){dx} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\frac{\pi^{\mathrm{4}} }{\mathrm{36}}−\frac{\pi^{\mathrm{2}} }{\mathrm{3}}−\mathrm{4}\zeta\left(\mathrm{3}\right)+\mathrm{6} \\ $$

Commented by mnjuly1970 last updated on 09/Jan/22

very nice sir kamel

$$\:\:\:\:{very}\:{nice}\:{sir}\:{kamel} \\ $$

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