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Question Number 123462 by mnjuly1970 last updated on 25/Nov/20
...nice calculus... prove that :: Ω=∫_0 ^( 1) ((ln(1−x)ln(1−x^2 ))/x) =^(??) ((11ζ( 3 ))/8) .................
$$\:\:\:\:\:\:\:\:\:\:\:\:\:…{nice}\:\:\:{calculus}… \\ $$$$\:\:\:\:\:\:{prove}\:\:{that}\::: \\ $$$$\:\:\:\:\:\:\:\Omega=\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{ln}\left(\mathrm{1}−{x}\right){ln}\left(\mathrm{1}−{x}^{\mathrm{2}} \right)}{{x}}\:\overset{??} {=}\frac{\mathrm{11}\zeta\left(\:\mathrm{3}\:\right)}{\mathrm{8}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:…………….. \\ $$
Answered by Lordose last updated on 25/Nov/20
Ω = ∫_( 0) ^( 1) ((ln(1−x)(ln(1−x)+ln(1+x)))/x)dx Ω = ∫_( 0) ^( 1) ((ln^2 (1−x))/x)dx + ∫_0 ^( 1) ((ln(1−x)ln(1+x))/x)dx Ω = 2ζ(3) − ((5ζ(3))/8) Ω = ((11ζ(3))/8)
$$ \\ $$$$\Omega\:=\:\int_{\:\mathrm{0}} ^{\:\mathrm{1}} \frac{\mathrm{ln}\left(\mathrm{1}−\mathrm{x}\right)\left(\mathrm{ln}\left(\mathrm{1}−\mathrm{x}\right)+\mathrm{ln}\left(\mathrm{1}+\mathrm{x}\right)\right)}{\mathrm{x}}\mathrm{dx} \\ $$$$\Omega\:=\:\int_{\:\mathrm{0}} ^{\:\mathrm{1}} \frac{\mathrm{ln}^{\mathrm{2}} \left(\mathrm{1}−\mathrm{x}\right)}{\mathrm{x}}\mathrm{dx}\:+\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\mathrm{ln}\left(\mathrm{1}−\mathrm{x}\right)\mathrm{ln}\left(\mathrm{1}+\mathrm{x}\right)}{\mathrm{x}}\mathrm{dx} \\ $$$$\Omega\:=\:\mathrm{2}\zeta\left(\mathrm{3}\right)\:−\:\frac{\mathrm{5}\zeta\left(\mathrm{3}\right)}{\mathrm{8}} \\ $$$$\Omega\:=\:\frac{\mathrm{11}\zeta\left(\mathrm{3}\right)}{\mathrm{8}} \\ $$
Commented by Dwaipayan Shikari last updated on 25/Nov/20
∫_0 ^1 ((log^2 (1−x))/x)dx =∫_0 ^1 ((log^2 x)/(1−x))dx =∫_0 ^1 log^2 x Σ_(n=0) ^∞ x^n =Σ_(n=0) ^∞ ∫_0 ^1 x^n log^2 (x) logx=t =Σ_(n=0) ^∞ ∫_(−∞) ^0 t^2 e^((n+1)x) dx (n+1)x=−Φ =−Σ_(n=0) ^∞ (1/((n+1)^3 ))∫_∞ ^0 Φ^2 e^(−Φ) dΦ =Σ_(n=0) ^∞ (1/((n+1)^3 )).Γ(3)=2ζ(3)
$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{log}^{\mathrm{2}} \left(\mathrm{1}−{x}\right)}{{x}}{dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{log}^{\mathrm{2}} {x}}{\mathrm{1}−{x}}{dx}\:\:=\int_{\mathrm{0}} ^{\mathrm{1}} {log}^{\mathrm{2}} {x}\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}{x}^{{n}} =\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{n}} {log}^{\mathrm{2}} \left({x}\right)\:\:\:\:\:\:{logx}={t} \\ $$$$=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\int_{−\infty} ^{\mathrm{0}} {t}^{\mathrm{2}} {e}^{\left({n}+\mathrm{1}\right){x}} {dx}\:\:\:\:\:\:\:\:\:\:\left({n}+\mathrm{1}\right){x}=−\Phi \\ $$$$=−\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)^{\mathrm{3}} }\int_{\infty} ^{\mathrm{0}} \Phi^{\mathrm{2}} {e}^{−\Phi} {d}\Phi\:=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)^{\mathrm{3}} }.\Gamma\left(\mathrm{3}\right)=\mathrm{2}\zeta\left(\mathrm{3}\right) \\ $$
Commented by mnjuly1970 last updated on 26/Nov/20
thank you
$${thank}\:{you} \\ $$
Commented by mnjuly1970 last updated on 26/Nov/20
thank you..
$${thank}\:{you}.. \\ $$

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