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Question Number 162298 by mathmax by abdo last updated on 28/Dec/21
find ∫_0 ^1 lnx ln(1−x^3 )dx
$$\mathrm{find}\:\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{lnx}\:\mathrm{ln}\left(\mathrm{1}−\mathrm{x}^{\mathrm{3}} \right)\mathrm{dx} \\ $$
Answered by Ar Brandon last updated on 28/Dec/21
I=∫_0 ^1 lnxln(1−x^3 )dx, x=u^(1/3) =(1/9)∫_0 ^1 u^(−(2/3)) lnu∙ln(1−u)du=−(1/9)Σ_(n=1) ^∞ (1/n)∫u^(n−(2/3)) lnudu =−(1/9)Σ_(n=1) ^∞ (1/n)∙(∂/∂α)∣_(α=n−(2/3)) ∫_0 ^1 x^α dx=(1/9)Σ_(n=1) ^∞ (1/n)∙(1/((n+(1/3))^2 )) =Σ_(n=1) ^∞ (1/(n(3n+1)^2 ))=Σ_(n=1) ^∞ ((1/n)−(3/((3n+1)))−(3/((3n+1)^2 ))) =Σ_(n=1) ^∞ ((1/n)−(1/(n+(1/3)))−(1/(3(n+(1/3))^2 )))=ψ((4/3))−ψ(1)−(1/3)ψ′((4/3)) =3+ψ((1/3))+γ−(1/3)ψ′((4/3))
$${I}=\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{ln}{x}\mathrm{ln}\left(\mathrm{1}−{x}^{\mathrm{3}} \right){dx},\:{x}={u}^{\frac{\mathrm{1}}{\mathrm{3}}} \\ $$$$\:\:\:=\frac{\mathrm{1}}{\mathrm{9}}\int_{\mathrm{0}} ^{\mathrm{1}} {u}^{−\frac{\mathrm{2}}{\mathrm{3}}} \mathrm{ln}{u}\centerdot\mathrm{ln}\left(\mathrm{1}−{u}\right){du}=−\frac{\mathrm{1}}{\mathrm{9}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}}\int{u}^{{n}−\frac{\mathrm{2}}{\mathrm{3}}} \mathrm{ln}{udu} \\ $$$$\:\:\:=−\frac{\mathrm{1}}{\mathrm{9}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}}\centerdot\frac{\partial}{\partial\alpha}\mid_{\alpha={n}−\frac{\mathrm{2}}{\mathrm{3}}} \int_{\mathrm{0}} ^{\mathrm{1}} {x}^{\alpha} {dx}=\frac{\mathrm{1}}{\mathrm{9}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}}\centerdot\frac{\mathrm{1}}{\left({n}+\frac{\mathrm{1}}{\mathrm{3}}\right)^{\mathrm{2}} } \\ $$$$\:\:\:=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}\left(\mathrm{3}{n}+\mathrm{1}\right)^{\mathrm{2}} }=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{{n}}−\frac{\mathrm{3}}{\left(\mathrm{3}{n}+\mathrm{1}\right)}−\frac{\mathrm{3}}{\left(\mathrm{3}{n}+\mathrm{1}\right)^{\mathrm{2}} }\right) \\ $$$$\:\:\:=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{{n}}−\frac{\mathrm{1}}{{n}+\frac{\mathrm{1}}{\mathrm{3}}}−\frac{\mathrm{1}}{\mathrm{3}\left({n}+\frac{\mathrm{1}}{\mathrm{3}}\right)^{\mathrm{2}} }\right)=\psi\left(\frac{\mathrm{4}}{\mathrm{3}}\right)−\psi\left(\mathrm{1}\right)−\frac{\mathrm{1}}{\mathrm{3}}\psi'\left(\frac{\mathrm{4}}{\mathrm{3}}\right) \\ $$$$\:\:\:=\mathrm{3}+\psi\left(\frac{\mathrm{1}}{\mathrm{3}}\right)+\gamma−\frac{\mathrm{1}}{\mathrm{3}}\psi'\left(\frac{\mathrm{4}}{\mathrm{3}}\right) \\ $$

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