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Question Number 162177 by mnjuly1970 last updated on 27/Dec/21

     𝛗 = ∫_0 ^( 1) (( ln^( 2) ( x ). Li_( 2)  (x ))/x^  ) dx =?

$$ \\ $$$$\:\:\:\boldsymbol{\phi}\:=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\:\mathrm{ln}^{\:\mathrm{2}} \left(\:{x}\:\right).\:\mathrm{Li}_{\:\mathrm{2}} \:\left({x}\:\right)}{{x}^{\:} }\:{dx}\:=? \\ $$

Answered by Ar Brandon last updated on 27/Dec/21

Ο†=∫_0 ^1 ((log^2 xLi_2 (x))/x)dx=Ξ£_(n=1) ^∞ (1/n^2 )∫_0 ^1 ((x^n log^2 x)/x)dx     =Ξ£_(n=1) ^∞ (1/n^2 )([(1/3)x^n log^3 x]_0 ^1 βˆ’(n/3)∫_0 ^1 x^(nβˆ’1) log^3 xdx)     =βˆ’(1/3)Ξ£_(n=1) ^∞ (1/n)βˆ™(βˆ‚^3 /βˆ‚Ξ±^3 )∣_(Ξ±=nβˆ’1) ∫_0 ^1 x^Ξ± dx     =βˆ’(1/3)Ξ£_(n=1) ^∞ (1/n)βˆ™(βˆ‚^3 /βˆ‚Ξ±^3 )∣_(Ξ±=nβˆ’1) (1/(Ξ±+1))     =βˆ’(1/3)Ξ£_(n=1) ^∞ (1/n)(((βˆ’6)/n^4 ))=2Ξ£_(n=1) ^∞ (1/n^5 )=2ΞΆ(5)

$$\phi=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{log}^{\mathrm{2}} {xLi}_{\mathrm{2}} \left({x}\right)}{{x}}{dx}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{{n}} {log}^{\mathrm{2}} {x}}{{x}}{dx} \\ $$$$\:\:\:=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\left(\left[\frac{\mathrm{1}}{\mathrm{3}}{x}^{{n}} {log}^{\mathrm{3}} {x}\right]_{\mathrm{0}} ^{\mathrm{1}} βˆ’\frac{{n}}{\mathrm{3}}\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{n}βˆ’\mathrm{1}} {log}^{\mathrm{3}} {xdx}\right) \\ $$$$\:\:\:=βˆ’\frac{\mathrm{1}}{\mathrm{3}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}}\centerdot\frac{\partial^{\mathrm{3}} }{\partial\alpha^{\mathrm{3}} }\mid_{\alpha={n}βˆ’\mathrm{1}} \int_{\mathrm{0}} ^{\mathrm{1}} {x}^{\alpha} {dx} \\ $$$$\:\:\:=βˆ’\frac{\mathrm{1}}{\mathrm{3}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}}\centerdot\frac{\partial^{\mathrm{3}} }{\partial\alpha^{\mathrm{3}} }\mid_{\alpha={n}βˆ’\mathrm{1}} \frac{\mathrm{1}}{\alpha+\mathrm{1}} \\ $$$$\:\:\:=βˆ’\frac{\mathrm{1}}{\mathrm{3}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}}\left(\frac{βˆ’\mathrm{6}}{{n}^{\mathrm{4}} }\right)=\mathrm{2}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{5}} }=\mathrm{2}\zeta\left(\mathrm{5}\right) \\ $$

Commented by mnjuly1970 last updated on 27/Dec/21

 grateful sir brandon

$$\:{grateful}\:{sir}\:{brandon} \\ $$

Commented by Ar Brandon last updated on 27/Dec/21

My pleasure, Sir.

$$\mathrm{My}\:\mathrm{pleasure},\:\mathrm{Sir}. \\ $$

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