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Question-by-M-N-July-0-1-ln-1-x-4-x-8-x-dx-x-x-1-2-1-2-0-1-ln-1-x-2-x-4-x-dx-1-2-0-1-ln-1-x-2-1-x-6-x-dx-1-2-0-1-ln-1-x-2-

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Question Number 165218 by Lordose last updated on 27/Jan/22
Question by M.N July Φ = ∫_0 ^( 1) ((ln(1 + x^4 + x^8 ))/x)dx Φ =^(x=x^(1/2) ) (1/2)∫_0 ^( 1) ((ln(1+x^2 +x^4 ))/x)dx Φ = (1/2)∫_0 ^( 1) ((ln(((1−x^2 )/(1−x^6 ))))/x)dx = (1/2)∫_0 ^( 1) ((ln(1−x^2 ))/x)dx − (1/2)∫_0 ^( 1) ((ln(1−x^6 ))/x)dx Φ = (1/2)(A − B) A =^(x=x^(1/2) ) (1/2)∫_0 ^( 1) ((ln(1−x))/x)dx = (1/2)Li_2 (1) B =^(x=x^(1/6) ) (1/6)∫_0 ^( 1) ((x^((1/6)−1) ln(1−x))/x^(1/6) )dx B = (1/6)∫_0 ^( 1) ((ln(1−x))/x)dx = (1/6)Li_2 (1) Φ = (1/2)((1/2)Li_2 (1)−(1/6)Li_2 (1)) = (1/6)Li_2 (1) 𝚽 = ((𝛇(2))/3) ▲▲▲
$$\mathrm{Question}\:\mathrm{by}\:\boldsymbol{\mathrm{M}}.\boldsymbol{\mathrm{N}}\:\boldsymbol{\mathrm{July}} \\ $$$$\Phi\:=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\mathrm{ln}\left(\mathrm{1}\:+\:\mathrm{x}^{\mathrm{4}} \:+\:\mathrm{x}^{\mathrm{8}} \right)}{\mathrm{x}}\mathrm{dx} \\ $$$$\Phi\:\overset{\mathrm{x}=\mathrm{x}^{\frac{\mathrm{1}}{\mathrm{2}}} } {=}\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\mathrm{ln}\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} +\mathrm{x}^{\mathrm{4}} \right)}{\mathrm{x}}\mathrm{dx} \\ $$$$\Phi\:=\:\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\mathrm{ln}\left(\frac{\mathrm{1}−\mathrm{x}^{\mathrm{2}} }{\mathrm{1}−\mathrm{x}^{\mathrm{6}} }\right)}{\mathrm{x}}\mathrm{dx}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\mathrm{ln}\left(\mathrm{1}−\mathrm{x}^{\mathrm{2}} \right)}{\mathrm{x}}\mathrm{dx}\:−\:\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\mathrm{ln}\left(\mathrm{1}−\mathrm{x}^{\mathrm{6}} \right)}{\mathrm{x}}\mathrm{dx} \\ $$$$\Phi\:=\:\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{A}\:−\:\mathrm{B}\right) \\ $$$$\mathrm{A}\:\overset{\mathrm{x}=\mathrm{x}^{\frac{\mathrm{1}}{\mathrm{2}}} } {=}\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\mathrm{ln}\left(\mathrm{1}−\mathrm{x}\right)}{\mathrm{x}}\mathrm{dx}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\boldsymbol{\mathrm{Li}}_{\mathrm{2}} \left(\mathrm{1}\right) \\ $$$$\mathrm{B}\:\overset{\mathrm{x}=\mathrm{x}^{\frac{\mathrm{1}}{\mathrm{6}}} } {=}\frac{\mathrm{1}}{\mathrm{6}}\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\mathrm{x}^{\frac{\mathrm{1}}{\mathrm{6}}−\mathrm{1}} \mathrm{ln}\left(\mathrm{1}−\mathrm{x}\right)}{\mathrm{x}^{\frac{\mathrm{1}}{\mathrm{6}}} }\mathrm{dx} \\ $$$$\mathrm{B}\:=\:\frac{\mathrm{1}}{\mathrm{6}}\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\mathrm{ln}\left(\mathrm{1}−\mathrm{x}\right)}{\mathrm{x}}\mathrm{dx}\:=\:\frac{\mathrm{1}}{\mathrm{6}}\boldsymbol{\mathrm{Li}}_{\mathrm{2}} \left(\mathrm{1}\right)\: \\ $$$$\Phi\:=\:\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{\mathrm{2}}\boldsymbol{\mathrm{Li}}_{\mathrm{2}} \left(\mathrm{1}\right)−\frac{\mathrm{1}}{\mathrm{6}}\boldsymbol{\mathrm{Li}}_{\mathrm{2}} \left(\mathrm{1}\right)\right)\:=\:\frac{\mathrm{1}}{\mathrm{6}}\boldsymbol{\mathrm{Li}}_{\mathrm{2}} \left(\mathrm{1}\right) \\ $$$$\boldsymbol{\Phi}\:=\:\frac{\boldsymbol{\zeta}\left(\mathrm{2}\right)}{\mathrm{3}}\:\blacktriangle\blacktriangle\blacktriangle \\ $$
Commented by mnjuly1970 last updated on 27/Jan/22
thx alot sir lordose
$$\:\:\:{thx}\:{alot}\:\:{sir}\:{lordose} \\ $$
Answered by Ar Brandon last updated on 27/Jan/22
Φ=∫_0 ^1 ((ln(1+x^4 +x^8 ))/x)dx=∫_0 ^1 (1/x)ln(((1−x^(12) )/(1−x^4 )))dx =Σ_(n=1) ^∞ (1/n)∫_0 ^1 (x^(4n−1) −x^(12n−1) )dx=Σ_(n=1) ^∞ (1/n)((1/(4n))−(1/(12n))) =(1/6)Σ_(n=1) ^∞ (1/n^2 )=((ζ(2))/6)=(π^2 /(36))
$$\Phi=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}\left(\mathrm{1}+{x}^{\mathrm{4}} +{x}^{\mathrm{8}} \right)}{{x}}{dx}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{{x}}\mathrm{ln}\left(\frac{\mathrm{1}−{x}^{\mathrm{12}} }{\mathrm{1}−{x}^{\mathrm{4}} }\right){dx} \\ $$$$\:\:\:=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}}\int_{\mathrm{0}} ^{\mathrm{1}} \left({x}^{\mathrm{4}{n}−\mathrm{1}} −{x}^{\mathrm{12}{n}−\mathrm{1}} \right){dx}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}}\left(\frac{\mathrm{1}}{\mathrm{4}{n}}−\frac{\mathrm{1}}{\mathrm{12}{n}}\right) \\ $$$$\:\:\:=\frac{\mathrm{1}}{\mathrm{6}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{2}} }=\frac{\zeta\left(\mathrm{2}\right)}{\mathrm{6}}=\frac{\pi^{\mathrm{2}} }{\mathrm{36}} \\ $$
Commented by amin96 last updated on 28/Jan/22
correct sir. great solution
$$\boldsymbol{\mathrm{correct}}\:\boldsymbol{\mathrm{sir}}.\:\boldsymbol{\mathrm{great}}\:\boldsymbol{\mathrm{solution}} \\ $$

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