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Question Number 154202 by mathdanisur last updated on 15/Sep/21

𝛀 =∫_( 0) ^( ∞) ((ln∙(1 + x))/(x∙(x^2 + x + 1))) dx = ?

$$\boldsymbol{\Omega}\:\:=\underset{\:\mathrm{0}} {\overset{\:\infty} {\int}}\:\frac{\mathrm{ln}\centerdot\left(\mathrm{1}\:+\:\mathrm{x}\right)}{\mathrm{x}\centerdot\left(\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{x}\:+\:\mathrm{1}\right)}\:\mathrm{dx}\:=\:? \\ $$

Answered by qaz last updated on 15/Sep/21

∫_0 ^∞ ((ln(1+x))/(x(x^2 +x+1)))dx =∫_1 ^∞ ((lnx)/((x−1)(x^2 −x+1)))dx =∫_1 ^∞ (((x+1)lnx)/((x−1)(x^3 +1)))dx =∫_1 ^0 ((x(1+x)lnx)/((1−x)(1+x^3 )))dx =∫_1 ^0 ((1/(1−x))+((x^2 −1)/(1+x^3 )))lnxdx =(π^2 /6)+∫_1 ^0 (x^2 /(1+x^3 ))lnxdx−∫_1 ^0 ((lnx)/(1+x^3 ))dx =(π^2 /6)+(1/3)∫_0 ^1 ((ln(1+x^3 ))/x)dx+Σ_(n=0) ^∞ (−1)^n ∫_0 ^1 x^(3n) lnxdx =(π^2 /6)+(1/9)∫_0 ^1 ((ln(1+x))/x)dx−Σ_(n=0) ^∞ (((−1)^n )/((3n+1)^2 )) =((19π^2 )/(108))+(1/(18))ψ′((2/3))−(1/9)ψ′((1/3)) =((19π^2 )/(108))+(1/6)ψ′((2/3))−(1/9)π^2 csc^2 (π/3) =(π^2 /(36))+(1/6)ψ′((2/3))

$$\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{ln}\left(\mathrm{1}+\mathrm{x}\right)}{\mathrm{x}\left(\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{1}\right)}\mathrm{dx} \\ $$$$=\int_{\mathrm{1}} ^{\infty} \frac{\mathrm{lnx}}{\left(\mathrm{x}−\mathrm{1}\right)\left(\mathrm{x}^{\mathrm{2}} −\mathrm{x}+\mathrm{1}\right)}\mathrm{dx} \\ $$$$=\int_{\mathrm{1}} ^{\infty} \frac{\left(\mathrm{x}+\mathrm{1}\right)\mathrm{lnx}}{\left(\mathrm{x}−\mathrm{1}\right)\left(\mathrm{x}^{\mathrm{3}} +\mathrm{1}\right)}\mathrm{dx} \\ $$$$=\int_{\mathrm{1}} ^{\mathrm{0}} \frac{\mathrm{x}\left(\mathrm{1}+\mathrm{x}\right)\mathrm{lnx}}{\left(\mathrm{1}−\mathrm{x}\right)\left(\mathrm{1}+\mathrm{x}^{\mathrm{3}} \right)}\mathrm{dx} \\ $$$$=\int_{\mathrm{1}} ^{\mathrm{0}} \left(\frac{\mathrm{1}}{\mathrm{1}−\mathrm{x}}+\frac{\mathrm{x}^{\mathrm{2}} −\mathrm{1}}{\mathrm{1}+\mathrm{x}^{\mathrm{3}} }\right)\mathrm{lnxdx} \\ $$$$=\frac{\pi^{\mathrm{2}} }{\mathrm{6}}+\int_{\mathrm{1}} ^{\mathrm{0}} \frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{1}+\mathrm{x}^{\mathrm{3}} }\mathrm{lnxdx}−\int_{\mathrm{1}} ^{\mathrm{0}} \frac{\mathrm{lnx}}{\mathrm{1}+\mathrm{x}^{\mathrm{3}} }\mathrm{dx} \\ $$$$=\frac{\pi^{\mathrm{2}} }{\mathrm{6}}+\frac{\mathrm{1}}{\mathrm{3}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}\left(\mathrm{1}+\mathrm{x}^{\mathrm{3}} \right)}{\mathrm{x}}\mathrm{dx}+\underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{\mathrm{n}} \int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{x}^{\mathrm{3n}} \mathrm{lnxdx} \\ $$$$=\frac{\pi^{\mathrm{2}} }{\mathrm{6}}+\frac{\mathrm{1}}{\mathrm{9}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}\left(\mathrm{1}+\mathrm{x}\right)}{\mathrm{x}}\mathrm{dx}−\underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\left(\mathrm{3n}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{19}\pi^{\mathrm{2}} }{\mathrm{108}}+\frac{\mathrm{1}}{\mathrm{18}}\psi'\left(\frac{\mathrm{2}}{\mathrm{3}}\right)−\frac{\mathrm{1}}{\mathrm{9}}\psi'\left(\frac{\mathrm{1}}{\mathrm{3}}\right) \\ $$$$=\frac{\mathrm{19}\pi^{\mathrm{2}} }{\mathrm{108}}+\frac{\mathrm{1}}{\mathrm{6}}\psi'\left(\frac{\mathrm{2}}{\mathrm{3}}\right)−\frac{\mathrm{1}}{\mathrm{9}}\pi^{\mathrm{2}} \mathrm{csc}^{\mathrm{2}} \frac{\pi}{\mathrm{3}} \\ $$$$=\frac{\pi^{\mathrm{2}} }{\mathrm{36}}+\frac{\mathrm{1}}{\mathrm{6}}\psi'\left(\frac{\mathrm{2}}{\mathrm{3}}\right) \\ $$

Commented by puissant last updated on 15/Sep/21

Mr Qaz, you are really strong.. ψ′(x)+ψ′(1−x)=π^2 (1+cotan^2 (πx)) x=(2/3) → ψ′((2/3))=π^2 +π^2 cotan(((2π)/3))−ψ′((1/3))..

$${Mr}\:{Qaz},\:{you}\:{are}\:{really}\:{strong}.. \\ $$$$ \\ $$$$\psi'\left({x}\right)+\psi'\left(\mathrm{1}−{x}\right)=\pi^{\mathrm{2}} \left(\mathrm{1}+{cotan}^{\mathrm{2}} \left(\pi{x}\right)\right) \\ $$$${x}=\frac{\mathrm{2}}{\mathrm{3}}\:\rightarrow\:\psi'\left(\frac{\mathrm{2}}{\mathrm{3}}\right)=\pi^{\mathrm{2}} +\pi^{\mathrm{2}} {cotan}\left(\frac{\mathrm{2}\pi}{\mathrm{3}}\right)−\psi'\left(\frac{\mathrm{1}}{\mathrm{3}}\right).. \\ $$

Commented by mathdanisur last updated on 15/Sep/21

Very nice Ser thank you

$$\mathrm{Very}\:\mathrm{nice}\:\mathrm{Ser}\:\mathrm{thank}\:\mathrm{you} \\ $$

Commented by mathdanisur last updated on 16/Sep/21

Ser, aanswer: (1/6)(((5π^2 )/6) - ψ^′ ((1/3)))

$$\mathrm{Ser},\:\mathrm{aanswer}:\:\frac{\mathrm{1}}{\mathrm{6}}\left(\frac{\mathrm{5}\pi^{\mathrm{2}} }{\mathrm{6}}\:-\:\psi\:^{'} \left(\frac{\mathrm{1}}{\mathrm{3}}\right)\right) \\ $$

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