solve-0-1-ln-2-x-tanh-1-x-x-dx-0-1-tanh-1-x-2-1-x-

Question Number 164653 by mnjuly1970 last updated on 20/Jan/22

solve 𝛗 = ∫_0 ^( 1) ((ln^( 2) ( x ). tanh^( −1) ( x ))/x)dx =? Ω= ∫_0 ^( 1) (( (tanh^(−1) (x))^( 2) )/(1+x)) = ? −−−−

$$ \\ $$$$\:\:\:\:\:\:\:\:{solve} \\ $$$$\:\:\boldsymbol{\phi}\:=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \:\frac{\mathrm{ln}^{\:\mathrm{2}} \left(\:{x}\:\right).\:{tanh}^{\:−\mathrm{1}} \left(\:{x}\:\:\right)}{{x}}{dx}\:=? \\ $$$$\:\:\:\Omega=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\:\left({tanh}^{−\mathrm{1}} \left({x}\right)\right)^{\:\mathrm{2}} }{\mathrm{1}+{x}}\:=\:? \\ $$$$\:\:\:\:\:\:−−−− \\ $$

Answered by Lordose last updated on 20/Jan/22

Ω = ∫_0 ^( 1) (((tanh^(−1) (x))^2 )/(1+x))dx Ω = (1/4)∫_0 ^( 1) ((ln^2 (((1−x)/(1+x))))/(1+x))dx Ω =^(x=((1−x)/(1+x))) (1/4)∫_0 ^( 1) ((ln^2 (x))/(1+x)) = (1/4)Σ_(n=1) ^∞ (−1)^(n−1) ∫_0 ^( 1) x^(n−1) ln^2 (x)dx Ω =^(IBP×2) (1/2)Σ_(n=1) ^∞ (((−1)^(n−1) )/n^3 ) = 𝛈(3) = (3/8)𝛇(3) 𝛀 = (3/8)𝛇(3)

$$ \\ $$$$\Omega\:=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\left(\mathrm{tanh}^{−\mathrm{1}} \left(\mathrm{x}\right)\right)^{\mathrm{2}} }{\mathrm{1}+\mathrm{x}}\mathrm{dx} \\ $$$$\Omega\:=\:\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\mathrm{ln}^{\mathrm{2}} \left(\frac{\mathrm{1}−\mathrm{x}}{\mathrm{1}+\mathrm{x}}\right)}{\mathrm{1}+\mathrm{x}}\mathrm{dx} \\ $$$$\Omega\:\overset{\mathrm{x}=\frac{\mathrm{1}−\mathrm{x}}{\mathrm{1}+\mathrm{x}}} {=}\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\mathrm{ln}^{\mathrm{2}} \left(\mathrm{x}\right)}{\mathrm{1}+\mathrm{x}}\:=\:\frac{\mathrm{1}}{\mathrm{4}}\underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{\mathrm{n}−\mathrm{1}} \int_{\mathrm{0}} ^{\:\mathrm{1}} \mathrm{x}^{\mathrm{n}−\mathrm{1}} \mathrm{ln}^{\mathrm{2}} \left(\mathrm{x}\right)\mathrm{dx} \\ $$$$\Omega\:\overset{\boldsymbol{\mathrm{IBP}}×\mathrm{2}} {=}\frac{\mathrm{1}}{\mathrm{2}}\underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{\mathrm{n}−\mathrm{1}} }{\mathrm{n}^{\mathrm{3}} }\:=\:\boldsymbol{\eta}\left(\mathrm{3}\right)\:=\:\frac{\mathrm{3}}{\mathrm{8}}\boldsymbol{\zeta}\left(\mathrm{3}\right) \\ $$$$\boldsymbol{\Omega}\:=\:\frac{\mathrm{3}}{\mathrm{8}}\boldsymbol{\zeta}\left(\mathrm{3}\right)\: \\ $$

Answered by Ar Brandon last updated on 20/Jan/22

φ=∫_0 ^1 ((ln^2 xtanh^(−1) (x))/x)dx =[((ln^3 x)/3)tanh^(−1) (x)]_0 ^1 −(1/3)∫_0 ^1 ((ln^3 x)/(1−x^2 ))dx =−(1/3)∙(1/2)∙(1/8)∫_0 ^1 ((u^(−(1/2)) ln^3 u)/(1−u))du=(1/(48))ψ^((3)) ((1/2)) =2(ζ(4)−(1/(16))ζ(4))=((15)/8)ζ(4)=((15)/(720))π^4

$$ \\ $$$$\phi=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}^{\mathrm{2}} {x}\mathrm{tanh}^{−\mathrm{1}} \left({x}\right)}{{x}}{dx} \\ $$$$\:\:\:=\left[\frac{\mathrm{ln}^{\mathrm{3}} {x}}{\mathrm{3}}\mathrm{tanh}^{−\mathrm{1}} \left({x}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} −\frac{\mathrm{1}}{\mathrm{3}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}^{\mathrm{3}} {x}}{\mathrm{1}−{x}^{\mathrm{2}} }{dx} \\ $$$$\:\:\:=−\frac{\mathrm{1}}{\mathrm{3}}\centerdot\frac{\mathrm{1}}{\mathrm{2}}\centerdot\frac{\mathrm{1}}{\mathrm{8}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{u}^{−\frac{\mathrm{1}}{\mathrm{2}}} \mathrm{ln}^{\mathrm{3}} {u}}{\mathrm{1}−{u}}{du}=\frac{\mathrm{1}}{\mathrm{48}}\psi^{\left(\mathrm{3}\right)} \left(\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$\:\:\:=\mathrm{2}\left(\zeta\left(\mathrm{4}\right)−\frac{\mathrm{1}}{\mathrm{16}}\zeta\left(\mathrm{4}\right)\right)=\frac{\mathrm{15}}{\mathrm{8}}\zeta\left(\mathrm{4}\right)=\frac{\mathrm{15}}{\mathrm{720}}\pi^{\mathrm{4}} \\ $$

Commented by mnjuly1970 last updated on 20/Jan/22

$$\:\:{verh}\:{nice}\:…{thank}\:{you} \\ $$$${sir}\:{brandon}… \\ $$

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