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

solve 𝛗 = ∫_0 ^( 1) ((ln^( 2) ( x ). tanh^( βˆ’1) ( x ))/x)dx =? Ξ©= ∫_0 ^( 1) (( (tanh^(βˆ’1) (x))^( 2) )/(1+x)) = ? βˆ’βˆ’βˆ’βˆ’

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

Answered by Lordose last updated on 20/Jan/22

Ξ© = ∫_0 ^( 1) (((tanh^(βˆ’1) (x))^2 )/(1+x))dx Ξ© = (1/4)∫_0 ^( 1) ((ln^2 (((1βˆ’x)/(1+x))))/(1+x))dx Ξ© =^(x=((1βˆ’x)/(1+x))) (1/4)∫_0 ^( 1) ((ln^2 (x))/(1+x)) = (1/4)Ξ£_(n=1) ^∞ (βˆ’1)^(nβˆ’1) ∫_0 ^( 1) x^(nβˆ’1) ln^2 (x)dx Ξ© =^(IBPΓ—2) (1/2)Ξ£_(n=1) ^∞ (((βˆ’1)^(nβˆ’1) )/n^3 ) = π›ˆ(3) = (3/8)𝛇(3) 𝛀 = (3/8)𝛇(3)

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

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

Ο†=∫_0 ^1 ((ln^2 xtanh^(βˆ’1) (x))/x)dx =[((ln^3 x)/3)tanh^(βˆ’1) (x)]_0 ^1 βˆ’(1/3)∫_0 ^1 ((ln^3 x)/(1βˆ’x^2 ))dx =βˆ’(1/3)βˆ™(1/2)βˆ™(1/8)∫_0 ^1 ((u^(βˆ’(1/2)) ln^3 u)/(1βˆ’u))du=(1/(48))ψ^((3)) ((1/2)) =2(ΞΆ(4)βˆ’(1/(16))ΞΆ(4))=((15)/8)ΞΆ(4)=((15)/(720))Ο€^4

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

Commented by mnjuly1970 last updated on 20/Jan/22

verh nice ...thank you sir brandon...

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

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