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Question Number 184029 by SEKRET last updated on 02/Jan/23

∫_0 ^( 1) (( x−1)/(x+1))∙(1/(ln(x))) dx =?

∫01x−1x+1⋅1ln(x)dx=?

Answered by ARUNG_Brandon_MBU last updated on 02/Jan/23

Ω(α)=∫_0 ^1 ((x^α −1)/(x+1))∙(1/(lnx))dx Ω′(α)=∫_0 ^1 (x^α /(x+1))dx=∫_0 ^1 ((x^α (1−x))/(1−x^2 ))dx=(1/2)∫_0 ^1 ((x^((α−1)/2) −x^(α/2) )/(1−x))dx =(1/2)(ψ(((α+2)/2))−ψ(((α+1)/2))) Ω(α)=ln(Γ(((α+2)/2)))−ln(Γ(((α+1)/2)))+C Ω(0)=0=ln(Γ(1))−ln(Γ((1/2)))+C ⇒C=ln(√π) ⇒Ω(α)=ln(Γ(((α+2)/2)))−ln(Γ(((α+1)/2)))+ln(√π) ∫_0 ^1 ((x−1)/(x+1))∙(1/(lnx))dx=Ω(1)=ln(Γ((3/2)))−ln(Γ(1))+ln(√π) ∫_0 ^1 ((x−1)/(x+1))∙(1/(lnx))dx=ln(((√π)/2))+ln(√π)=ln((π/2))

Ω(α)=∫01xα−1x+1⋅1lnxdxΩ′(α)=∫01xαx+1dx=∫01xα(1−x)1−x2dx=12∫01xα−12−xα21−xdx=12(ψ(α+22)−ψ(α+12))Ω(α)=ln(Γ(α+22))−ln(Γ(α+12))+CΩ(0)=0=ln(Γ(1))−ln(Γ(12))+C⇒C=lnπ⇒Ω(α)=ln(Γ(α+22))−ln(Γ(α+12))+lnπ∫01x−1x+1⋅1lnxdx=Ω(1)=ln(Γ(32))−ln(Γ(1))+lnπ∫01x−1x+1⋅1lnxdx=ln(π2)+lnπ=ln(π2)

Commented by SEKRET last updated on 02/Jan/23

good thank you sir

goodthankyousir

Commented by ARUNG_Brandon_MBU last updated on 02/Jan/23

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