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Question Number 116667 by mnjuly1970 last updated on 05/Oct/20

... nice calculus... very nice integral:: demonstrate::: Ω = ∫_0 ^( 1) ((1−x)/((1+x+x^2 +x^3 )log(x))) dx=^(???) log((√(1/2))) .m.n.1970.

...nicecalculus...veryniceintegral::demonstrate:::Ω=∫011−x(1+x+x2+x3)log(x)dx=???log(12).m.n.1970.

Answered by maths mind last updated on 06/Oct/20

Hello after many tries Ω=∫_0 ^1 ((1−x)/((1+x+x^2 +x^3 ))).(dx/(log(x))) let f(a)=∫_0 ^1 ((1−x)/((1+x+x^2 +x^3 ))).(x^a /(log(x)))dx,a∈[0,1] Ω=f(0) f(a)=∫_0 ^1 (((1−x)^2 )/(1−x^4 ))..(x^a /(log(x)))dx f′(a)=∫_0 ^1 (((1−x)^2 )/((1−x^4 ))).∂_a x^a .(dx/(log(x)))=∫_0 ^1 (((x^2 −2x+1))/(1−x^4 )).x^a dx x^4 =t⇒dx=(t^(−(3/4)) /4)dt f′(a)=∫_0 ^1 ((t^(2/4) −2t^(1/4) +1)/(1−t))t^(a/4) .t^(−(3/4)) (dt/4) =−∫_0 ^1 ((1−t^((a/4)−(1/4)) )/(1−t))dt+2∫((1−t^((a/4)−(2/4)) )/(1−t))dt−∫((1−t^((a/4)−(3/4)) )/(1−t))dt we Have one of definition of Digamma Ψ(s+1)=−γ+∫_0 ^1 ((1−x^s )/(1−x))dx =(lnΓ(s+1))′ 4f′(a)=−Ψ(((a+3)/4))+2Ψ(((a+2)/4))−Ψ(((a+1)/4)) 4f(a)=−4lnΓ(((a+3)/4))−4lnΓ(((a+1)/4))+8lnΓ(((a+2)/4))+c lim_(a→∞) f(a)=0⇒c=0 f(0)=Ω 4f(0)=4Ω=−4lnΓ((3/4))−4lnΓ((1/4))+8lnΓ((1/2)) =−4ln(((Γ((1/4))Γ((3/4)))/(Γ^2 ((1/2))))):Γ((1/4))Γ(1−(1/4))=(π/(sin((π/4))))=π(√2) Γ((1/2))=(√π)⇒4Ω=−4ln(((π(√2))/(((√π))^2 )))=−4ln((√2))=4ln((√(1/2))) Ω=ln((√(1/2))) ∫_0 ^1 ((1−x)/(1+x+x^2 +x^3 )).(dx/(log(x)))=ln((√(1/2)))

HelloaftermanytriesΩ=∫011−x(1+x+x2+x3).dxlog(x)letf(a)=∫011−x(1+x+x2+x3).xalog(x)dx,a∈[0,1]Ω=f(0)f(a)=∫01(1−x)21−x4..xalog(x)dxf′(a)=∫01(1−x)2(1−x4).∂axa.dxlog(x)=∫01(x2−2x+1)1−x4.xadxx4=t⇒dx=t−344dtf′(a)=∫01t24−2t14+11−tta4.t−34dt4=−∫011−ta4−141−tdt+2∫1−ta4−241−tdt−∫1−ta4−341−tdtweHaveoneofdefinitionofDigammaΨ(s+1)=−γ+∫011−xs1−xdx=(lnΓ(s+1))′4f′(a)=−Ψ(a+34)+2Ψ(a+24)−Ψ(a+14)4f(a)=−4lnΓ(a+34)−4lnΓ(a+14)+8lnΓ(a+24)+clima→∞f(a)=0⇒c=0f(0)=Ω4f(0)=4Ω=−4lnΓ(34)−4lnΓ(14)+8lnΓ(12)=−4ln(Γ(14)Γ(34)Γ2(12)):Γ(14)Γ(1−14)=πsin(π4)=π2Γ(12)=π⇒4Ω=−4ln(π2(π)2)=−4ln(2)=4ln(12)Ω=ln(12)∫011−x1+x+x2+x3.dxlog(x)=ln(12)

Commented by maths mind last updated on 06/Oct/20

sam idea worck for ∫_0 ^1 ((1−x^s )/(Σ_(k=0) ^n x^k )).(dx/(log(x)))

samideaworckfor∫011−xs∑nk=0xk.dxlog(x)

Commented by mnjuly1970 last updated on 07/Oct/20

thank you good for you your effort is admirable...

thankyougoodforyouyoureffortisadmirable...

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