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Question Number 135821 by mnjuly1970 last updated on 16/Mar/21
....nice ..... calculus.... prove that :: 𝛗=∫_0 ^( 1) (((ln(1−x))/(1−(√(1−x)))))dx=4(1−ζ(2))
….nice…..calculus….provethat::ϕ=∫01(ln(1−x)1−1−x)dx=4(1−ζ(2))
Answered by mathmax by abdo last updated on 16/Mar/21
Φ=∫_0 ^1 ((log(1−x))/(1−(√(1−x))))dx we do the changement(√(1−x))=t ⇒1−x=t^2 ⇒x=1−t^2 Φ =−∫_0 ^1 ((log(1−1+t^2 ))/(1−t))(−2t)dt =4∫_0 ^1 ((tlog(t))/(1−t))dt =4∫_0 ^1 tlogtΣ_(n=0) ^∞ t^n dt =4Σ_(n=0) ^∞ ∫_0 ^1 t^(n+1) logt dt =4Σ U_n U_n =∫_0 ^1 t^(n+1) logt dt =[(t^(n+2) /(n+2))logt]_0 ^1 −(1/(n+2))∫_0 ^1 t^(n+1) dt =−(1/((n+2)^2 )) ⇒Φ =−4 Σ_(n=0) ^∞ (1/((n+2)^2 )) =−4Σ_(n=2) ^∞ (1/n^2 ) =−4{Σ_(n=1) ^∞ (1/n^2 )−1} =4−4.(π^2 /6) =4−((2π^2 )/3)
Φ=∫01log(1−x)1−1−xdxwedothechangement1−x=t⇒1−x=t2⇒x=1−t2Φ=−∫01log(1−1+t2)1−t(−2t)dt=4∫01tlog(t)1−tdt=4∫01tlogt∑n=0∞tndt=4∑n=0∞∫01tn+1logtdt=4ΣUnUn=∫01tn+1logtdt=[tn+2n+2logt]01−1n+2∫01tn+1dt=−1(n+2)2⇒Φ=−4∑n=0∞1(n+2)2=−4∑n=2∞1n2=−4{∑n=1∞1n2−1}=4−4.π26=4−2π23
Answered by Dwaipayan Shikari last updated on 16/Mar/21
∫_0 ^1 ((log(1−x))/(1−(√(1−x))))dx=∫_0 ^1 ((log(x))/(1−(√x)))dx =∫_0 ^1 ((4tlog(t))/(1−t))dt =4Φ′(2)=4(1−(π^2 /6)) Φ(α)=∫_0 ^1 ((1−x^(α−1) )/(1−x))dx=−γ+ψ(α) Φ′(α)=−∫_0 ^1 ((x^(α−1) log(x))/(1−x))dx=ψ^1 (α)⇒∫_0 ^1 ((xlog(x))/(1−x))=−ψ^1 (2) Φ′(2)=−Σ_(n=1) ^∞ (1/((n+1)^2 ))=−((π^2 /6)−1)
∫01log(1−x)1−1−xdx=∫01log(x)1−xdx=∫014tlog(t)1−tdt=4Φ′(2)=4(1−π26)Φ(α)=∫011−xα−11−xdx=−γ+ψ(α)Φ′(α)=−∫01xα−1log(x)1−xdx=ψ1(α)⇒∫01xlog(x)1−x=−ψ1(2)Φ′(2)=−∑∞n=11(n+1)2=−(π26−1)
Commented by mnjuly1970 last updated on 16/Mar/21
thanks alot mr payan...
thanksalotmrpayan…
Answered by Ñï= last updated on 16/Mar/21
∫_0 ^1 ((ln(1−x))/(1−(√(1−x))))dx=∫_0 ^1 ((lnx)/(1−(√x)))dx=∫_0 ^1 (((1−(√x))lnx)/(1−x))dx =∫_0 ^1 ((lnx)/(1−x))dx−∫_0 ^1 (((√x)lnx)/(1−x))dx =∫_0 ^1 ((lnx)/(1−x))dx−∫_0 ^1 ((4x^2 lnx)/(1−x^2 ))dx ((√x)→x) =−Li_2 (1)−4∫_0 ^1 (1−(1/(1−x^2 )))lnxdx =−Li_2 (1)−4∫_0 ^1 lnxdx+4∫_0 ^1 ((lnx)/((1−x)(1+x)))dx =−Li_2 (1)+4+2∫_0 ^1 ((1/(1−x))+(1/(1+x)))lnxdx =−Li_2 (1)+4−2Li_2 (1)+2Li_2 (−1) =−4Li_2 (1)+4 =4(1−ζ(2))
∫01ln(1−x)1−1−xdx=∫01lnx1−xdx=∫01(1−x)lnx1−xdx=∫01lnx1−xdx−∫01xlnx1−xdx=∫01lnx1−xdx−∫014x2lnx1−x2dx(x→x)=−Li2(1)−4∫01(1−11−x2)lnxdx=−Li2(1)−4∫01lnxdx+4∫01lnx(1−x)(1+x)dx=−Li2(1)+4+2∫01(11−x+11+x)lnxdx=−Li2(1)+4−2Li2(1)+2Li2(−1)=−4Li2(1)+4=4(1−ζ(2))
Commented by mnjuly1970 last updated on 16/Mar/21
thank you do much...
thankyoudomuch…
Answered by mnjuly1970 last updated on 16/Mar/21
solution: y^2 =1−x 𝛗=2∫_0 ^( 1) ((ln((√(1−x)) ))/(1−(√(1−x))))dx 𝛗=4∫_0 ^( 1) ((yln(y))/(1−y))dy =−4∫_0 ^( 1) ln(y)dy−4li_2 (y) =4−4ζ(2)=4(1−ζ(2))... we know that: 1: li_2 (x)=−∫_0 ^( x) ((ln(1−z))/z)dz=Σ_(n=1) ^∞ (x^n /n^2 ) 1^∗ : li_2 (1)=ζ(2) 1^(∗∗) :∫_0 ^( 1) ln(t)dt=−1
solution:y2=1−xϕ=2∫01ln(1−x)1−1−xdxϕ=4∫01yln(y)1−ydy=−4∫01ln(y)dy−4li2(y)=4−4ζ(2)=4(1−ζ(2))…weknowthat:1:li2(x)=−∫0xln(1−z)zdz=∑∞n=1xnn21∗:li2(1)=ζ(2)1∗∗:∫01ln(t)dt=−1

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