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0-lnx-x-1-2-dx-2-3-pi-2-

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Question Number 140202 by qaz last updated on 05/May/21
∫_0 ^∞ (((lnx)/(x−1)))^2 dx=(2/3)π^2
∫0∞(lnxx−1)2dx=23π2
Answered by mathmax by abdo last updated on 05/May/21
Φ =∫_0 ^∞ ((log^2 x)/((1−x)^2 ))dx ⇒Φ=∫_0 ^1 ((log^2 x)/((1−x)^2 ))dx +∫_1 ^∞ ((log^2 x)/((1−x)^2 ))dx(→x=(1/t)) =∫_0 ^1 ((log^2 x)/((1−x)^2 ))dx−∫_0 ^1 ((log^2 t)/((1−(1/t))^2 ))(−(dt/t^2 )) =2∫_0 ^1 ((log^2 x)/((1−x)^2 ))dx we have (1/(1−x)) =Σ_(n=0) ^∞ x^n ⇒(1/((1−x)^2 )) =Σ_(n=1) ^∞ nx^(n−1) (by derivation) ⇒ (1/((1−x)^2 ))=Σ_(n=1) ^∞ nx^(n−1) ⇒Φ =2 ∫_0 ^1 Σ_(n=1) ^∞ nx^(n−1) log^2 x dx =2 Σ_(n=1) ^∞ n U_n with U_n =∫_0 ^1 x^(n−1) log^2 xdx by parts U_n =[(x^n /n)log^2 x]_0 ^1 −∫_0 ^1 (x^n /n)((2logx)/x)dx =−(2/n) ∫_0 ^1 x^(n−1) logx dx =−(2/n)( [(x^n /n)logx]_0 ^1 −∫_0 ^1 (x^n /n)(dx/x)) =−(2/n)(−(1/n) ∫_0 ^1 x^(n−1) dx) =(2/n^3 ) ⇒Φ =2 Σ_(n=1) ^∞ n×(2/n^3 ) =4 Σ_(n=1) ^∞ (1/n^2 ) =4×(π^2 /6) =((2π^2 )/3) ⇒ ∫_0 ^∞ (((logx)/(x−1)))^2 dx =(2/3)π^2
Φ=∫0∞log2x(1−x)2dx⇒Φ=∫01log2x(1−x)2dx+∫1∞log2x(1−x)2dx(→x=1t)=∫01log2x(1−x)2dx−∫01log2t(1−1t)2(−dtt2)=2∫01log2x(1−x)2dxwehave11−x=∑n=0∞xn⇒1(1−x)2=∑n=1∞nxn−1(byderivation)⇒1(1−x)2=∑n=1∞nxn−1⇒Φ=2∫01∑n=1∞nxn−1log2xdx=2∑n=1∞nUnwithUn=∫01xn−1log2xdxbypartsUn=[xnnlog2x]01−∫01xnn2logxxdx=−2n∫01xn−1logxdx=−2n([xnnlogx]01−∫01xnndxx)=−2n(−1n∫01xn−1dx)=2n3⇒Φ=2∑n=1∞n×2n3=4∑n=1∞1n2=4×π26=2π23⇒∫0∞(logxx−1)2dx=23π2
Answered by Ar Brandon last updated on 05/May/21
=∫_0 ^1 ((ln^2 x)/((1−x)^2 ))dx+∫_0 ^1 ((ln^2 x)/((1−x)^2 ))dx=2∫_0 ^1 ((ln^2 x)/((1−x)^2 ))dx =2[−((ln^2 x)/(1−x))+2∫_0 ^1 ((lnx)/(x(1−x)))dx]_0 ^1 =((2ln^2 x)/(x−1))+4∫_0 ^1 [((lnx)/x)+((lnx)/(1−x))]dx =((2ln^2 x)/(x−1))+2ln^2 x+4∫_0 ^1 ((lnx)/(1−x))dx=−4ψ′(1) =4Σ_(n=0) ^∞ (1/((n+1)^2 ))=4ζ(2)=4×(π^2 /6)=((2π^2 )/3)
=∫01ln2x(1−x)2dx+∫01ln2x(1−x)2dx=2∫01ln2x(1−x)2dx=2[−ln2x1−x+2∫01lnxx(1−x)dx]01=2ln2xx−1+4∫01[lnxx+lnx1−x]dx=2ln2xx−1+2ln2x+4∫01lnx1−xdx=−4ψ′(1)=4∑∞n=01(n+1)2=4ζ(2)=4×π26=2π23

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