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

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Question Number 140200 by qaz last updated on 05/May/21
∫_0 ^∞ (((lnx)/(x−1)))^3 dx=π^2
∫0∞(lnxx−1)3dx=π2
Commented by Ar Brandon last updated on 05/May/21
Φ=∫_0 ^∞ (((lnx)/(x−1)))^3 dx=∫_0 ^1 (((lnx)/(x−1)))^3 dx+∫_1 ^∞ (((lnx)/(x−1)))^3 dx =∫_0 ^1 ((ln^3 x)/((x−1)^3 ))dx+∫_0 ^1 ((xln^3 x)/((x−1)^3 ))dx =∫_0 ^1 ((ln^3 x)/((x−1)^2 ))dx+2∫_0 ^1 ((ln^3 x)/((x−1)^3 ))dx =[−((ln^3 x)/(x−1))+3∫((ln^2 x)/(x(x−1)))dx]_0 ^1 +2[−((ln^3 x)/(2(x−1)^2 ))+(3/2)∫((ln^2 x)/(x(x−1)^2 ))dx]_0 ^1 =[−((ln^3 x)/(x−1))−((ln^3 x)/((x−1)^2 ))]_0 ^1 −3[∫((ln^2 x)/x)dx−∫((ln^2 x)/(x−1))dx]_0 ^1 −3[∫((ln^2 x)/(x(x−1)))dx−∫((ln^2 x)/((x−1)^2 ))dx]_0 ^1 =−ln^3 x+3ψ′′(1)+3[∫((ln^2 x)/x)dx−∫((ln^2 x)/(x−1))dx]−3[−((ln^2 x)/(x−1))+2∫((lnx)/(x(x−1)))dx] =−ln^3 x+3ψ′′(1)+ln^3 x−3ψ′′(1)+((3ln^2 x)/(x−1))−6[∫((lnx)/x)dx+∫((lnx)/(x−1))dx] =((3ln^2 x)/(x−1))−3ln^2 x−6ψ′(1)=6Σ_(n=0) ^∞ (1/((n+1)^2 ))=6ζ(2)=π^2
Φ=∫0∞(lnxx−1)3dx=∫01(lnxx−1)3dx+∫1∞(lnxx−1)3dx=∫01ln3x(x−1)3dx+∫01xln3x(x−1)3dx=∫01ln3x(x−1)2dx+2∫01ln3x(x−1)3dx=[−ln3xx−1+3∫ln2xx(x−1)dx]01+2[−ln3x2(x−1)2+32∫ln2xx(x−1)2dx]01=[−ln3xx−1−ln3x(x−1)2]01−3[∫ln2xxdx−∫ln2xx−1dx]01−3[∫ln2xx(x−1)dx−∫ln2x(x−1)2dx]01=−ln3x+3ψ″(1)+3[∫ln2xxdx−∫ln2xx−1dx]−3[−ln2xx−1+2∫lnxx(x−1)dx]=−ln3x+3ψ″(1)+ln3x−3ψ″(1)+3ln2xx−1−6[∫lnxxdx+∫lnxx−1dx]=3ln2xx−1−3ln2x−6ψ′(1)=6∑∞n=01(n+1)2=6ζ(2)=π2
Answered by mathmax by abdo last updated on 05/May/21
I =∫_0 ^∞ ((log^3 x)/((1−x)^3 ))dx ⇒I =−∫_0 ^1 ((log^3 x)/((1−x)^3 ))dx−∫_1 ^∞ ((log^3 x)/((1−x)^3 ))dx(→x=(1/t)) =−∫_0 ^1 ((log^3 x)/((1−x)^3 ))+∫_0 ^1 ((−log^3 t)/((1−(1/t))^3 ))(−(dt/t^2 )) =−∫_0 ^1 ((log^3 x)/((1−x)^3 ))dx−∫_0 ^1 ((tlog^3 t)/((1−t)^3 )) dt =−∫_0 ^1 (((1+x)log^3 x)/((1−x)^3 ))dx we have (1/(1−x))=Σ_(n=0) ^∞ x^n ⇒(1/((1−x)^2 ))=Σ_(n=1) ^∞ nx^(n−1) ⇒((2(1−x))/((1−x)^4 )) =Σ_(n=2) ^∞ n(n−1)x^(n−2) ⇒(2/((1−x)^3 )) =Σ_(n=2) ^∞ n(n−1)x^(n−2) ⇒ I =−(1/2)∫_0 ^1 Σ_(n=2) ^∞ n(n−1)x^(n−2) (1+x)log^3 x dx =−(1/2)Σ_(n=2) ^∞ n(n−1)∫_0 ^1 (x^(n−2) +x^(n−1) )log^3 xdx =−(1/2)Σ_(n=2) ^∞ n(n−1) X_n X_n =∫_0 ^1 (x^(n−2) +x^(n−1) )log^3 xdx =[(x^(n−1) /(n−1))+(x^n /n))log^3 x]_0 ^1 −∫_0 ^1 ((x^(n−1) /(n−1))+(x^n /n))3log^2 x (dx/x) =−3 ∫_0 ^1 ((x^(n−2) /(n−1)) +(x^(n−1) /n))log^2 x dx =−3{ [(x^(n−1) /((n−1)^2 ))+(x^n /n^2 ))log^2 x]_0 ^1 −∫_0 ^1 ((x^(n−1) /((n−1)^2 ))+(x^n /n^2 ))((2logx)/x)dx =6 ∫_0 ^1 ((x^(n−2) /((n−1)^2 )) +(x^(n−1) /n^2 ))log xdx =6{[((x^(n−1) /((n−1)^3 )) +(x^n /n^3 ))logx]_0 ^1 −∫_0 ^1 ((x^(n−1) /((n−1)^3 ))+(x^n /n^3 ))(dx/x) =−6 ∫_0 ^1 ( (x^(n−2) /((n−1)^3 )) +(x^(n−1) /n^3 ))dx =−6[(1/((n−1)^4 ))x^(n−1) +(1/n^4 )x^n ]_0 ^1 =−6((1/((n−1)^4 )) +(1/n^4 )) ⇒ I =3 Σ_(n=2) ^∞ n(n−1)((1/((n−1)^4 ))+(1/n^4 )) =3 Σ_(n=2) ^∞ (n/((n−1)^3 )) +3 Σ_(n=2) ^∞ ((n−1)/n^3 ) =3 Σ_(n=1) ^∞ ((n+1)/n^3 ) +3 Σ_(n=1) ^∞ ((n−1)/n^3 ) =6 Σ_(n=1) ^∞ (1/n^2 ) =6.(π^2 /6)=π^2 ⇒ ∫_0 ^∞ (((logx)/(x−1)))^3 =π^2
I=∫0∞log3x(1−x)3dx⇒I=−∫01log3x(1−x)3dx−∫1∞log3x(1−x)3dx(→x=1t)=−∫01log3x(1−x)3+∫01−log3t(1−1t)3(−dtt2)=−∫01log3x(1−x)3dx−∫01tlog3t(1−t)3dt=−∫01(1+x)log3x(1−x)3dxwehave11−x=∑n=0∞xn⇒1(1−x)2=∑n=1∞nxn−1⇒2(1−x)(1−x)4=∑n=2∞n(n−1)xn−2⇒2(1−x)3=∑n=2∞n(n−1)xn−2⇒I=−12∫01∑n=2∞n(n−1)xn−2(1+x)log3xdx=−12∑n=2∞n(n−1)∫01(xn−2+xn−1)log3xdx=−12∑n=2∞n(n−1)XnXn=∫01(xn−2+xn−1)log3xdx=[xn−1n−1+xnn)log3x]01−∫01(xn−1n−1+xnn)3log2xdxx=−3∫01(xn−2n−1+xn−1n)log2xdx=−3{[xn−1(n−1)2+xnn2)log2x]01−∫01(xn−1(n−1)2+xnn2)2logxxdx=6∫01(xn−2(n−1)2+xn−1n2)logxdx=6{[(xn−1(n−1)3+xnn3)logx]01−∫01(xn−1(n−1)3+xnn3)dxx=−6∫01(xn−2(n−1)3+xn−1n3)dx=−6[1(n−1)4xn−1+1n4xn]01=−6(1(n−1)4+1n4)⇒I=3∑n=2∞n(n−1)(1(n−1)4+1n4)=3∑n=2∞n(n−1)3+3∑n=2∞n−1n3=3∑n=1∞n+1n3+3∑n=1∞n−1n3=6∑n=1∞1n2=6.π26=π2⇒∫0∞(logxx−1)3=π2

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