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advanced-mathematics-please-demonstrate-that-0-1-xlog-1-x-log-1-x-1-4-log-2-m-n-july-1970-

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Question Number 111719 by mnjuly1970 last updated on 04/Sep/20
....advanced mathematics.... please demonstrate that:: Φ =∫_0 ^( 1) xlog(1−x).log(1+x)= (1/4) − log(2) ... m.n.july 1970 #
….advancedmathematics….pleasedemonstratethat::Φ=∫01xlog(1−x).log(1+x)=14−log(2)…You can't use 'macro parameter character #' in math mode
Answered by mathmax by abdo last updated on 05/Sep/20
let take a try let A =∫_0 ^1 xln(1−x)ln(1+x)dx we have (d/dx)ln(1+x) =(1/(1+x)) =Σ_(n=0) ^(∞ ) (−1)^n x^n ⇒ln(1+x) =Σ_(n=0) ^∞ (((−1)^n x^(n+1) )/(n+1)) +c(c=0) =Σ_(n=1) ^∞ (((−1)^(n−1) )/n) x^n ⇒ A =∫_0 ^1 xln(1−x)Σ_(n=1) ^∞ (((−1)^(n−1) )/n) x^n dx =Σ_(n=1) ^∞ (((−1)^(n−1) )/n) ∫_0 ^1 x^(n+1) ln(1−x)dx let U_n =∫_0 ^1 x^(n+1) ln(1−x)dx by parts U_n =[((x^(n+2) −1)/(n+2)) ln(1−x)]_0 ^1 −∫_0 ^1 ((x^(n+2) −1)/(n+2))(((−1)/(1−x)))dx =0 +(1/(n+2)) ∫_0 ^1 ((x^(n+2) −1)/(1−x)) dx =−(1/(n+2)) ∫_0 ^1 ((x^(n+2) −1)/(x−1)) dx =−(1/(n+2)) ∫_0 ^1 (((x−1)(1+x+x^2 +...+x^(n+1) )/(x−1)) dx =−(1/(n+2)) ∫ (1+x+x^2 +...+x^(n+1) )dx =−(1/(n+2))[x+(x^2 /2) +(x^3 /3)+....+(x^(n+2) /(n+2))]_0 ^1 =−(1/(n+2))(1+(1/2)+(1/3)+....+(1/(n+2))) =−(H_(n+2) /(n+2)) (H_n =Σ_(k=1) ^n (1/k)) ⇒ A =−Σ_(n=1) ^∞ (((−1)^(n−1) )/n).(H_(n+2) /(n+2)) =Σ_(n=1) ^∞ (−1)^n ((1/n)−(1/(n+2)))H_(n+2) =Σ_(n=1) ^∞ (((−1)^(n ) H_(n+2) )/n) −Σ_(n=1) ^∞ (((−1)^n H_(n+2) )/(n+2)) Σ_(n=1) ^∞ (((−1)^n H_(n+2) )/n) =_(n+2=p) Σ_(p=3) ^∞ (((−1)^(p−2) H_p )/(p−2)) =Σ_(n=3) ^∞ (((−1)^n H_n )/(n−2)) Σ_(n=1) ^∞ (((−1)^n H_(n+2) )/(n+2)) =Σ_(n=3) ^∞ (((−1)^n H_n )/n) ⇒ A =Σ_(n=3) ^∞ (((−1)^n )/(n−2)) H_n −Σ_(n=3) ^∞ (((−1)^n )/n) H_n rest to find vslues of those series ....be continued...
lettakeatryletA=∫01xln(1−x)ln(1+x)dxwehaveddxln(1+x)=11+x=∑n=0∞(−1)nxn⇒ln(1+x)=∑n=0∞(−1)nxn+1n+1+c(c=0)=∑n=1∞(−1)n−1nxn⇒A=∫01xln(1−x)∑n=1∞(−1)n−1nxndx=∑n=1∞(−1)n−1n∫01xn+1ln(1−x)dxletUn=∫01xn+1ln(1−x)dxbypartsUn=[xn+2−1n+2ln(1−x)]01−∫01xn+2−1n+2(−11−x)dx=0+1n+2∫01xn+2−11−xdx=−1n+2∫01xn+2−1x−1dx=−1n+2∫01(x−1)(1+x+x2+…+xn+1x−1dx=−1n+2∫(1+x+x2+…+xn+1)dx=−1n+2[x+x22+x33+….+xn+2n+2]01=−1n+2(1+12+13+….+1n+2)=−Hn+2n+2(Hn=∑k=1n1k)⇒A=−∑n=1∞(−1)n−1n.Hn+2n+2=∑n=1∞(−1)n(1n−1n+2)Hn+2=∑n=1∞(−1)nHn+2n−∑n=1∞(−1)nHn+2n+2∑n=1∞(−1)nHn+2n=n+2=p∑p=3∞(−1)p−2Hpp−2=∑n=3∞(−1)nHnn−2∑n=1∞(−1)nHn+2n+2=∑n=3∞(−1)nHnn⇒A=∑n=3∞(−1)nn−2Hn−∑n=3∞(−1)nnHnresttofindvsluesofthoseseries….becontinued…
Answered by maths mind last updated on 07/Sep/20
log(1−x)log(1+x)=(1/4)[(log(1−x)+log(1+x))^2 −(log(1−x)−log(1+x))^2 ] =((log^2 (1−x^2 ))/4)−((log^2 (((1−x)/(1+x))))/4) Φ=(1/4)∫_0 ^1 xlog^2 (1−x^2 )dx−(1/4)∫_0 ^1 xlog^2 (((1−x)/(1+x)))dx Φ=(1/4)(I−J) I=∫_0 ^1 xlog^2 (1−x^2 )dx let u=x^2 ⇒xdx=(du/2) I=(1/2)∫_0 ^1 log^2 (1−u)du β(x,y)=∫_0 ^1 t^(x−1) (1−t)^(y−1) ⇒∂^2 yβ(1,1)=2I J=∫_0 ^1 xlog^2 (((1−x)/(1+x)))dx u=((1−x)/(1+x))⇒x=((1−u)/(1+u)) dx=−(2/((1+u)^2 ))du =2∫_0 ^1 (((1−u))/((1+u)^3 )).log^2 (u)du =2∫_0 ^1 (−(1/((1+u)^2 ))+(2/((1+u)^3 )))log^2 (u)du =lim_(y→0) 2[_y ^1 ((1/(1+u))−(1/((1+u)^2 ))).]log^2 (u)_(=0) −4∫_0 ^1 [((log(u))/(u(1+u)))−((log(u))/((1+u)^2 u))]du =−4∫_0 ^1 ((log(u))/((1+u)^2 ))du=4lim_(x→0) ([((log(u))/(1+u))]_x ^1 −∫_x ^1 (du/(u(1+u)))) =4lim_(x→0) [−((log(x))/(1+x))+log(x)+log(2)−log(1+x)] =4log(2) ∂_y β(x,y)=β(x,y){Ψ(y)−Ψ(x+y)} ∂^2 yβ(x,y)=β(x,y)[{Ψ(y)−Ψ(x+y)}^2 +(Ψ′(y)−Ψ′(x+y))] ={Ψ(1)−Ψ(2)}^2 +(Ψ′(1)−Ψ′(2))=(−γ−(−γ−1))+(ζ(2)−(ζ(2)−1))^2 =2 I=(1/2)∂^2 yβ(1,1)=1 Φ=(1/4)(I−J)=(1/4)(1−4log(2))=(1/4)−log(2)
log(1−x)log(1+x)=14[(log(1−x)+log(1+x))2−(log(1−x)−log(1+x))2]=log2(1−x2)4−log2(1−x1+x)4Φ=14∫01xlog2(1−x2)dx−14∫01xlog2(1−x1+x)dxΦ=14(I−J)I=∫01xlog2(1−x2)dxletu=x2⇒xdx=du2I=12∫01log2(1−u)duβ(x,y)=∫01tx−1(1−t)y−1⇒∂2yβ(1,1)=2IJ=∫01xlog2(1−x1+x)dxu=1−x1+x⇒x=1−u1+udx=−2(1+u)2du=2∫01(1−u)(1+u)3.log2(u)du=2∫01(−1(1+u)2+2(1+u)3)log2(u)duMissing \left or extra \right−4∫01[log(u)u(1+u)−log(u)(1+u)2u]du=−4∫01log(u)(1+u)2du=4limx→0([log(u)1+u]x1−∫x1duu(1+u))=4limx→0[−log(x)1+x+log(x)+log(2)−log(1+x)]=4log(2)∂yβ(x,y)=β(x,y){Ψ(y)−Ψ(x+y)}∂2yβ(x,y)=β(x,y)[{Ψ(y)−Ψ(x+y)}2+(Ψ′(y)−Ψ′(x+y))]={Ψ(1)−Ψ(2)}2+(Ψ′(1)−Ψ′(2))=(−γ−(−γ−1))+(ζ(2)−(ζ(2)−1))2=2I=12∂2yβ(1,1)=1Φ=14(I−J)=14(1−4log(2))=14−log(2)

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