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Advanced-Calculus-Calculate-i-I-0-1-ln-x-ln-1-x-dx-ii-J-0-1-Li-2-1-x-2-Note-Li-2-x-

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Question Number 147287 by mnjuly1970 last updated on 19/Jul/21
...Advanced Calculus... Calculate :: { (( i :: I := ∫_0 ^( 1) ln(x).ln(1+x) dx)),(( ii :: J := ∫_0 ^( 1) Li_( 2) ( 1− x^( 2) ) =?)) :} Note:: Li_2 (x) = Σ_(n=1) ^( ∞) (x^( n) /n^( 2) ) ........ ■ .... m.n....
…AdvancedCalculus…Calculate::{i::I:=∫01ln(x).ln(1+x)dxii::J:=∫01Li2(1−x2)=?Note::Li2(x)=∑∞n=1xnn2……..◼….m.n….
Answered by mindispower last updated on 19/Jul/21
I=[(xln(x)−x)ln(1+x)]_0 ^1 −∫_0 ^1 ((xln(x))/(1+x))−(x/(1+x))dx =−ln(2)−∫_0 ^1 (((x+1−1)ln(x))/(1+x))−1+(1/(1+x))dx =−ln(2)−∫_0 ^1 ln(x)dx+∫_0 ^1 ((ln(x))/(1+x))dx+∫dx−∫(dx/(1+x)) =−ln(2)+1−∫_0 ^1 ((ln(1+x))/x)dx+1−ln(2) =−ln(2)−∫_0 ^1 ((ln(1−(−x)))/(−x))d(−x)+2 =ln(2)+[Li_2 (−x)]_0 ^1 =−2ln(2)+Li_2 (−1)+2=−ln(4)−(π^2 /(12))+2 J=[xli_2 (1−x^2 )]−∫_0 ^1 ((2x^2 ln(x^2 ))/(1−x^2 )) =−4∫_0 ^1 ((ln(x))/(1−x^2 ))dx+4∫_0 ^1 ln(x)dx =−2∫_0 ^1 ((ln(x))/(1−x))dx−2∫_0 ^1 ((ln(x))/(1+x))dx−4 =2Li_2 (1−x)]_0 ^1 +2∫_0 ^1 ((ln(1+x))/x)dx−4 =(π^2 /3)−2Li(−1)−4 =(π^2 /3)+(π^2 /6)−4=(π^2 /2)−4=(1/2)(π^2 −8)
I=[(xln(x)−x)ln(1+x)]01−∫01xln(x)1+x−x1+xdx=−ln(2)−∫01(x+1−1)ln(x)1+x−1+11+xdx=−ln(2)−∫01ln(x)dx+∫01ln(x)1+xdx+∫dx−∫dx1+x=−ln(2)+1−∫01ln(1+x)xdx+1−ln(2)=−ln(2)−∫01ln(1−(−x))−xd(−x)+2=ln(2)+[Li2(−x)]01=−2ln(2)+Li2(−1)+2=−ln(4)−π212+2J=[xli2(1−x2)]−∫012x2ln(x2)1−x2=−4∫01ln(x)1−x2dx+4∫01ln(x)dx=−2∫01ln(x)1−xdx−2∫01ln(x)1+xdx−4=2Li2(1−x)]01+2∫01ln(1+x)xdx−4=π23−2Li(−1)−4=π23+π26−4=π22−4=12(π2−8)
Answered by mathmax by abdo last updated on 19/Jul/21
Υ=∫_0 ^1 ln(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))=Σ_(n=1) ^∞ (((−1)^(n−1) x^n )/n) ⇒Υ=∫_0 ^1 (Σ_(n=1) ^∞ (((−1)^(n−1) x^n )/n))ln(x)dx =Σ_(n=1) ^∞ (((−1)^(n−1) )/n)∫_0 ^1 x^n ln(x)dx U_n =∫_0 ^1 x^n ln(x)dx=[(x^(n+1) /(n+1))ln(x)]_0 ^1 −∫_0 ^1 (x^n /(n+1))dx =−(1/((n+1)^2 )) ⇒Υ=−Σ_(n=1) ^∞ (((−1)^(n−1) )/(n(n+1)^2 ))=Σ_(n=1) ^∞ (((−1)^n )/(n(n+1)^2 )) (1/(x(x+1)^2 ))=(a/x)+(b/(x+1))+(c/((x+1)^2 ))=h(x) a=1 ,c=−1 ,lim_(x→+∞) xh(x)=0=a+b ⇒b=−1 ⇒ h(x)=(1/x)−(1/(x+1))−(1/((x+1)^2 )) ⇒Υ=Σ_(n.1) ^∞ (((−1)^n )/n)−Σ_(n=1) ^∞ (((−1)^n )/(n+1)) −Σ_(n=1) ^∞ (((−1)^n )/((n+1)^2 )) we have Σ_(n=1) ^∞ (((−1)^n )/n)=−ln2 Σ_(n=1) ^∞ (((−1)^n )/(n+1))=Σ_(n=2) ^∞ (((−1)^(n−1) )/n)=ln(2)−1 Σ_(n=1) ^∞ (((−1)^n )/((n+1)^2 ))=Σ_(n=2) ^∞ (((−1)^(n−1) )/n^2 )=−Σ_(n=2) ^∞ (((−1)^n )/n^2 ) =−(Σ_(n=1) ^∞ (((−1)^n )/n^2 )+1)=−(2^(1−2) −1)ξ(2)−1=(1/2)(π^2 /6)−1=(π^2 /(12))−1 ⇒Υ=−ln2−ln2+1−(π^2 /(12))+1 =2−2ln2−(π^2 /(12))
Υ=∫01ln(x)ln(1+x)dxwehaveddxln(1+x)=11+x=∑n=0∞(−1)nxn⇒ln(1+x)=∑n=0∞(−1)nxn+1n+1=∑n=1∞(−1)n−1xnn⇒Υ=∫01(∑n=1∞(−1)n−1xnn)ln(x)dx=∑n=1∞(−1)n−1n∫01xnln(x)dxUn=∫01xnln(x)dx=[xn+1n+1ln(x)]01−∫01xnn+1dx=−1(n+1)2⇒Υ=−∑n=1∞(−1)n−1n(n+1)2=∑n=1∞(−1)nn(n+1)21x(x+1)2=ax+bx+1+c(x+1)2=h(x)a=1,c=−1,limx→+∞xh(x)=0=a+b⇒b=−1⇒h(x)=1x−1x+1−1(x+1)2⇒Υ=∑n.1∞(−1)nn−∑n=1∞(−1)nn+1−∑n=1∞(−1)n(n+1)2wehave∑n=1∞(−1)nn=−ln2∑n=1∞(−1)nn+1=∑n=2∞(−1)n−1n=ln(2)−1∑n=1∞(−1)n(n+1)2=∑n=2∞(−1)n−1n2=−∑n=2∞(−1)nn2=−(∑n=1∞(−1)nn2+1)=−(21−2−1)ξ(2)−1=12π26−1=π212−1⇒Υ=−ln2−ln2+1−π212+1=2−2ln2−π212

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