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Question Number 136921 by mnjuly1970 last updated on 27/Mar/21
evaluation of :: 𝛗=∫_0 ^( 1) ((xln(1+x))/(1+x^2 ))dx solution: 𝛗=^(I.B.P ) [(1/2)ln(1+x^2 )ln(1+x)]_0 ^1 −(1/2){∫_0 ^( 1) ((ln(1+x^2 ))/(1+x))dx=𝚽} 𝛗=(1/2)ln^2 (2)−(1/2) 𝚽 ........✓ 𝚽=∫_0 ^( 1) ((ln(1+x^2 ))/(1+x))dx =??? h(a)=∫_0 ^( 1) ((ln(1+ax^2 ))/(1+x))dx h′(a)=∫_0 ^( 1) (∂/∂_a )(((ln(1+ax^2 ))/(1+x)))dx h′(a)=∫_0 ^( 1) (x^2 /((1+x)(1+ax^2 )))dx h′(a)=(1/(1+a))∫_0 ^( 1) (1/(1+x))dx+(1/(1+a))∫_0 ^( 1) (x/(1+ax^2 ))dx−(1/(1+a))∫_0 ^( 1) (1/(1+ax^2 ))dx =(1/(1+a))ln(2)+(1/2).(1/(a(1+a)))ln(1+a)−((√a)/(a(1+a)))[(tan^(−1) (x(√a) )]_0 ^1 =((ln(2))/(1+a))+(1/2).(1/(a(1+a))) ln(1+a)−((√a)/(a(1+a)))tan^(−1) ((√a) ) ∫_0 ^( 1) h′(a)da=ln^2 (2)+(1/2)∫_0 ^( 1) ((ln(1+a))/a)da−(1/4)ln^2 (2)−((π^2 /(16))) =ln^2 (2)+(π^2 /(24))−(1/4)ln^2 (2)−(π^2 /(16)) =(3/4)ln^2 (2)−(π^2 /(48)) ∴ h(1)=𝚽=(3/4)ln^2 (2)−(π^2 /(48)) ..✓ 𝛗=(1/2)ln^2 (2)−(3/8)ln^2 (2)+(π^2 /(96)) 𝛗=(1/8)ln^2 (2)+(π^2 /(96))
evaluationof::ϕ=∫01xln(1+x)1+x2dxsolution:ϕ=I.B.P[12ln(1+x2)ln(1+x)]01−12{∫01ln(1+x2)1+xdx=Φ}ϕ=12ln2(2)−12Φ……..✓Φ=∫01ln(1+x2)1+xdx=???h(a)=∫01ln(1+ax2)1+xdxh′(a)=∫01∂∂a(ln(1+ax2)1+x)dxh′(a)=∫01x2(1+x)(1+ax2)dxh′(a)=11+a∫0111+xdx+11+a∫01x1+ax2dx−11+a∫0111+ax2dx=11+aln(2)+12.1a(1+a)ln(1+a)−aa(1+a)[(tan−1(xa)]01=ln(2)1+a+12.1a(1+a)ln(1+a)−aa(1+a)tan−1(a)∫01h′(a)da=ln2(2)+12∫01ln(1+a)ada−14ln2(2)−(π216)=ln2(2)+π224−14ln2(2)−π216=34ln2(2)−π248∴h(1)=Φ=34ln2(2)−π248..✓ϕ=12ln2(2)−38ln2(2)+π296ϕ=18ln2(2)+π296
Answered by mathmax by abdo last updated on 27/Mar/21
Φ=∫_0 ^1 ((log(1+x^2 ))/(1+x))dx =ϕ(1) with ϕ(t)=∫_0 ^1 ((log(1+tx^2 ))/(1+x)) (t>0) we have ϕ^′ (t)=∫_0 ^1 (x^2 /((x+1)(tx^2 +1)))dx =(1/t)∫_0 ^1 ((tx^2 +1−1)/((x+1)(tx^2 +1)))dx =(1/t)∫_0 ^1 (dx/(x+1))−(1/t)∫_0 ^1 (dx/((x+1)(tx^2 +1))) but ∫_0 ^1 (dx/(x+1))=ln2 let decompose F(x)=(1/((x+1)(tx^2 +1)))=(a/(x+1)) +((bx+c)/(tx^2 +1)) a=(1/(t+1)) , lim_(x→∞) xF(x)=0 =a+(b/t) ⇒0=at+b ⇒b=−(t/(t+1)) F(0)=1=a+c ⇒c=1−(1/(t+1))=(t/(t+1)) ⇒ F(x)=(1/((t+1)(x+1)))+((−(t/(t+1))x+(t/(t+1)))/(tx^2 +1)) =(1/(t+1)){(1/(x+1))−((tx−t)/(tx^2 +1))} ⇒∫_0 ^1 F(x)dx=(1/(t+1)){ln2−∫_0 ^1 ((tx−t)/(tx^2 +1))dx} and ∫_0 ^1 ((tx−t)/(tx^2 +1))dx =(1/2)∫_0 ^1 ((2tx)/(tx^2 +1))dx−∫_0 ^1 (t/(tx^2 +1))dx((√t)x=y) =(1/2)[ln(tx^2 +1)]_0 ^1 −∫_0 ^(√t) (t/(y^2 +1))(dy/( (√t))) =(1/2)ln(t+1)−(√t)arctan((√t)) ⇒ ϕ^′ (t)=((log2)/t)−(1/(t(t+1))){ln2−(1/2)log(t+1)+(√t)arctan((√t))} =((log2)/t)−((log2)/(t(t+1)))−((log(t+1))/(t(t+1)))−(((√t)arctan((√t)))/(t(t+1))) =((log2)/t)(1−(1/(t+1)))−...=((log2)/t).(t/(t+1))−((log(t+1))/(t(t+1)))−(((√t)arctan((√t)))/(t(t+1))) =((log2)/(t+1))−((1/t)−(1/(t+1)))log(t+1)−(((√t)arctan((√t)))/(t(t+1))) ⇒∫_0 ^1 ϕ^′ (t)dt =log^2 (2)−∫_0 ^1 ((log(t+1))/t)dt+∫_0 ^1 ((log(t+1))/(t+1))dt −∫_0 ^1 (((√t)arctan((√t)))/(t(t+1)))dt ∫_0 ^1 ((log(t+1))/t)dt =[logt.log(t+1)]_0 ^1 −∫_0 ^1 ((logt)/(t+1))dt =−∫_0 ^1 log(t)Σ_(n=0) ^∞ (−1)^n t^n dt=−Σ_(n=0) ^∞ (−1)^n ∫_0 ^1 t^n logt dt U_n =∫_0 ^1 t^n logt dt =[(t^(n+1) /(n+1))logt]_0 ^1 −(1/(n+1))∫_0 ^1 t^n dt =−(1/((n+1)^2 )) ⇒∫_0 ^1 ((log(t+1))/t)dt =Σ_(n=0) ^∞ (((−1)^n )/((n+1)^2 )) =−Σ_(n=1) ^∞ (((−1)^n )/n^2 ) =−(2^(1−2) −1)ξ(2)=−(−(1/2)).(π^2 /6)=(π^2 /(12)) ∫_0 ^1 ((log(t+1))/(t+1))dt =[log^2 (t+1)]_0 ^1 −∫_0 ^1 ((log(t+1))/(t+1))dt ⇒ ∫_0 ^1 ((log(t+1))/(t+1))dt =((ln^2 (2))/2) J=∫_0 ^1 (((√t) arctan((√t)))/(t(t+1)))dt =_((√t)=y) ∫_0 ^1 ((yarctany)/(y^2 (y^2 +1)))(2y)dy =2∫_0 ^1 ((arctan(y))/(y^2 +1))dy =2{ [arctan^2 y]_0 ^1 −∫_0 ^1 ((arctany)/(y^2 +1))} =2×(π^2 /(16))−2∫(...)dy ⇒4∫_0 ^1 ((arctany)/(y^2 +1))dy =(π^2 /8) ⇒ ∫_0 ^1 ((arctany)/(y^2 +1))dy =(π^2 /(32)) ⇒ Φ =ϕ(1)=log^2 (2)−(π^2 /(12)) +((ln^2 (2))/2)−(π^2 /(16))=(3/2)ln^2 (2)−((4π^2 )/(48))−((3π^2 )/(48)) =(3/2)ln^2 (2)−((7π^2 )/(48)) ??
Φ=∫01log(1+x2)1+xdx=φ(1)withφ(t)=∫01log(1+tx2)1+x(t>0)wehaveφ′(t)=∫01x2(x+1)(tx2+1)dx=1t∫01tx2+1−1(x+1)(tx2+1)dx=1t∫01dxx+1−1t∫01dx(x+1)(tx2+1)but∫01dxx+1=ln2letdecomposeF(x)=1(x+1)(tx2+1)=ax+1+bx+ctx2+1a=1t+1,limx→∞xF(x)=0=a+bt⇒0=at+b⇒b=−tt+1F(0)=1=a+c⇒c=1−1t+1=tt+1⇒F(x)=1(t+1)(x+1)+−tt+1x+tt+1tx2+1=1t+1{1x+1−tx−ttx2+1}⇒∫01F(x)dx=1t+1{ln2−∫01tx−ttx2+1dx}and∫01tx−ttx2+1dx=12∫012txtx2+1dx−∫01ttx2+1dx(tx=y)=12[ln(tx2+1)]01−∫0tty2+1dyt=12ln(t+1)−tarctan(t)⇒φ′(t)=log2t−1t(t+1){ln2−12log(t+1)+tarctan(t)}=log2t−log2t(t+1)−log(t+1)t(t+1)−tarctan(t)t(t+1)=log2t(1−1t+1)−…=log2t.tt+1−log(t+1)t(t+1)−tarctan(t)t(t+1)=log2t+1−(1t−1t+1)log(t+1)−tarctan(t)t(t+1)⇒∫01φ′(t)dt=log2(2)−∫01log(t+1)tdt+∫01log(t+1)t+1dt−∫01tarctan(t)t(t+1)dt∫01log(t+1)tdt=[logt.log(t+1)]01−∫01logtt+1dt=−∫01log(t)∑n=0∞(−1)ntndt=−∑n=0∞(−1)n∫01tnlogtdtUn=∫01tnlogtdt=[tn+1n+1logt]01−1n+1∫01tndt=−1(n+1)2⇒∫01log(t+1)tdt=∑n=0∞(−1)n(n+1)2=−∑n=1∞(−1)nn2=−(21−2−1)ξ(2)=−(−12).π26=π212∫01log(t+1)t+1dt=[log2(t+1)]01−∫01log(t+1)t+1dt⇒∫01log(t+1)t+1dt=ln2(2)2J=∫01tarctan(t)t(t+1)dt=t=y∫01yarctanyy2(y2+1)(2y)dy=2∫01arctan(y)y2+1dy=2{[arctan2y]01−∫01arctanyy2+1}=2×π216−2∫(…)dy⇒4∫01arctanyy2+1dy=π28⇒∫01arctanyy2+1dy=π232⇒Φ=φ(1)=log2(2)−π212+ln2(2)2−π216=32ln2(2)−4π248−3π248=32ln2(2)−7π248??

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