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Question Number 98256 by  M±th+et+s last updated on 12/Jun/20
∫_0 ^∞ ((log(x))/( (√x)(x+1)^2 ))dx
∫0∞log(x)x(x+1)2dx
Answered by mathmax by abdo last updated on 12/Jun/20
A =∫_0 ^∞ ((ln(x))/( (√x)(x+1)^2 ))dx changement (√x)=t give x=t^2 ⇒ A =∫_0 ^∞ ((ln(t^2 ))/(t(t^2 +1)^2 )) (2t)dt =4 ∫_0 ^∞ ((ln(t))/((1+t^2 )^2 ))dt A =4{ ∫_0 ^1 ((ln(t))/((1+t^2 )^2 ))dt +∫_1 ^(+∞) ((lnt)/((1+t^2 )^2 ))dt(→t=(1/u))} =4{ ∫_0 ^1 ((ln(t))/((1+t^2 )^2 ))dt −∫_0 ^1 ((−lnu)/((1+(1/u^2 ))^2 ))(−(du/u^2 ))} =4{ ∫_0 ^1 ((ln(t))/((1+t^2 )^2 ))dt−∫_0 ^1 ((ln(u)u^2 )/((1+u^2 )^2 ))} =4 ∫_0 ^1 (((1−t^2 )lnt)/((1+t^2 )^2 )) dt we have (1/(1+u)) =Σ_(n=0) ^∞ (−1)^n u^n ⇒−(1/((1+u)^2 )) =Σ_(n=1) ^∞ n(−1)^n u^(n−1) ⇒ (1/((1+u)^2 )) =−Σ_(n=1) ^∞ n(−1)^n u^(n−1) ⇒(1/((1+t^2 )^2 )) =−Σ_(n=1) ^∞ n(−1)^n t^(2n−2) ⇒ A =−4 ∫_0 ^1 (1−t^2 )ln(t)(Σ_(n=1) ^∞ n(−1)^n t^(2n−2) )dt =4 Σ_(n=1) ^∞ n(−1)^n ∫_0 ^1 (t^2 −1)t^(2n−2) ln(t)dt =4 Σ_(n=1) ^∞ n(−1)^n w_n W_n =∫_0 ^1 (t^(2n) −t^(2n−2) )ln(t)dt by parts W_n =[((t^(2n+1) /(2n+1)) −(t^(2n−1) /(2n−1)))ln(t)]_0 ^1 −∫_0 ^1 ((t^(2n+1) /(2n+1))−(t^(2n−1) /(2n−1)))(dt/t) =−(1/((2n+1)^2 )) +(1/((2n−1)^2 )) ⇒ A =4 Σ_(n=1) ^∞ n(−1)^n ((1/((2n−1)^2 ))−(1/((2n+1)^2 ))) =4 Σ_(n=1) ^∞ (−1)^n (n/((2n−1)^2 )) −4 Σ_(n=1) ^∞ (−1)^n (n/((2n+1)^2 )) Σ_(n=1) ^∞ ((n(−1)^n )/((2n−1)^2 )) =_(n=p+1) Σ_(p=0) ^∞ (((p+1)(−1)^(p+1) )/((2p+1)^2 )) =−1 −Σ_(n=1) ^∞ (((n+1)(−1)^n )/((2n+1)^2 )) =−1−Σ_(n=1) ^∞ ((n(−1)^n )/((2n+1)^2 )) −Σ_(n=1) ^∞ (((−1)^n )/((2n+1)^2 )) ⇒ A =−4−4Σ_(n=1) ^∞ ((n(−1)^n )/((2n+1)^2 )) −4 Σ_(n=1) ^∞ (((−1)^n )/((2n+1)^2 ))−4 Σ_(n=1) ^∞ ((n(−1)^n )/((2n+1)^2 )) =−4 −4 Σ_(n=1) ^∞ (((−1)^n )/((2n+1)^2 )) −8 Σ_(n=1) ^∞ ((n(−1)^n )/((2n+1)^2 )) Σ_(n=1) ^∞ (((−1)^n )/((2n+1)^2 )) =k−1 (k catalan constsnt) Σ_(n=1) ^∞ ((n(−1)^n )/((2n+1)^2 )) =(1/2)Σ_(n=1) ^∞ (((2n+1−1)(−1)^n )/((2n+1)^2 )) =(1/2) Σ_(n=1) ^∞ (((−1)^n )/((2n+1)))−(1/2)Σ_(n=1) ^∞ (((−1)^n )/((2n+1)^2 )) =(1/2)((π/4)−1)−(1/2)(k−1) ⇒ A =−4 −4(k−1)−8{(π/8)−(1/2) −(1/2)(k−1)} =−4 −4(k−1)−π +4 +4(k−1) =−π ⇒ A =−π
A=∫0∞ln(x)x(x+1)2dxchangementx=tgivex=t2⇒A=∫0∞ln(t2)t(t2+1)2(2t)dt=4∫0∞ln(t)(1+t2)2dtA=4{∫01ln(t)(1+t2)2dt+∫1+∞lnt(1+t2)2dt(→t=1u)}=4{∫01ln(t)(1+t2)2dt−∫01−lnu(1+1u2)2(−duu2)}=4{∫01ln(t)(1+t2)2dt−∫01ln(u)u2(1+u2)2}=4∫01(1−t2)lnt(1+t2)2dtwehave11+u=∑n=0∞(−1)nun⇒−1(1+u)2=∑n=1∞n(−1)nun−1⇒1(1+u)2=−∑n=1∞n(−1)nun−1⇒1(1+t2)2=−∑n=1∞n(−1)nt2n−2⇒A=−4∫01(1−t2)ln(t)(∑n=1∞n(−1)nt2n−2)dt=4∑n=1∞n(−1)n∫01(t2−1)t2n−2ln(t)dt=4∑n=1∞n(−1)nwnWn=∫01(t2n−t2n−2)ln(t)dtbypartsWn=[(t2n+12n+1−t2n−12n−1)ln(t)]01−∫01(t2n+12n+1−t2n−12n−1)dtt=−1(2n+1)2+1(2n−1)2⇒A=4∑n=1∞n(−1)n(1(2n−1)2−1(2n+1)2)=4∑n=1∞(−1)nn(2n−1)2−4∑n=1∞(−1)nn(2n+1)2∑n=1∞n(−1)n(2n−1)2=n=p+1∑p=0∞(p+1)(−1)p+1(2p+1)2=−1−∑n=1∞(n+1)(−1)n(2n+1)2=−1−∑n=1∞n(−1)n(2n+1)2−∑n=1∞(−1)n(2n+1)2⇒A=−4−4∑n=1∞n(−1)n(2n+1)2−4∑n=1∞(−1)n(2n+1)2−4∑n=1∞n(−1)n(2n+1)2=−4−4∑n=1∞(−1)n(2n+1)2−8∑n=1∞n(−1)n(2n+1)2∑n=1∞(−1)n(2n+1)2=k−1(kcatalanconstsnt)∑n=1∞n(−1)n(2n+1)2=12∑n=1∞(2n+1−1)(−1)n(2n+1)2=12∑n=1∞(−1)n(2n+1)−12∑n=1∞(−1)n(2n+1)2=12(π4−1)−12(k−1)⇒A=−4−4(k−1)−8{π8−12−12(k−1)}=−4−4(k−1)−π+4+4(k−1)=−π⇒A=−π
Commented by  M±th+et+s last updated on 12/Jun/20
well done . correct solution
welldone.correctsolution
Answered by maths mind last updated on 13/Jun/20
=∫_0 ^(+∞) ((2log(t))/(t(t^2 +1)^2 )).2tdt=4∫_0 ((log(t))/((t^2 +1)^2 )) ⇔4∫_0 ^(π/2) log(tg(z))cos^2 (z) =4∫_0 ^(π/2) log(sin(z))cos^2 (z)−log(cos(z))cos^2 (z)dz...1 2∫_0 ^(π/2) sin^(2x−1) (t)cos^(2y−1) (t)dt=β(x,y) 1=∂_x β((1/2),(3/2))−∂_y β((1/2),(3/2))=β((1/2),(3/2))(Ψ((1/2))−Ψ((3/2))) =β((1/2),(3/2))(Ψ((1/2))−Ψ((1/2))−2) =−2.((Γ((1/2))Γ((3/2)))/(Γ(2)))=−2Γ((1/2)).2^(−1) .(√π).Γ(2)=−π =
=∫0+∞2log(t)t(t2+1)2.2tdt=4∫0log(t)(t2+1)2⇔4∫0π2log(tg(z))cos2(z)=4∫0π2log(sin(z))cos2(z)−log(cos(z))cos2(z)dz…12∫0π2sin2x−1(t)cos2y−1(t)dt=β(x,y)1=∂xβ(12,32)−∂yβ(12,32)=β(12,32)(Ψ(12)−Ψ(32))=β(12,32)(Ψ(12)−Ψ(12)−2)=−2.Γ(12)Γ(32)Γ(2)=−2Γ(12).2−1.π.Γ(2)=−π=
Commented by maths mind last updated on 13/Jun/20
i used Γ(z)Γ(z+(1/2))=2^(1−2z) (√π).Γ(z) for Γ((3/2)),z=(1/2),Γ((1/2))=(√π)
iusedΓ(z)Γ(z+12)=21−2zπ.Γ(z)forΓ(32),z=12,Γ(12)=π
Commented by  M±th+et+s last updated on 13/Jun/20
very nice sir thanx
verynicesirthanx

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