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Question Number 138710 by bramlexs22 last updated on 17/Apr/21
∫^( ∞) _1 ((ln x)/((1+x)(1+x^2 ))) dx =?
∫1∞lnx(1+x)(1+x2)dx=?
Answered by mathmax by abdo last updated on 17/Apr/21
Φ=∫_1 ^∞ ((logx)/((1+x)(1+x^2 )))dx ⇒Φ=_(x=(1/t)) −∫_0 ^1 ((−logt)/((1+(1/t))(1+(1/t^2 ))))(−(dt/t^2 )) =−∫_0 ^1 ((tlogt)/((t+1)(t^2 +1)))dt =−∫_0 ^1 (((t+1−1)logt)/((t+1)(t^2 +1)))dt =−∫_0 ^1 ((logt)/(t^2 +1))dt +∫_0 ^1 ((logt)/((t+1)(t^2 +1)))dt we have ∫_0 ^1 ((logt)/(t^2 +1))dt =∫_0 ^1 logtΣ_(n0) ^∞ (−1)^n t^(2n) dt =Σ_(n=0) ^∞ (−1)^n ∫_0 ^1 t^(2n) logt dt =Σ_(n=0) ^∞ (−1)^n u_n u_n =[(t^(2n+1) /(2n+1))logt]_0 ^1 −∫_0 ^1 (t^(2n) /(2n+1))dt =−(1/((2n+1)^2 )) ⇒∫_0 ^1 ((logt)/(t^2 +1))dt =−Σ_(n=0) ^∞ (((−1)^n )/((2n+1)^2 )) =−K(katalan constant) ∫_0 ^1 ((logt)/((t+1)(t^2 +1)))dt =(1/2)∫_0 ^1 ((1/(t+1))−((t−1)/(t^2 +1)))logt dt =(1/2)∫_0 ^1 ((logt)/(t+1))dt −(1/2)∫_0 ^1 (((t−1)logt)/(t^2 +1))dt ∫_0 ^1 ((logt)/(t+1))dt =[log(t+1)logt]_0 ^1 −∫_0 ^1 ((log(t+1))/t)dt we have (d/dt)log(1+t)=(1/(1+t)) =Σ_(n=0) ^∞ (−1)^n t^n ⇒ log(1+t)=Σ_(n=0) ^∞ (−1)^n (t^(n+1) /(n+1))+c(c=0) =Σ_(n=1) ^∞ (((−1)^(n−1) t^n )/n) ⇒ ∫_0 ^1 ((log(1+t))/t)dt =Σ_(n=1) ^∞ ∫_0 ^1 (((−1)^(n−1) t^(n−1) )/n)dt =Σ_(n=1) ^∞ (((−1)^(n−1) )/n^2 ) =−Σ_(n=1) ^∞ (((−1)^n )/n^2 ) =−(2^(1−2) −1)ξ(2) =−((1/2)−1)(π^2 /6) =(π^2 /(12)) ∫_0 ^1 (((t−1)logt)/(t^2 +1))dt =∫_0 ^1 (t−1)logtΣ_(n=0) ^∞ (−1)^n t^(2n) dt =Σ_(n=0) ^∞ (−1)^n ∫_0 ^1 (t^(2n+1) −t^(2n) )logt dt =Σ_(n=0) ^∞ (−1)^n v_n with v_n =∫_0 ^1 (t^(2n+1) −t^(2n) )logt dt =[((t^(2n+2) /(2n+2))−(t^(2n+1) /(2n+1)))logt]_0 ^1 −∫_0 ^1 ((t^(2n+1) /(2n+2))−(t^(2n) /(2n+1)))dt =−(1/((2n+2)^2 ))+(1/((2n+1)^2 )) ⇒ ∫_0 ^1 (((t−1)logt)/(t^2 +1))dt =−Σ_(n=0) ^∞ (((−1)^n )/(4(n+1)^2 ))+Σ_(n=0) ^∞ (((−1)^n )/((2n+1)^2 )) =−(1/4)Σ_(n=1) ^∞ (((−1)^(n−1) )/n^2 )+K =(1/4)Σ_(n=1) ^∞ (((−1)^n )/n^2 )+K =(1/4)(2^(1−2) −1)ξ(2)=(1/4)(−(1/2)).(π^2 /6) =−(π^2 /(48)) rest to collect the values...
Φ=∫1∞logx(1+x)(1+x2)dx⇒Φ=x=1t−∫01−logt(1+1t)(1+1t2)(−dtt2)=−∫01tlogt(t+1)(t2+1)dt=−∫01(t+1−1)logt(t+1)(t2+1)dt=−∫01logtt2+1dt+∫01logt(t+1)(t2+1)dtwehave∫01logtt2+1dt=∫01logt∑n0∞(−1)nt2ndt=∑n=0∞(−1)n∫01t2nlogtdt=∑n=0∞(−1)nunun=[t2n+12n+1logt]01−∫01t2n2n+1dt=−1(2n+1)2⇒∫01logtt2+1dt=−∑n=0∞(−1)n(2n+1)2=−K(katalanconstant)∫01logt(t+1)(t2+1)dt=12∫01(1t+1−t−1t2+1)logtdt=12∫01logtt+1dt−12∫01(t−1)logtt2+1dt∫01logtt+1dt=[log(t+1)logt]01−∫01log(t+1)tdtwehaveddtlog(1+t)=11+t=∑n=0∞(−1)ntn⇒log(1+t)=∑n=0∞(−1)ntn+1n+1+c(c=0)=∑n=1∞(−1)n−1tnn⇒∫01log(1+t)tdt=∑n=1∞∫01(−1)n−1tn−1ndt=∑n=1∞(−1)n−1n2=−∑n=1∞(−1)nn2=−(21−2−1)ξ(2)=−(12−1)π26=π212∫01(t−1)logtt2+1dt=∫01(t−1)logt∑n=0∞(−1)nt2ndt=∑n=0∞(−1)n∫01(t2n+1−t2n)logtdt=∑n=0∞(−1)nvnwithvn=∫01(t2n+1−t2n)logtdt=[(t2n+22n+2−t2n+12n+1)logt]01−∫01(t2n+12n+2−t2n2n+1)dt=−1(2n+2)2+1(2n+1)2⇒∫01(t−1)logtt2+1dt=−∑n=0∞(−1)n4(n+1)2+∑n=0∞(−1)n(2n+1)2=−14∑n=1∞(−1)n−1n2+K=14∑n=1∞(−1)nn2+K=14(21−2−1)ξ(2)=14(−12).π26=−π248resttocollectthevalues…
Answered by Kamel last updated on 17/Apr/21
Ω=∫_1 ^(+∞) ((Ln(x))/((1+x)(1+x^2 )))dx=−∫_0 ^1 ((tLn(t)dt)/((1+t)(1+t^2 ))) (t/((1+t)(1+t^2 )))=(1/2)(((t(1−t))/((1−t^2 )))+((t(1−t))/(1+t^2 )))=−(1/2)((1/(1+t))−((1+t)/(1+t^2 ))) ∴ Ω=^(u=t^2 ) (1/2)(∫_0 ^1 ((Ln(t))/(1+t))dt−(1/4)∫_0 ^1 ((Ln(u))/(1+u))du−∫_0 ^1 ((Ln(t))/(1+t^2 ))dt)=−(3/8)∫_0 ^1 ((Ln(1+t))/t)dt+(G/2) =(3/8)Li_2 (−1)+(G/2)=(G/2)−(π^2 /(32))
Ω=∫1+∞Ln(x)(1+x)(1+x2)dx=−∫01tLn(t)dt(1+t)(1+t2)t(1+t)(1+t2)=12(t(1−t)(1−t2)+t(1−t)1+t2)=−12(11+t−1+t1+t2)∴Ω=u=t212(∫01Ln(t)1+tdt−14∫01Ln(u)1+udu−∫01Ln(t)1+t2dt)=−38∫01Ln(1+t)tdt+G2=38Li2(−1)+G2=G2−π232
Answered by qaz last updated on 17/Apr/21
∫_1 ^∞ ((ln x)/((1+x)(1+x^2 )))dx=∫_0 ^1 ((−xlnx)/((1+x)(1+x^2 )))dx =(1/2)∫_0 ^1 ((1/(1+x))−((1+x)/(1+x^2 )))lnxdx =−(1/2)∫_0 ^1 ((ln(1+x))/x)dx−(1/2)Σ_(n=0) ^∞ (−1)^n ∫_0 ^1 x^(2n) lnxdx+(1/4)∫_0 ^1 ((ln(1+x^2 ))/x)dx =(1/2)Li_2 (−1)+(1/2)Σ_(n=0) ^∞ (((−1)^n )/((2n+1)^2 ))+(1/4)Σ_(n=1) ^∞ (((−1)^(n−1) )/n)∫_0 ^1 x^(2n−1) dx =(1/2)Li_2 (−1)+(1/2)G+(1/8)Σ_(n=1) ^∞ (((−1)^(n−1) )/n^2 ) =(1/2)Li_2 (−1)+(1/2)G−(1/8)Li_2 (−1) =(1/2)G+(3/8)(2^(1−2) −1)ζ(2) =(1/2)G−(π^2 /(32))
∫1∞lnx(1+x)(1+x2)dx=∫01−xlnx(1+x)(1+x2)dx=12∫01(11+x−1+x1+x2)lnxdx=−12∫01ln(1+x)xdx−12∑∞n=0(−1)n∫01x2nlnxdx+14∫01ln(1+x2)xdx=12Li2(−1)+12∑∞n=0(−1)n(2n+1)2+14∑∞n=1(−1)n−1n∫01x2n−1dx=12Li2(−1)+12G+18∑∞n=1(−1)n−1n2=12Li2(−1)+12G−18Li2(−1)=12G+38(21−2−1)ζ(2)=12G−π232
Answered by Ar Brandon last updated on 17/Apr/21
Ω=∫_1 ^∞ ((lnx)/((1+x)(1+x^2 )))dx , x=(1/t) =∫_1 ^0 ((ln((1/t)))/((1+(1/t))(1+(1/t^2 ))))∙(−(dt/t^2 ))=−∫_0 ^1 ((tlnt)/((1+t)(1+t^2 )))dt (t/((t+1)(t^2 +1)))=(a/(t+1))+((bt+c)/(t^2 +1))=((a(t^2 +1)+(bt+c)(t+1))/((t+1)(t^2 +1))) a=−(1/2) , a+b=0 , b=(1/2), a+c=0, c=(1/2) (t/((t+1)(t^2 +1)))=−(1/(2(1+t)))+((1+t)/(2(1+t^2 ))) Ω=(1/2)∫_0 ^1 ((lnt)/(1+t))dt−(1/2)∫_0 ^1 ((lnt)/(1+t^2 ))dt−(1/2)∫_0 ^1 ((tlnt)/(1+t^2 ))dt =(1/2)∫_0 ^1 (((1−t)lnt)/(1−t^2 ))dt−(1/2)∫_0 ^(π/4) ln(tanθ)dθ−(1/8)∫_0 ^1 ((ln(u))/(1+u))du =(1/8)∫_0 ^1 (((v^(−(1/2)) −1)ln(v))/(1−v))dv+(G/2)−(1/8)∫_0 ^1 (((1−u)ln(u))/(1−u^2 ))du =(1/8)∫_0 ^1 (((v^(−(1/2)) −1)ln(v))/(1−v))dv+(G/2)−(1/(32))∫_0 ^1 (((p^(−(1/2)) −1)ln(p))/(1−p))dp =(1/8){ψ′(1)−ψ′((1/2))}+(G/2)−(1/(32)){ψ′(1)−ψ′((1/2))} =(1/8)(ζ(2)−3ζ(2))+(G/2)−(1/(32))(ζ(2)−3ζ(2)) =(G/2)+(3/(32))((π^2 /6)−(π^2 /2))=(G/2)−(π^2 /(32))
Ω=∫1∞lnx(1+x)(1+x2)dx,x=1t=∫10ln(1t)(1+1t)(1+1t2)⋅(−dtt2)=−∫01tlnt(1+t)(1+t2)dtt(t+1)(t2+1)=at+1+bt+ct2+1=a(t2+1)+(bt+c)(t+1)(t+1)(t2+1)a=−12,a+b=0,b=12,a+c=0,c=12t(t+1)(t2+1)=−12(1+t)+1+t2(1+t2)Ω=12∫01lnt1+tdt−12∫01lnt1+t2dt−12∫01tlnt1+t2dt=12∫01(1−t)lnt1−t2dt−12∫0π4ln(tanθ)dθ−18∫01ln(u)1+udu=18∫01(v−12−1)ln(v)1−vdv+G2−18∫01(1−u)ln(u)1−u2du=18∫01(v−12−1)ln(v)1−vdv+G2−132∫01(p−12−1)ln(p)1−pdp=18{ψ′(1)−ψ′(12)}+G2−132{ψ′(1)−ψ′(12)}=18(ζ(2)−3ζ(2))+G2−132(ζ(2)−3ζ(2))=G2+332(π26−π22)=G2−π232

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