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Find-out-0-1-ln-1-t-t-2-dt-Then-deduce-the-value-of-A-n-1-1-n-n-1-2n-1-n-

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Question Number 79128 by ~blr237~ last updated on 22/Jan/20
Find out ∫_0 ^1 ln(1−t+t^2 )dt Then deduce the value of A=Σ_(n=1) ^∞ (1/(n(n+1) (((2n+1)),(n) )))
Findout∫01ln(1−t+t2)dtThendeducethevalueofA=∑∞n=11n(n+1)(2n+1n)
Commented by mathmax by abdo last updated on 24/Jan/20
let I =∫_0 ^1 ln(1−t +t^2 )dt by parts I =[t ln(1−t +t^2 )]_0 ^1 −∫_0 ^1 t ×((2t−1)/(t^2 −t+1))dt =−∫_0 ^1 ((2t^2 −t)/(t^2 −t +1))dt =−∫_0 ^1 ((2(t^2 −t +1)+2t−2−t)/(t^2 −t +1))dt =−2 −∫_0 ^1 ((t−2)/(t^2 −t +1))dt =−2−(1/2) ∫_0 ^1 ((2t−1−3)/(t^2 −t +1))dt =−2 −(1/2)[ln(t^2 −t+1)]_0 ^1 +(3/2) ∫_0 ^1 (dt/((t−(1/2))^2 +(3/4))) =−2+(3/2) ∫_0 ^1 (dt/((t−(1/2))^2 +(3/4))) changement t−(1/2) =((√3)/2)u give ∫_0 ^1 (dt/((t−(1/2))^2 +(3/4))) =(4/3)∫_(−(1/( (√3)))) ^(1/( (√3))) (1/(u^2 +1))×((√3)/2)du =(2/( (√3))) ∫_(−(1/( (√3)))) ^(1/( (√3))) (du/(1+u^2 )) =(4/( (√3)))[arctan(u)]_0 ^(1/( (√3))) =(4/( (√3)))×(π/6) =((2π)/(3(√3))) ⇒ I =−2 +(3/2)×((2π)/(3(√3))) =−2+(π/3)
letI=∫01ln(1−t+t2)dtbypartsI=[tln(1−t+t2)]01−∫01t×2t−1t2−t+1dt=−∫012t2−tt2−t+1dt=−∫012(t2−t+1)+2t−2−tt2−t+1dt=−2−∫01t−2t2−t+1dt=−2−12∫012t−1−3t2−t+1dt=−2−12[ln(t2−t+1)]01+32∫01dt(t−12)2+34=−2+32∫01dt(t−12)2+34changementt−12=32ugive∫01dt(t−12)2+34=43∫−13131u2+1×32du=23∫−1313du1+u2=43[arctan(u)]013=43×π6=2π33⇒I=−2+32×2π33=−2+π3
Commented by mathmax by abdo last updated on 24/Jan/20
I =−2+(π/( (√3)))
I=−2+π3
Commented by mathmax by abdo last updated on 24/Jan/20
we have I =∫_0 ^1 ln(1−(t−t^2 ))dt we have ∣t−t^2 ∣=∣t∣.∣1−t^2 ∣<1 ⇒ ln^′ (1−u) =−(1/(1−u)) =−Σ_(n=0) ^∞ u^n ⇒ln(1−u)=−Σ_(n=0) ^∞ (u^(n+1) /(n+1)) =−Σ_(n=1) ^∞ (u^n /n) ⇒ln(1−(t−t^2 )) =−Σ_(n=1) ^∞ (((t−t^2 )^n )/n) ⇒ I =−Σ_(n=1) ^∞ (1/n) ∫_0 ^1 t^n (1−t)^n dt let remember that B(p,q)=∫_0 ^1 x^(p−1) (1−x)^(q−1) dx =((Γ(p).Γ(q))/(Γ(p+q))) ⇒ ∫_0 ^1 t^n (1−t)^n dt =B(n+1,n+1) =((Γ(n+1)Γ(n+1))/(Γ(2n+2))) =(((n!)^2 )/((2n+1)!)) ( Γ(m)=(m−1)!) we have so I =−Σ_(n=1) ^∞ (1/n)×(((n!)^2 )/((2n+1)!)) A =Σ_(n=1) ^∞ (1/(n(n+1)C_(2n+1) ^n )) =Σ_(n=1) ^∞ (1/(n(n+1)(((2n+1)!)/(n!(n+1)!)))) =Σ_(n=1) ^∞ (((n!)^2 )/(n(2n+1)!)) ⇒ A =−I
wehaveI=∫01ln(1−(t−t2))dtwehave∣t−t2∣=∣t∣.∣1−t2∣<1⇒ln′(1−u)=−11−u=−∑n=0∞un⇒ln(1−u)=−∑n=0∞un+1n+1=−∑n=1∞unn⇒ln(1−(t−t2))=−∑n=1∞(t−t2)nn⇒I=−∑n=1∞1n∫01tn(1−t)ndtletrememberthatB(p,q)=∫01xp−1(1−x)q−1dx=Γ(p).Γ(q)Γ(p+q)⇒∫01tn(1−t)ndt=B(n+1,n+1)=Γ(n+1)Γ(n+1)Γ(2n+2)=(n!)2(2n+1)!(Γ(m)=(m−1)!)wehavesoI=−∑n=1∞1n×(n!)2(2n+1)!A=∑n=1∞1n(n+1)C2n+1n=∑n=1∞1n(n+1)(2n+1)!n!(n+1)!=∑n=1∞(n!)2n(2n+1)!⇒A=−I
Commented by mathmax by abdo last updated on 24/Jan/20
⇒ A =2−(π/( (√3))) .
⇒A=2−π3.
Answered by mind is power last updated on 22/Jan/20
=∫_0 ^1 ln((1−t+t^2 )dt by part=[tln(1−t+t^2 )]_0 ^1 −∫_0 ^1 (((2t−1)t)/(1−t+t^2 ))dt =−∫_0 ^1 ((2t^2 −t)/(1−t+t^2 ))dt−∫_0 ^1 (((2t^2 −2t+2)−1+t)/(t^2 −t+1))dt =−2+∫_0 ^1 ((t−1)/(t^2 −t+1))=−2+∫_0 ^1 ((t−(1/2))/(t^2 −t+1))dt−(1/2)∫_0 ^1 (dt/((t−(1/2))^2 +(3/4))) =−2−[_0 ^1 (1/( (√3)))arctan(((2t−1)/( (√3))))]=−2−(2/( (√3)))arctan((1/( (√3))))=−2−(π/(3(√3))). ln(1−t+t^2 )=ln(1−(t−t^2 ))=−Σ_(n≥1) (((t−t^2 )^n )/n) ∫_0 ^1 ln(1−t+t^2 )dt=∫_0 ^1 −Σ_(n≥1) (((t−t^2 )^n )/n)dt =−∫_0 ^1 Σ_(n≥1) (((t−t^2 )^n )/n)dt=−Σ_(n≥1) ∫_0 ^1 ((t^n (1−t)^n )/n)dt =−Σ_(n≥1) (1/n)β(n+1,n+1)=−Σ_(n≥1) ((n!.n!)/(n(2n+1)!)) =−Σ_(n≥1) (1/(n.(((2n+1)!(n+1))/(n!.(2n+1−n)!.n!))))=−Σ_(n≥1) (1/(n(n+1).(((2n+1)!)/(n!(n+1)!)))) =−Σ_(n≥1) (1/(n(n+1) (((2n+1)),(n) )))=−2−(π/(3(√3))) by first ⇒A=2+(π/(3(√3)))
=∫01ln((1−t+t2)dtbypart=[tln(1−t+t2)]01−∫01(2t−1)t1−t+t2dt=−∫012t2−t1−t+t2dt−∫01(2t2−2t+2)−1+tt2−t+1dt=−2+∫01t−1t2−t+1=−2+∫01t−12t2−t+1dt−12∫01dt(t−12)2+34=−2−[0113arctan(2t−13)]=−2−23arctan(13)=−2−π33.ln(1−t+t2)=ln(1−(t−t2))=−∑n⩾1(t−t2)nn∫01ln(1−t+t2)dt=∫01−∑n⩾1(t−t2)nndt=−∫01∑n⩾1(t−t2)nndt=−∑n⩾1∫01tn(1−t)nndt=−∑n⩾11nβ(n+1,n+1)=−∑n⩾1n!.n!n(2n+1)!=−∑n⩾11n.(2n+1)!(n+1)n!.(2n+1−n)!.n!=−∑n⩾11n(n+1).(2n+1)!n!(n+1)!=−∑n⩾11n(n+1)(2n+1n)=−2−π33byfirst⇒A=2+π33
Commented by ~blr237~ last updated on 23/Jan/20
nice sir
nicesir
Commented by mind is power last updated on 23/Jan/20
thanx
thanx

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