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Question Number 65925 by mathmax by abdo last updated on 05/Aug/19
calculate ∫_0 ^1 e^(−2t) ln(1−t)dt
calculate∫01e−2tln(1−t)dt
Commented by mathmax by abdo last updated on 07/Aug/19
let I =∫_0 ^1 e^(−2t) ln(1−t)dt we have ln^′ (1−t) =((−1)/(1−t))=−Σ_(n=0) ^∞ t^n for ∣t∣<1 ⇒ln(1−t) =−Σ_(n=0) ^∞ (t^(n+1) /(n+1)) =−Σ_(n=1) ^∞ (t^n /n) ⇒ I =−∫_0 ^1 e^(−2t) (Σ_(n=1) ^∞ (t^n /n))dt =−Σ_(n=1) ^∞ (1/n) ∫_0 ^1 t^n e^(−2t) dt let A_n =∫_0 ^1 t^n e^(−2t) dt by psrts u^′ =t^n and v=e^(−2t) A_n =[(1/(n+1))t^(n+1) e^(−2t) ]_0 ^1 −∫_0 ^1 (t^(n+1) /(n+1))(−2)e^(−2t) dt =(e^(−2) /(n+1)) +(2/(n+1)) ∫_0 ^1 t^(n+1) e^(−2t) dt⇒(n+1)A_n =e^(−2) +2A_(n+1) ⇒ 2A_(n+1) =(n+1)A_n −e^(−2) ⇒A_(n+1) =(((n+1))/2)A_n −(e^(−2) /2) ⇒ A_n =(n/2)A_(n−1) −(e^(−2) /2) let V_n =2(A_n /(n!)) ⇒ V_(n+1) −V_n =2(A_(n+1) /((n+1)!))−2(A_n /(n!)) =2(((n+1)A_n )/(2(n+1)!))−2(e^(−2) /2)−2(A_n /(n!)) =(A_n /(n!))−2(A_n /(n!)) −e^(−2) =−(1/(n!))A_n −e^(−2) ⇒Σ_(k=0) ^(n−1) (V_(k+1) −V_k ) =−Σ_(k=0) ^(n−1) (A_k /(k!))−ne^(−2) ⇒V_n −V_0 =−Σ_(k=0) ^(n−1) (A_k /(k!)) −ne^(−2) ⇒ V_n =V_0 −ne^(−2) −Σ_(k=0) ^(n−1) (A_k /(k!)) ...be continued...
letI=∫01e−2tln(1−t)dtwehaveln′(1−t)=−11−t=−∑n=0∞tnfor∣t∣<1⇒ln(1−t)=−∑n=0∞tn+1n+1=−∑n=1∞tnn⇒I=−∫01e−2t(∑n=1∞tnn)dt=−∑n=1∞1n∫01tne−2tdtletAn=∫01tne−2tdtbypsrtsu′=tnandv=e−2tAn=[1n+1tn+1e−2t]01−∫01tn+1n+1(−2)e−2tdt=e−2n+1+2n+1∫01tn+1e−2tdt⇒(n+1)An=e−2+2An+1⇒2An+1=(n+1)An−e−2⇒An+1=(n+1)2An−e−22⇒An=n2An−1−e−22letVn=2Ann!⇒Vn+1−Vn=2An+1(n+1)!−2Ann!=2(n+1)An2(n+1)!−2e−22−2Ann!=Ann!−2Ann!−e−2=−1n!An−e−2⇒∑k=0n−1(Vk+1−Vk)=−∑k=0n−1Akk!−ne−2⇒Vn−V0=−∑k=0n−1Akk!−ne−2⇒Vn=V0−ne−2−∑k=0n−1Akk!…becontinued…
Commented by mathmax by abdo last updated on 07/Aug/19
at form of serie we have e^(−2t) =Σ_(n=0) ^∞ (((−2t)^n )/(n!)) and ln(1−t) =−Σ_(n=1) ^∞ (t^n /n) ⇒e^(−2t) ln(1−t) =−(Σ_(n=0) ^∞ (((−2)^n t^n )/(n!)))(Σ_(n=1) ^∞ (t^n /n)) =−(1+Σ_(n=1) ^∞ (((−2)^n t^n )/(n!)))Σ_(n=1) ^∞ (t^n /n) =−Σ_(n=1) ^∞ (t^n /n) −(Σ_(n=1) ^∞ (((−2)^n t^n )/(n!)))(Σ_(n=1) ^∞ (t^n /n)) =ln(1−t)−Σ_(n=1) ^∞ c_n t^n with c_n =Σ_(i+j=1) a_i b_j =Σ_(i=1) ^(n−1) a_i b_(n−i) =Σ_(i=1) ^(n−1) (((−2)^i )/(i!))(1/(n−i)) ⇒ e^(−2t) ln(1−t) =ln(1−t)−Σ_(n=1) ^∞ (Σ_(i=1) ^(n−1) (((−2)^i )/((n−i)i!)))t^n ⇒ ∫_0 ^1 e^(−2t) ln(1−t)dt =∫_0 ^1 ln(1−t)dt −Σ_(n=1) ^∞ (Σ_(i=1) ^(n−1) (((−2)^i )/((n−i)i!)))(1/(n+1)) ∫_0 ^1 ln(1−t)dt =_(1−t =u) −∫_0 ^1 ln(u)(−du) =∫_0 ^1 ln(u)du =[ulnu−u]_0 ^1 =−1 ⇒ ∫_0 ^1 ln(1−t)dt =−1−Σ_(n=1) ^∞ (1/(n+1))(Σ_(i=1) ^(n−1) (((−2)^i )/((n−i)i!)))
atformofseriewehavee−2t=∑n=0∞(−2t)nn!andln(1−t)=−∑n=1∞tnn⇒e−2tln(1−t)=−(∑n=0∞(−2)ntnn!)(∑n=1∞tnn)=−(1+∑n=1∞(−2)ntnn!)∑n=1∞tnn=−∑n=1∞tnn−(∑n=1∞(−2)ntnn!)(∑n=1∞tnn)=ln(1−t)−∑n=1∞cntnwithcn=∑i+j=1aibj=∑i=1n−1aibn−i=∑i=1n−1(−2)ii!1n−i⇒e−2tln(1−t)=ln(1−t)−∑n=1∞(∑i=1n−1(−2)i(n−i)i!)tn⇒∫01e−2tln(1−t)dt=∫01ln(1−t)dt−∑n=1∞(∑i=1n−1(−2)i(n−i)i!)1n+1∫01ln(1−t)dt=1−t=u−∫01ln(u)(−du)=∫01ln(u)du=[ulnu−u]01=−1⇒∫01ln(1−t)dt=−1−∑n=1∞1n+1(∑i=1n−1(−2)i(n−i)i!)
Commented by mathmax by abdo last updated on 07/Aug/19
∫_0 ^1 e^(−2t) ln(1−t)dt =−1−....
∫01e−2tln(1−t)dt=−1−….

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