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Question Number 40887 by prof Abdo imad last updated on 28/Jul/18
calculate ∫_0 ^1 ((tln(t))/(t^2 −1))dt
calculate∫01tln(t)t2−1dt
Commented by prof Abdo imad last updated on 30/Jul/18
let I = ∫_0 ^1 ((tln(t))/(t^2 −1))dt I =−∫_0 ^1 tln(t)(Σ_(n=0) ^∞ t^(2n) )dt =−Σ_(n=0) ^∞ ∫_0 ^1 t^(2n+1) ln(t)dt by parts A_n =∫_0 ^1 t^(2n+1) ln(t)dt =[(1/(2n+2))t^(2n+2) ln(t)]_0 ^1 −∫_0 ^1 (1/(2n+2)) t^(2n+1) dt =−(1/((2n+2)^2 )) ⇒ I =Σ_(n=0) ^∞ (1/((2n+2)^2 )) =(1/4) Σ_(n=0) ^∞ (1/((n+1)^2 )) =(1/4)Σ_(n=1) ^∞ (1/n^2 ) =(1/4) .(π^2 /6) ⇒ I =(π^2 /(24))
letI=∫01tln(t)t2−1dtI=−∫01tln(t)(∑n=0∞t2n)dt=−∑n=0∞∫01t2n+1ln(t)dtbypartsAn=∫01t2n+1ln(t)dt=[12n+2t2n+2ln(t)]01−∫0112n+2t2n+1dt=−1(2n+2)2⇒I=∑n=0∞1(2n+2)2=14∑n=0∞1(n+1)2=14∑n=1∞1n2=14.π26⇒I=π224

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