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Question Number 40885 by prof Abdo imad last updated on 28/Jul/18
prove that 1) ∫_0 ^1 ((t^p ln(t))/(t−1))dt =(π^2 /6) −Σ_(k=1) ^p (1/k^2 ) 2) ∫_0 ^1 ((t^(2p) ln(t))/(t^2 −1))dt =(π^2 /8) −Σ_(k=0) ^(p−1) (1/((2k+1)^2 ))
provethat1)∫01tpln(t)t−1dt=π26−∑k=1p1k22)∫01t2pln(t)t2−1dt=π28−∑k=0p−11(2k+1)2
Commented by math khazana by abdo last updated on 02/Aug/18
1) ∫_0 ^1 ((t^p ln(t))/(t−1)) dt =−∫_0 ^1 t^p ln(t)(Σ_(n=0) ^∞ t^n ))dt =−Σ_(n=0) ^∞ ∫_0 ^1 t^(n+p) ln(t) dt =−Σ_(n=0) ^∞ A_n and by parts A_n =∫_0 ^1 t^(n+p) ln(t)dt =[(1/(n+p+1))t^(n+p+1) ln(t)]_0 ^1 −∫_0 ^1 (1/(n+p+1)) t^(n+p) dt =−(1/((n+p+1)^2 )) ⇒−Σ_(n=0) ^∞ A_n =Σ_(n=0) ^∞ (1/((n+p+1)^2 )) =_(n+p+1=k) Σ_(k=p+1) ^∞ (1/k^2 ) but Σ_(k=1) ^∞ (1/k^2 ) =(π^2 /6) =Σ_(k=1) ^p (1/k^2 ) +Σ_(k=p+1) ^∞ (1/k^2 ) ⇒ Σ_(k=p+1) ^∞ =(π^2 /6) −Σ_(k=1) ^p (1/k^2 ) = ∫_0 ^1 ((t^p ln(t))/(t−1)) dt.
1)∫01tpln(t)t−1dt=−∫01tpln(t)(∑n=0∞tn))dt=−∑n=0∞∫01tn+pln(t)dt=−∑n=0∞AnandbypartsAn=∫01tn+pln(t)dt=[1n+p+1tn+p+1ln(t)]01−∫011n+p+1tn+pdt=−1(n+p+1)2⇒−∑n=0∞An=∑n=0∞1(n+p+1)2=n+p+1=k∑k=p+1∞1k2but∑k=1∞1k2=π26=∑k=1p1k2+∑k=p+1∞1k2⇒∑k=p+1∞=π26−∑k=1p1k2=∫01tpln(t)t−1dt.

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