please help me prouve that ∫_0 ^1 ((lnt)/(t^2 −1))dt=(π^2 /8)


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Question Number 164419 by akornes last updated on 16/Jan/22

$p l e a s e h e l p m e$ $p r o u v e t h a t \int_{0}^{1} \frac{l n t}{t^{2} - 1} d t = \frac{π^{2}}{8}$
	Answered by Ar Brandon last updated on 17/Jan/22
	$I=∫_0 ^1 ((lnt)/(t^2 −1))dt=−Σ_(n=0) ^∞ ∫_0 ^1 t^(2n) lntdt , (∵(1/(1−α))=Σ_(n=0) ^∞ α^n ) { ((u(t)=lnt)),((v′(t)=t^(2n) )) :} ⇒ { ((u′(t)=(1/t))),((v(t)=(t^(2n+1) /(2n+1)))) :} I=−Σ_(n=0) ^∞ {[(t^(2n+1) /(2n+1))lnt]_0 ^1 −(1/(2n+1))∫_0 ^1 t^(2n) dt} =Σ_(n=0) ^∞ (1/(2n+1))[(t^(2n+1) /(2n+1))]_0 ^1 =Σ_(n=0) ^∞ (1/((2n+1)^2 ))=(3/4)ζ(2)=(π^2 /8)$
	$I = \int_{0}^{1} \frac{\ln t}{t^{2} - 1} d t = - \underset{n = 0}{\sum^{\infty}} \int_{0}^{1} t^{2 n} \ln t d t, (∵ \frac{1}{1 - α} = \underset{n = 0}{\sum^{\infty}} α^{n})$ ${\begin{cases} u (t) = \ln t \\ v^{'} (t) = t^{2 n} \end{cases} \Rightarrow {\begin{cases} u^{'} (t) = \frac{1}{t} \\ v (t) = \frac{t^{2 n + 1}}{2 n + 1} \end{cases}$ $I = - \underset{n = 0}{\sum^{\infty}} {{[\frac{t^{2 n + 1}}{2 n + 1} \ln t]}_{0}^{1} - \frac{1}{2 n + 1} \int_{0}^{1} t^{2 n} d t}$ $= \underset{n = 0}{\sum^{\infty}} \frac{1}{2 n + 1} {[\frac{t^{2 n + 1}}{2 n + 1}]}_{0}^{1} = \underset{n = 0}{\sum^{\infty}} \frac{1}{{(2 n + 1)}^{2}} = \frac{3}{4} ζ (2) = \frac{π^{2}}{8}$
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