please-help-me-prouve-that-0-1-lnt-t-2-1-dt-pi-2-8-

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 = \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}