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Question Number 216742 by Tawa11 last updated on 17/Feb/25

$$\frac{1}{2}{\int }_0^1\frac{\ln (\mathrm{a}-1)}{\mathrm{a}}\mathrm{da}$$

Answered by sniper237 last updated on 17/Feb/25

$$Notdefined!$$$$But\frac{1}{2}{\int }_0^1\frac{ln(1-a)}{a}dais.LetnameditI$$$$Letrecallthat\frac{1}{1-a}=\sum _{n⩾0}{a}^{n}for\mid a\mid <1$$$$Sobyintegrating-ln(1-a)=\sum _{n⩾1}\frac{a^n}{n}$$$$Then-2I=\sum _{n⩾1}\frac{1}{n}{\int }_0^1{a}^{n-1}da=\sum _{n⩾1}\frac{1}{n^2}=\zeta \left(2\right)$$$$SoI=-\frac{{\pi }^2}{12}$$

Commented by Tawa11 last updated on 17/Feb/25

$$\mathrm{Thanks}\mathrm{sir},\mathrm{I}\mathrm{really}\mathrm{appreciate}.$$

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