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Question Number 141017 by 777316 last updated on 14/May/21

Commented by mnjuly1970 last updated on 15/May/21

$$hellomr:$$$$what‘‘font‘‘and‘‘app‘‘didyou$$$$usetowritemath??$$$$$$

Answered by mnjuly1970 last updated on 14/May/21

$$a_n:\stackrel{x^n=t}{=}{\int }_0^1\frac{ln(1-t)}{t^{\frac{1}{n}}}\left(\frac{1}{n}{t}^{\frac{1}{n}-1}\right)dt$$$$=\frac{1}{n}{\int }_0^1\frac{ln(1-t)}{t}dt=\frac{-{li}_2\left(1\right)}{n}=\frac{-{\pi }^2}{6n}$$$$\mathit{\varphi }:=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{a_n}{n}=\frac{-{\pi }^2}{6}\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n^2}=\frac{-{\pi }^4}{36}...✓$$

Answered by Dwaipayan Shikari last updated on 14/May/21

$${\int }_0^1\frac{log(1-x^n)}{x}dx{x}^n=u$$$$=\frac{1}{n}{\int }_0^1\frac{log(1-u)}{u}du=-\frac{{\pi }^2}{6n}$$$$\underset{n=1}{\sum ^{\mathrm{\infty }}}-\frac{{\pi }^2}{6n^2}=-\frac{{\pi }^4}{36}$$

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