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Question Number 137829 by mnjuly1970 last updated on 07/Apr/21

$$.......nice..........calculus.....$$$$provethat::::$$$$\mathit{\varphi }={\int }_0^1{\left(\frac{log(1-x)}{x}\right)}^{2}dx=2\zeta \left(2\right)....$$$$$$

Answered by EnterUsername last updated on 07/Apr/21

$${\int }_0^1{\left(\frac{ln(1-x)}{x}\right)}^{2}dx$$$$={\int }_0^1\frac{{ln}^{2}x}{{(1-x)}^2}dx={[\frac{{ln}^{2}x}{1-x}-2\int \frac{lnx}{x(1-x)}dx]}_0^1$$$$={[\frac{{ln}^{2}x}{1-x}-2\int (\frac{lnx}{x}+\frac{lnx}{1-x})dx]}_0^1$$$$={[\frac{{ln}^{2}x}{1-x}-{ln}^{2}x]}_0^1-2{\int }_0^1\frac{lnx}{1-x}dx$$$$=2{\psi }^{\prime }\left(1\right)=2\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{1}{{(n+1)}^2}=2\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n^2}=2\zeta \left(2\right)=\frac{{\pi }^2}{3}$$

Commented by mnjuly1970 last updated on 07/Apr/21

$$thanksalot...$$

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