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Question Number 141346 by mnjuly1970 last updated on 17/May/21

Answered by qaz last updated on 17/May/21

$${\int }_0^1\frac{{ln}^{2}x}{1-x^2}dx$$$$={\int }_0^{\mathrm{\infty }}\frac{y^2{e}^{-y}}{1-e^{-2y}}dy$$$$=\underset{n=0}{\sum ^{\mathrm{\infty }}}{\int }_0^{\mathrm{\infty }}{y}^2{e}^{-(2n+1)y}dy$$$$=2\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{1}{{(2n+1)}^3}$$$$=2(1-2^{-3})\zeta \left(3\right)$$$$=\frac{7}{4}\zeta \left(3\right)$$

Commented by mnjuly1970 last updated on 18/May/21

$$thanksalot...$$

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