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Question Number 166373 by mnjuly1970 last updated on 19/Feb/22

p r o v e

ϕ = \int_{0}^{1} \frac{{l n}^{2} (1 - x^{2})}{x^{2}} d x = \frac{π^{2}}{3} - 4 {l n}^{2} (2)

- - - s o l u t i o n (t e c h n i c a l m e t h o d) - - -

ϕ = \int_{0}^{1} {l n}^{2} (1 - x^{2}) d (1 - \frac{1}{x})

= {[(1 - \frac{1}{x}) {l n}^{2} (1 - x^{2})]}_{0}^{1} + 4 \int_{0}^{1} (1 - \frac{1}{x}) \frac{x l n (1 - x^{2})}{1 - x^{2}} d x

= - 4 \int_{0}^{1} \frac{l n (1 - x^{2})}{1 + x} d x

= - 4 \int_{0}^{1} \frac{l n (1 + x)}{1 + x} d x - 4 \int_{0}^{1} \frac{l n (1 - x) d x}{1 + x}

= - 2 {l n}^{2} (2) - 4 (- \frac{π^{2}}{12} + \frac{1}{2} {l n}^{2} (2))

∴ ϕ = \frac{π^{2}}{3} - 4 {l n}^{2} (2) m . n