prove that 𝛗=∫_0 ^( 1) (( ln^( 2) (1−x ))/x^( 2) ) dx = 2 ζ (2) −−−proof−−− 𝛗= [((−1)/x) ln^( 2) (1−x) ]_0 ^1 −∫_0 ^( 1) ((2ln(1−x))/(x(1−x)))dx =−lim_( ξ→1^− ) (1/ξ)ln^( 2) (1−ξ)−2{ ∫_0 ^( 1) ((ln(1−x))/(1−x))dx+∫_0 ^( 1) ((ln(1−x))/x)dx} = −lim_( ξ→1^− ) {(1/ξ)ln^( 2) (1−ξ)+ln^( 2) (1 −ξ)}+2 ζ(2) =lim_(ξ→1^− ) (((ξ−1)/ξ))ln^( 2) (1−ξ) +2ζ(2) =_(ξ→1^− , δ→0^( +) ) ^(1−ξ= δ) [lim_( δ→0^( +) ) (((−δ)/(1−δ)))ln^2 (δ)=0] +2ζ(2) ■ m.n ∴ 𝛗 = 2 ζ(2)


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Question Number 166180 by mnjuly1970 last updated on 15/Feb/22
$prove that 𝛗=∫_0 ^( 1) (( ln^( 2) (1−x ))/x^( 2) ) dx = 2 ζ (2) −−−proof−−− 𝛗= [((−1)/x) ln^( 2) (1−x) ]_0 ^1 −∫_0 ^( 1) ((2ln(1−x))/(x(1−x)))dx =−lim_( ξ→1^− ) (1/ξ)ln^( 2) (1−ξ)−2{ ∫_0 ^( 1) ((ln(1−x))/(1−x))dx+∫_0 ^( 1) ((ln(1−x))/x)dx} = −lim_( ξ→1^− ) {(1/ξ)ln^( 2) (1−ξ)+ln^( 2) (1 −ξ)}+2 ζ(2) =lim_(ξ→1^− ) (((ξ−1)/ξ))ln^( 2) (1−ξ) +2ζ(2) =_(ξ→1^− , δ→0^( +) ) ^(1−ξ= δ) [lim_( δ→0^( +) ) (((−δ)/(1−δ)))ln^2 (δ)=0] +2ζ(2) ■ m.n ∴ 𝛗 = 2 ζ(2)$
$p r o v e t h a t$ $ϕ = \int_{0}^{1} \frac{{l n}^{2} (1 - x)}{x^{2}} d x = 2 ζ (2)$ $- - - p r o o f - - -$ $ϕ = {[\frac{- 1}{x} {l n}^{2} (1 - x)]}_{0}^{1} - \int_{0}^{1} \frac{2 l n (1 - x)}{x (1 - x)} d x$ $= - {l i m}_{ξ \to 1^{-}} \frac{1}{ξ} {l n}^{2} (1 - ξ) - 2 {\int_{0}^{1} \frac{l n (1 - x)}{1 - x} d x + \int_{0}^{1} \frac{l n (1 - x)}{x} d x}$ $= - {l i m}_{ξ \to 1^{-}} {\frac{1}{ξ} {l n}^{2} (1 - ξ) + {l n}^{2} (1 - ξ)} + 2 ζ (2)$ $= {l i m}_{ξ \to 1^{-}} (\frac{ξ - 1}{ξ}) {l n}^{2} (1 - ξ) + 2 ζ (2)$ $\underset{ξ \to 1^{-}, δ \to 0^{+}}{\overset{1 - ξ = δ}{=}} [{l i m}_{δ \to 0^{+}} (\frac{- δ}{1 - δ}) {l n}^{2} (δ) = 0] + 2 ζ (2) ◼ m . n$ $∴ ϕ = 2 ζ (2)$
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