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Question Number 107831 by mathdave last updated on 12/Aug/20

	Answered by hgrocks last updated on 12/Aug/20

	Commented by mathdave last updated on 12/Aug/20

	$c o r r e c t m a n t h a n k s$
	Answered by mathmax by abdo last updated on 12/Aug/20
	$I =∫_0 ^1 ((ln^2 (1−x^2 ))/x)dx we do the changement x^2 =t ⇒ I =∫_0 ^1 ((ln^2 (1−t))/(√t))×(dt/(2(√t))) =(1/2)∫_0 ^1 ((ln^2 (1−t))/t) dt =_(by parts) (1/2){ [ lnt ln^2 (1−t)]_0 ^1 −∫_0 ^1 ln(t).((2ln(1−t)(−1))/(1−t)) dt =∫_0 ^1 ((ln(t)ln(1−t))/(1−t)) dt =_(1−t =u) ∫_0 ^1 ((ln(1−u)lnu)/u) du we have ln^′ (1−u) =−(1/(1−u)) =−Σ_(n=0) ^∞ u^n ⇒ln(1−u) =−Σ_(n=0) ^∞ (u^(n+1) /(n+1)) =−Σ_(n=1) ^∞ (u^n /n) ⇒((ln(1−u))/u) =−Σ_(n=1) ^∞ (u^(n−1) /n) ⇒ I =−∫_0 ^1 lnu(Σ_(n=1) ^∞ (u^(n−1) /n))du =−Σ_(n=1) ^∞ (1/n) ∫_0 ^1 u^(n−1) lnu du but ∫_0 ^1 u^(n−1) ln(u)du =[(u^n /n)lnu]_0 ^1 −∫_0 ^1 (u^n /n)(du/u) =−(1/n)∫_0 ^1 u^(n−1) dy =−(1/n^2 ) ⇒ ★I =Σ_(n=1) ^∞ (1/n^3 ) =ξ(3)★$
	$I = \int_{0}^{1} \frac{\ln^{2} (1 - x^{2})}{x} dx we do the changement x^{2} = t \Rightarrow$ $I = \int_{0}^{1} \frac{\ln^{2} (1 - t)}{\sqrt{t}} \times \frac{dt}{2 \sqrt{t}} = \frac{1}{2} \int_{0}^{1} \frac{\ln^{2} (1 - t)}{t} dt$ $=_{by parts} \frac{1}{2} {{[lnt \ln^{2} (1 - t)]}_{0}^{1} - \int_{0}^{1} \ln (t) . \frac{2 \ln (1 - t) (- 1)}{1 - t} dt$ $= \int_{0}^{1} \frac{\ln (t) \ln (1 - t)}{1 - t} dt =_{1 - t = u} \int_{0}^{1} \frac{\ln (1 - u) lnu}{u} du$ $we have \ln^{^{'}} (1 - u) = - \frac{1}{1 - u} = - \sum_{n = 0}^{\infty} u^{n} \Rightarrow \ln (1 - u) = - \sum_{n = 0}^{\infty} \frac{u^{n + 1}}{n + 1}$ $= - \sum_{n = 1}^{\infty} \frac{u^{n}}{n} \Rightarrow \frac{\ln (1 - u)}{u} = - \sum_{n = 1}^{\infty} \frac{u^{n - 1}}{n} \Rightarrow$ $I = - \int_{0}^{1} lnu (\sum_{n = 1}^{\infty} \frac{u^{n - 1}}{n}) du = - \sum_{n = 1}^{\infty} \frac{1}{n} \int_{0}^{1} u^{n - 1} lnu du but$ $\int_{0}^{1} u^{n - 1} \ln (u) du = {[\frac{u^{n}}{n} lnu]}_{0}^{1} - \int_{0}^{1} \frac{u^{n}}{n} \frac{du}{u} = - \frac{1}{n} \int_{0}^{1} u^{n - 1} dy = - \frac{1}{n^{2}} \Rightarrow$ $★ I = \sum_{n = 1}^{\infty} \frac{1}{n^{3}} = ξ (3) ★$
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