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Question Number 126003 by mnjuly1970 last updated on 16/Dec/20

$. . . n i c e i n t e g r a l . . .$ $p r o v e t h a t ::$ $ϕ = \int_{0}^{\frac{π}{2}} t a n (x) l n (s i n (x)) l n (c o s (x)) d x = \frac{ζ (3}{8}$ $G o o d l u c k$
	Answered by Olaf last updated on 16/Dec/20
	$φ = ∫_0 ^(π/2) tanxln(cosx)ln(sinx)dx φ = −∫_0 ^(π/2) [−tanxln(cosx)]ln(sinx)dx φ = −[(1/2)ln^2 (cosx)ln(sinx)]_0 ^(π/2) +∫_0 ^(π/2) [(1/2)ln^2 (cosx)]cotxdx φ = (1/2)∫_0 ^(π/2) cotx.ln^2 (cosx)dx Let u = cosx du = −sinxdx = −(√(1−u^2 ))dx φ = (1/2)∫_1 ^0 (u/( (√(1−u^2 ))))ln^2 u(−(du/( (√(1−u^2 ))))) φ = (1/2)∫_0 ^1 (u/( 1−u^2 ))ln^2 udu φ = (1/2)∫_0 ^1 u.ln^2 uΣ_(n=0) ^∞ u^(2n) du φ = (1/2)Σ_(n=0) ^∞ ∫_0 ^1 u^(2n+1) ln^2 udu φ = (1/2)Σ_(n=0) ^∞ {[(u^(2n+2) /(2n+2))ln^2 u]_0 ^1 −∫_0 ^1 (u^(2n+2) /(2n+2))(2((lnu)/u))du} φ = (1/2)Σ_(n=0) ^∞ {−∫_0 ^1 (u^(2n+2) /(2n+2))(2((lnu)/u))du} φ = −Σ_(n=0) ^∞ ∫_0 ^1 (u^(2n+1) /(2n+2))lnudu φ = −Σ_(n=0) ^∞ {[(u^(2n+2) /((2n+2)^2 ))lnu]_0 ^1 −∫_0 ^1 (u^(2n+2) /((2n+2)^2 )).(du/u)} φ = Σ_(n=0) ^∞ ∫_0 ^1 (u^(2n+2) /((2n+2)^2 )).(du/u) φ = Σ_(n=0) ^∞ ∫_0 ^1 (u^(2n+1) /((2n+2)^2 )).du φ = Σ_(n=0) ^∞ [(u^(2n+2) /((2n+2)^3 ))]_0 ^1 φ = Σ_(n=0) ^∞ (1/((2n+2)^3 )) φ = (1/8)Σ_(n=0) ^∞ (1/((n+1)^3 )) φ = (1/8)Σ_(n=1) ^∞ (1/n^3 ) φ = ((ζ(3))/8)$
	$ϕ = \int_{0}^{\frac{π}{2}} \tan x \ln (\cos x) \ln (\sin x) d x$ $ϕ = - \int_{0}^{\frac{π}{2}} [- \tan x \ln (\cos x)] \ln (\sin x) d x$ $ϕ = - {[\frac{1}{2} \ln^{2} (\cos x) \ln (\sin x)]}_{0}^{\frac{π}{2}}$ $+ \int_{0}^{\frac{π}{2}} [\frac{1}{2} \ln^{2} (\cos x)] \cot x d x$ $ϕ = \frac{1}{2} \int_{0}^{\frac{π}{2}} \cot x . \ln^{2} (\cos x) d x$ $Let u = \cos x$ $d u = - \sin x d x = - \sqrt{1 - u^{2}} d x$ $ϕ = \frac{1}{2} \int_{1}^{0} \frac{u}{\sqrt{1 - u^{2}}} \ln^{2} u (- \frac{d u}{\sqrt{1 - u^{2}}})$ $ϕ = \frac{1}{2} \int_{0}^{1} \frac{u}{1 - u^{2}} \ln^{2} u d u$ $ϕ = \frac{1}{2} \int_{0}^{1} u . \ln^{2} u \underset{n = 0}{\sum^{\infty}} u^{2 n} d u$ $ϕ = \frac{1}{2} \underset{n = 0}{\sum^{\infty}} \int_{0}^{1} u^{2 n + 1} \ln^{2} u d u$ $ϕ = \frac{1}{2} \underset{n = 0}{\sum^{\infty}} {{[\frac{u^{2 n + 2}}{2 n + 2} \ln^{2} u]}_{0}^{1} - \int_{0}^{1} \frac{u^{2 n + 2}}{2 n + 2} (2 \frac{\ln u}{u}) d u}$ $ϕ = \frac{1}{2} \underset{n = 0}{\sum^{\infty}} {- \int_{0}^{1} \frac{u^{2 n + 2}}{2 n + 2} (2 \frac{\ln u}{u}) d u}$ $ϕ = - \underset{n = 0}{\sum^{\infty}} \int_{0}^{1} \frac{u^{2 n + 1}}{2 n + 2} \ln u d u$ $ϕ = - \underset{n = 0}{\sum^{\infty}} {{[\frac{u^{2 n + 2}}{{(2 n + 2)}^{2}} \ln u]}_{0}^{1} - \int_{0}^{1} \frac{u^{2 n + 2}}{{(2 n + 2)}^{2}} . \frac{d u}{u}}$ $ϕ = \underset{n = 0}{\sum^{\infty}} \int_{0}^{1} \frac{u^{2 n + 2}}{{(2 n + 2)}^{2}} . \frac{d u}{u}$ $ϕ = \underset{n = 0}{\sum^{\infty}} \int_{0}^{1} \frac{u^{2 n + 1}}{{(2 n + 2)}^{2}} . d u$ $ϕ = \underset{n = 0}{\sum^{\infty}} {[\frac{u^{2 n + 2}}{{(2 n + 2)}^{3}}]}_{0}^{1}$ $ϕ = \underset{n = 0}{\sum^{\infty}} \frac{1}{{(2 n + 2)}^{3}}$ $ϕ = \frac{1}{8} \underset{n = 0}{\sum^{\infty}} \frac{1}{{(n + 1)}^{3}}$ $ϕ = \frac{1}{8} \underset{n = 1}{\sum^{\infty}} \frac{1}{n^{3}}$ $ϕ = \frac{ζ (3)}{8}$
	Commented by mnjuly1970 last updated on 16/Dec/20

	$t h a n k s a l o t s i r o l a f . . .$
	Answered by Lordose last updated on 16/Dec/20

	$Ω = \int_{0}^{\frac{π}{2}} \frac{\sin (x) \ln (\sqrt{1 - \cos^{2} (x)}) \ln (\cos (x))}{\cos (x)} dx \overset{u = \cos (x)}{=} \int_{1}^{0} \frac{- \ln (\sqrt{1 - u^{2}}) \ln (u)}{u} du$ $Ω = \frac{1}{2} \int_{0}^{1} \frac{\ln (1 - u^{2}) \ln (u)}{u} du = \frac{1}{2} \int_{0}^{1} \frac{- \ln (u) \underset{n = 1}{\sum^{\infty}} \frac{u^{2 n}}{n}}{u} du$ $Ω = - \frac{1}{2} \underset{n = 1}{\sum^{\infty}} \frac{1}{n} \int_{0}^{1} u^{2 n - 1} \ln (u) du = - \frac{1}{2} \underset{n = 1}{\sum^{\infty}} \frac{1}{n} ∣ \frac{u^{2 n} (2 nlnu - 1)}{{4 n}^{2}} ∣_{0}^{1}$ $Ω = - \frac{1}{2} \underset{n = 1}{\sum^{\infty}} \frac{1}{n} \cdot - \frac{1}{{4 n}^{2}} = \frac{1}{8} \underset{n = 1}{\sum^{\infty}} \frac{1}{n^{3}} = \frac{ζ (3)}{8}$
	Commented by mnjuly1970 last updated on 16/Dec/20

	$e x c e l l e n t s i r l o r d o s e . . .$
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