Prove that: ∫_( 0) ^( (𝛑/4)) ((4 ln (cotx))/(cos(2x + 2022𝛑))) dx = 3𝛇(2)


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Question Number 162416 by HongKing last updated on 29/Dec/21

$Prove that :$ $\underset{0}{\int^{\frac{π}{4}}} \frac{4 \ln (\cot x)}{\cos (2 x + 2022 π)} dx = 3 ζ (2)$
Commented by smallEinstein last updated on 29/Dec/21

	Answered by mnjuly1970 last updated on 29/Dec/21
	$−−− solution Ω= −4∫_0 ^( (π/4)) (((1+tan^( 2) (x))ln( tan(x)))/(1+tan^( 2) (x)))dx =−4 ∫_0 ^( 1) (( ln (x )dx)/(1−x^( 2) )) dx = −2 ∫_0 ^( 1) (( ln(x ))/(1−x)) dx −2 ∫_0 ^( 1) ((ln(x))/(1+x))dx =− 2 ∫_0 ^( 1) ((( ln (1−x))/x)) dx −2 Φ = 2 Li_( 2) (1 ) − 2 Φ = 2 ζ(2) −2 Φ Φ = ∫_0 ^( 1) ln(x)Σ_(n=0) ^∞ (−1)^( n) x^( n) = Σ_(n=0) ^∞ (−1)^( n) { [ (x^( n+1) /(n+1)) ln(x) ]_0 ^1 −(1/(n+1)) ∫_0 ^( 1) x^( n) dx } =Σ_(n=0) ^∞ (( (−1)^(n+1) )/((n+1)^( 2) )) =− η (2) =− ((ζ (2))/2) −−− Ω = 2ζ (2) +ζ (2) =3 ζ (2) −−−■ m.n$
	$- - -$ $s o l u t i o n$ $Ω = - 4 \int_{0}^{\frac{π}{4}} \frac{(1 + {t a n}^{2} (x)) l n (t a n (x))}{1 + {t a n}^{2} (x)} d x = - 4 \int_{0}^{1} \frac{l n (x) d x}{1 - x^{2}} d x$ $= - 2 \int_{0}^{1} \frac{l n (x)}{1 - x} d x - 2 \int_{0}^{1} \frac{l n (x)}{1 + x} d x$ $= - 2 \int_{0}^{1} (\frac{l n (1 - x)}{x}) d x - 2 Φ$ $= 2 {Li}_{2} (1) - 2 Φ = 2 ζ (2) - 2 Φ$ $Φ = \int_{0}^{1} l n (x) \underset{n = 0}{\sum^{\infty}} {(- 1)}^{n} x^{n}$ $= \underset{n = 0}{\sum^{\infty}} {(- 1)}^{n} {{[\frac{x^{n + 1}}{n + 1} l n (x)]}_{0}^{1} - \frac{1}{n + 1} \int_{0}^{1} x^{n} d x}$ $= \underset{n = 0}{\sum^{\infty}} \frac{{(- 1)}^{n + 1}}{{(n + 1)}^{2}} = - η (2) = - \frac{ζ (2)}{2}$ $- - - Ω = 2 ζ (2) + ζ (2) = 3 ζ (2) - - - ◼ m . n$
	Commented by HongKing last updated on 31/Dec/21

	$thank you so much cool my dear Sir$
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