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

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

$$\mathrm{Prove}\:\mathrm{that}: \\ $$$$\underset{\:\mathrm{0}} {\overset{\:\frac{\boldsymbol{\pi}}{\mathrm{4}}} {\int}}\:\frac{\mathrm{4}\:\mathrm{ln}\:\left(\mathrm{cot}\boldsymbol{\mathrm{x}}\right)}{\mathrm{cos}\left(\mathrm{2x}\:+\:\mathrm{2022}\boldsymbol{\pi}\right)}\:\mathrm{dx}\:=\:\mathrm{3}\boldsymbol{\zeta}\left(\mathrm{2}\right) \\ $$

Commented by smallEinstein last updated on 29/Dec/21

Answered by mnjuly1970 last updated on 29/Dec/21

βˆ’βˆ’βˆ’ solution Ξ©= βˆ’4∫_0 ^( (Ο€/4)) (((1+tan^( 2) (x))ln( tan(x)))/(1+tan^( 2) (x)))dx =βˆ’4 ∫_0 ^( 1) (( ln (x )dx)/(1βˆ’x^( 2) )) dx = βˆ’2 ∫_0 ^( 1) (( ln(x ))/(1βˆ’x)) dx βˆ’2 ∫_0 ^( 1) ((ln(x))/(1+x))dx =βˆ’ 2 ∫_0 ^( 1) ((( ln (1βˆ’x))/x)) dx βˆ’2 Ξ¦ = 2 Li_( 2) (1 ) βˆ’ 2 Ξ¦ = 2 ΞΆ(2) βˆ’2 Ξ¦ Ξ¦ = ∫_0 ^( 1) ln(x)Ξ£_(n=0) ^∞ (βˆ’1)^( n) x^( n) = Ξ£_(n=0) ^∞ (βˆ’1)^( n) { [ (x^( n+1) /(n+1)) ln(x) ]_0 ^1 βˆ’(1/(n+1)) ∫_0 ^( 1) x^( n) dx } =Ξ£_(n=0) ^∞ (( (βˆ’1)^(n+1) )/((n+1)^( 2) )) =βˆ’ Ξ· (2) =βˆ’ ((ΞΆ (2))/2) βˆ’βˆ’βˆ’ Ξ© = 2ΞΆ (2) +ΞΆ (2) =3 ΞΆ (2) βˆ’βˆ’βˆ’β–  m.n

$$\:\:\:βˆ’βˆ’βˆ’ \\ $$$$\:\:\:\:{solution} \\ $$$$\:\:\:\Omega=\:βˆ’\mathrm{4}\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{4}}} \frac{\left(\mathrm{1}+{tan}^{\:\mathrm{2}} \left({x}\right)\right){ln}\left(\:{tan}\left({x}\right)\right)}{\mathrm{1}+{tan}^{\:\mathrm{2}} \left({x}\right)}{dx}\:=βˆ’\mathrm{4}\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\:{ln}\:\left({x}\:\right){dx}}{\mathrm{1}βˆ’{x}^{\:\mathrm{2}} }\:{dx} \\ $$$$\:\:\:\:\:\:\:\:\:\:=\:βˆ’\mathrm{2}\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\:{ln}\left({x}\:\right)}{\mathrm{1}βˆ’{x}}\:{dx}\:βˆ’\mathrm{2}\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{ln}\left({x}\right)}{\mathrm{1}+{x}}{dx} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:=βˆ’\:\mathrm{2}\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \left(\frac{\:{ln}\:\left(\mathrm{1}βˆ’{x}\right)}{{x}}\right)\:{dx}\:βˆ’\mathrm{2}\:\Phi \\ $$$$\:\:\:\:\:\:\:\:\:=\:\mathrm{2}\:\mathrm{Li}_{\:\mathrm{2}} \:\left(\mathrm{1}\:\right)\:βˆ’\:\mathrm{2}\:\Phi\:=\:\mathrm{2}\:\zeta\left(\mathrm{2}\right)\:βˆ’\mathrm{2}\:\Phi \\ $$$$\:\:\:\:\:\:\:\:\:\Phi\:=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} {ln}\left({x}\right)\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(βˆ’\mathrm{1}\right)^{\:{n}} {x}^{\:{n}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(βˆ’\mathrm{1}\right)^{\:{n}} \:\left\{\:\left[\:\frac{{x}^{\:{n}+\mathrm{1}} }{{n}+\mathrm{1}}\:{ln}\left({x}\right)\:\right]_{\mathrm{0}} ^{\mathrm{1}} βˆ’\frac{\mathrm{1}}{{n}+\mathrm{1}}\:\int_{\mathrm{0}} ^{\:\mathrm{1}} {x}^{\:{n}} {dx}\:\right\} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\:\left(βˆ’\mathrm{1}\right)^{{n}+\mathrm{1}} }{\left({n}+\mathrm{1}\right)^{\:\mathrm{2}} }\:=βˆ’\:\eta\:\left(\mathrm{2}\right)\:=βˆ’\:\frac{\zeta\:\left(\mathrm{2}\right)}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:βˆ’βˆ’βˆ’\:\:\:\Omega\:=\:\mathrm{2}\zeta\:\left(\mathrm{2}\right)\:+\zeta\:\left(\mathrm{2}\right)\:=\mathrm{3}\:\zeta\:\left(\mathrm{2}\right)\:βˆ’βˆ’βˆ’\blacksquare\:{m}.{n}\:\:\:\:\: \\ $$$$\:\:\:\:\:\: \\ $$$$ \\ $$

Commented by HongKing last updated on 31/Dec/21

thank you so much cool my dear Sir

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{so}\:\mathrm{much}\:\mathrm{cool}\:\mathrm{my}\:\mathrm{dear}\:\mathrm{Sir} \\ $$

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