Prove-the-folowing-result-0-pi-2-cot-log-sec-3-d-pi-4-240-I-need-your-help-if-possible-please-

Question Number 140056 by Ndala last updated on 03/May/21

Prove the folowing result :

\int_{0}^{\frac{π}{2}} \cot θ \cdot {(\log \sec θ)}^{3} d θ = \frac{π^{4}}{240}

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I need your help, if possible please .

Answered by Ar Brandon last updated on 04/May/21

\int_{0}^{\frac{π}{2}} \cot θ {(\ln \sec θ)}^{3} d θ = - \int_{0}^{\frac{π}{2}} \sin^{- 1} θ \cos θ \cdot \ln^{3} \cos θ d θ

β (m, n) = 2 \int_{0}^{\frac{π}{2}} \sin^{2 m - 1} θ \cos^{2 n - 1} θ d θ = \frac{Γ (m) Γ (n)}{Γ (m + n)}

\ln β (m, n) = \ln Γ (m) + \ln Γ (n) - \ln Γ (m + n)

\frac{β^{'} (m, n)}{β (m, n)} = ψ (n) - ψ (m + n)

β^{″} (m, n) = β (m, n) [ψ^{'} (n) - ψ^{'} (m + n)] + β^{'} (m, n) [ψ (n) - ψ (m + n)]

β^{‴} (m, n) = β (m, n) [ψ^{″} (n) - ψ^{″} (m + n)] + β^{'} (m, n) [ψ^{'} (n) - ψ^{'} (m + n)]

+ β^{″} (m, n) [ψ (n) - ψ (m + n)] + β^{'} (m, n) [ψ^{'} (n) - ψ^{'} (m + n)]

Commented by Ndala last updated on 07/May/21

Can you complete ?

Answered by qaz last updated on 07/May/21

\int_{0}^{π / 2} (\cot θ) \cdot {l n}^{3} \sec θ d θ \dots \dots . . \cos θ \to x

= \int_{0}^{1} \frac{x}{1 - x^{2}} {l n}^{3} (\frac{1}{x}) d x

= \int_{0}^{\infty} \frac{e^{- 2 y}}{1 - e^{- 2 y}} y^{3} d y \dots \dots \dots \dots \dots . x \to e^{- y}

= \int_{0}^{\infty} (\frac{1}{1 - e^{- 2 y}} - 1) y^{3} d y

= \underset{n = 1}{\sum^{\infty}} \int_{0}^{\infty} y^{3} e^{- 2 n y} d y

= \underset{n = 1}{\sum^{\infty}} \frac{3!}{{(2 n)}^{4}}

= \frac{π^{4}}{240}

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