calculate-0-logx-x-6-1-dx-

Question Number 143545 by mathmax by abdo last updated on 15/Jun/21

calculate \int_{0}^{\infty} \frac{logx}{x^{6} + 1} dx

Answered by Pagnol last updated on 15/Jun/21

I = \int_{0}^{\infty} \frac{logx}{x^{6} + 1} dx, u = x^{6} \Rightarrow x = u^{\frac{1}{6}} \Rightarrow dx = \frac{1}{6} u^{- \frac{5}{6}} du

= \frac{1}{36} \int_{0}^{\infty} \frac{u^{- \frac{5}{6}} logu}{u + 1} du

I (α) = \int_{0}^{\infty} \frac{u^{α}}{u + 1} du = β (α + 1, - α) = \frac{Γ (α + 1) Γ (- α)}{Γ (1)}

I^{'} (α) = \int_{0}^{\infty} \frac{u^{α} lnu}{u + 1} du = - Γ (α + 1) Γ^{'} (- α) + Γ^{'} (α + 1) Γ (- α)

I^{'} (- \frac{5}{6}) = - Γ (\frac{1}{6}) Γ^{'} (\frac{5}{6}) + Γ^{'} (\frac{1}{6}) Γ (\frac{5}{6})

= Γ (\frac{1}{6}) Γ (\frac{5}{6}) [ψ (\frac{1}{6}) - ψ (\frac{5}{6})]

= \frac{π}{\sin (\frac{π}{6})} [- π \cot (\frac{5}{6} π)] = 2 π (π \sqrt{3}) = 2 \sqrt{3} π^{2}

I = \frac{1}{36} I^{'} (- \frac{5}{6}) = \frac{\sqrt{3}}{18} π^{2}

Answered by mathmax by abdo last updated on 15/Jun/21

Φ = \int_{0}^{\infty} \frac{logx}{x^{6} + 1} dx changement x^{6} = t give

Φ = \frac{1}{6} \int_{0}^{\infty} \frac{\log (t^{\frac{1}{6}})}{1 + t} t^{\frac{1}{6} - 1} dt = \frac{1}{36} \int_{0}^{\infty} \frac{t^{\frac{1}{6} - 1} logt}{1 + t} dt

let f (a) = \int_{0}^{\infty} \frac{t^{a - 1}}{1 + t} dt \Rightarrow f^{^{'}} (a) = \int_{0}^{\infty} \frac{t^{a - 1} logt}{1 + t} dt \Rightarrow f^{^{'}} (\frac{1}{6}) = 36 Φ \Rightarrow

Φ = \frac{1}{36} f^{^{'}} (\frac{1}{6}) we know f (a) = \frac{π}{\sin (π a)} \Rightarrow f^{^{'}} (a) = - \frac{π^{2} \cos (π a)}{\sin^{2} (π a)} \Rightarrow

f^{^{'}} (\frac{1}{6}) = - \frac{π^{2} \times \frac{\sqrt{3}}{2}}{{(\frac{1}{2})}^{2}} = - 4 π^{2} . \frac{\sqrt{3}}{2} = - 2 π^{2} \sqrt{3} \Rightarrow

Φ = \frac{1}{36} (- 2 π^{2} \sqrt{3}) = - \frac{π^{2}}{18} \sqrt{3}