calculate-0-logx-1-x-4-dx-

Question Number 141220 by mathmax by abdo last updated on 16/May/21

calculate \int_{0}^{\infty} \frac{logx}{1 + x^{4}} dx

Answered by Ar Brandon last updated on 16/May/21

Φ = \int_{0}^{\infty} \frac{lnx}{1 + x^{4}} dx, x^{4} = u \Rightarrow dx = \frac{1}{4} u^{- 3 / 4} du

= \frac{1}{16} \int_{0}^{\infty} \frac{u^{- 3 / 4} lnu}{1 + u} du

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

= \frac{Γ (α) Γ (1 - α)}{Γ (1)} = \frac{π}{\sin (π α)}

f^{'} (α) = \int_{0}^{\infty} \frac{u^{α - 1} \ln (u)}{1 + u} du = - \frac{π^{2}}{2} \cdot \cos (π α) {cosec}^{2} (π α)

f^{'} (\frac{1}{4}) = \int_{0}^{\infty} \frac{u^{- 3 / 4} \ln (u)}{1 + u} du = - \frac{π^{2}}{2} \cos (\frac{π}{4}) {cosec}^{2} (\frac{π}{4})

Answered by mathmax by abdo last updated on 16/May/21

let f (a) = \int_{0}^{\infty} \frac{x^{a}}{1 + x^{4}} dx \Rightarrow f (a) = \int_{0}^{\infty} \frac{e^{alogx}}{1 + x^{4}} dx \Rightarrow

f^{^{'}} (a) = \int_{0}^{\infty} \frac{x^{a} logx}{1 + x^{4}} dx \Rightarrow f^{^{'}} (0) = \int_{0}^{\infty} \frac{logx}{1 + x^{4}} dx

changement x = t^{\frac{1}{4}} give f (a) = \frac{1}{4} \int_{0}^{\infty} \frac{t^{\frac{a}{4}}}{1 + t} t^{\frac{1}{4} - 1} dt

= \frac{1}{4} \int_{0}^{\infty} \frac{t^{\frac{a + 1}{4} - 1}}{1 + t} = \frac{π}{4 \sin (\frac{π (a + 1)}{4})} \Rightarrow f^{^{'}} (a) = - \frac{π}{4} \times \frac{\frac{π}{4} \cos (\frac{π (a + 1)}{4})}{\sin^{2} (\frac{π (a + 1)}{4})}

\Rightarrow f^{^{'}} (0) = - \frac{π^{2}}{16} . \frac{\cos (\frac{π}{4})}{\sin^{2} (\frac{π}{4})} = - \frac{π^{2}}{16} \times \frac{\frac{\sqrt{2}}{2}}{\frac{1}{2}} = - \frac{π^{2} \sqrt{2}}{16} \Rightarrow

\int_{0}^{\infty} \frac{logx}{1 + x^{4}} dx = - \frac{π^{2}}{16} \sqrt{2}