calculate-0-logx-x-2-x-1-dx-

Question Number 148570 by mathmax by abdo last updated on 29/Jul/21

calculate \int_{0}^{\infty} \frac{logx}{x^{2} + x + 1} dx

Answered by Ar Brandon last updated on 29/Jul/21

Φ = \int_{0}^{\infty} \frac{\log x}{x^{2} + x + 1} d x

= \int_{0}^{1} \frac{\log x}{x^{2} + x + 1} d x + \int_{1}^{\infty} \frac{\log x}{x^{2} + x + 1} d x

= \int_{0}^{1} \frac{\log x}{x^{2} + x + 1} d x - \int_{0}^{1} \frac{\log x}{x^{2} + x + 1} d x = 0

Answered by mathmax by abdo last updated on 29/Jul/21

\int_{0}^{\infty} \frac{logx}{x^{2} + x + 1} dx = - \frac{1}{2} Re (Σ Res (f (z) \log^{2} z, a_{i})

f (z) = \frac{1}{z^{2} + z + 1} poles of f!

Δ = 1 - 4 = - 3 \Rightarrow z_{1} = \frac{- 1 + i \sqrt{3}}{2} = e^{\frac{2 i π}{3}} and z_{2} = \frac{- 1 - i \sqrt{3}}{2}

φ (z) = f (z) \log^{2} z \Rightarrow φ (z) = \frac{\log^{2} z}{(z - z_{1}) (z - z_{2})}

Res (φ, e^{\frac{2 i π}{3}}) = \frac{{(\frac{2 i π}{3})}^{2}}{i \sqrt{3}} = - \frac{4 π^{2}}{9 (i \sqrt{3})} = \frac{4 i π^{2}}{9 \sqrt{3}}

Res (φ, e^{- \frac{2 i π}{3}}) = \frac{{(- \frac{2 i π}{3})}^{2}}{- i \sqrt{3}} = \frac{4 π^{2}}{9 (i \sqrt{3})} = \frac{- 4 i π^{2}}{9 \sqrt{3}} \Rightarrow

Σ Res (φ . .) = 0 \Rightarrow \int_{0}^{\infty} \frac{logx}{x^{2} + x + 1} dx = 0