calculate-0-log-2-x-x-2-x-1-dx-

Question Number 139402 by mathmax by abdo last updated on 26/Apr/21

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

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

let f (a) = \int_{0}^{\infty} \frac{x^{a} logx}{x^{2} + x + 1} dx \Rightarrow f (a) = \int_{0}^{\infty} \frac{e^{alogx} logx}{x^{2} + x + 1} dx

\Rightarrow f^{^{'}} (a) = \int_{0}^{\infty} \frac{x^{a} \log^{2} x}{x^{2} + x + 1} dx \Rightarrow f^{^{'}} (o) = \int_{0}^{\infty} \frac{\log^{2} x}{x^{2} + x + 1} dx

f (a) = - \frac{1}{2} Re (Σ Res (φ a_{i})) with φ (z) = \frac{z^{a}}{z^{2} + z + 1} \log^{2} z

we have φ (z) = \frac{z^{a} \log^{2} z}{(z - e^{\frac{2 i π}{3}}) (z - e^{- \frac{2 i π}{3}})}

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

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

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

= - \frac{4 π^{2}}{9 i \sqrt{3}} (2 i \sin (\frac{2 π a}{3}) = - \frac{8 π^{2}}{9 \sqrt{3}} \sin (\frac{2 π a}{3}) \Rightarrow

f (a) = \frac{4 π^{2}}{9 \sqrt{3}} \sin (\frac{2 π a}{3}) \Rightarrow f^{^{'}} (a) = \frac{4 π^{2}}{9 \sqrt{3}} \times \frac{2 π}{3} \cos (\frac{2 π a}{3})

= \frac{8 π^{3}}{27 \sqrt{3}} \cos (\frac{2 π a}{3}) \Rightarrow f^{^{'}} (0) = \frac{8 π^{3}}{27 \sqrt{3}} \Rightarrow

\int_{0}^{\infty} \frac{\log^{2} x}{x^{2} + x + 1} dx = \frac{8 π^{3}}{27 \sqrt{3}}