calculate-0-lnx-x-2-x-2-2-dx-

Question Number 115725 by Bird last updated on 28/Sep/20

c a l c u l a t e \int_{0}^{\infty} \frac{l n x}{{(x^{2} + x + 2)}^{2}} d x

Answered by mathmax by abdo last updated on 28/Sep/20

let f (a) = \int_{0}^{\infty} \frac{lnx}{x^{2} + x + a} dx with a > \frac{1}{4}

we have f^{^{'}} (a) = - \int_{0}^{\infty} \frac{lnx}{{(x^{2} + x + a)}^{2}} and \int_{0}^{\infty} \frac{lnx}{{(x^{2} + x + a)}^{2}} = - f^{^{'}} (2)

let explicite f (a) by using \int_{0}^{\infty} q (x) \ln (x) dx = - \frac{1}{2} Re (\sum_{i} Res (q (z) \ln^{2} z, a_{i})

φ (z) = q (z) \ln^{2} z = \frac{\ln^{2} z}{z^{2} + z + a} poles of φ ?

Δ = 1 - 4 a < 0 \Rightarrow z_{1} = \frac{- 1 + i \sqrt{4 a - 1}}{2} and z_{2} = \frac{- 1 - i \sqrt{4 a - 1}}{2}

∣ z_{1} ∣ = \frac{1}{2} \sqrt{1 + 4 a - 1} = \sqrt{a} \Rightarrow z_{1} = \sqrt{a} e^{- iarctan \sqrt{4 a - 1}}

and z_{2} = \sqrt{a} e^{iarctan \sqrt{4 a - 1}} we have φ (z) = \frac{\ln^{2} z}{(z - z_{1}) (z - z_{2})}

Res (φ, z_{1}) = \lim_{z \to z_{1}} (z - z_{1}) φ (z) = \frac{\ln^{2} z_{1}}{z_{1} - z_{2}}

= \frac{{(\ln \sqrt{a} + iarctan \sqrt{4 a - 1})}^{2}}{i \sqrt{4 a - 1}} = \frac{\ln^{2} \sqrt{a} + 2 iln \sqrt{a} \arctan \sqrt{4 a - 1} - \arctan^{2} \sqrt{4 a - 1}}{i \sqrt{4 a - 1}}

= - i \frac{\ln^{2} \sqrt{a} + 2 iln (\sqrt{a}) \arctan \sqrt{4 a - 1} - \arctan^{2} \sqrt{4 a - 1}}{\sqrt{4 a - 1}}

= \frac{- {iln}^{2} \sqrt{a} + lna \arctan \sqrt{4 a - 1} + {iarctan}^{2} \sqrt{4 a - 1}}{\sqrt{4 a - 1}}

Res (φ, z_{2}) = \lim_{z \to z_{2}} (z - z_{2}) φ (z) = \frac{\ln^{2} z_{2}}{z_{2} - z_{1}} = \frac{{(\ln \sqrt{a} + iarctan \sqrt{4 a - 1})}^{2}}{- i \sqrt{4 a - 1}}

= \frac{\ln^{2} \sqrt{a} + 2 iln \sqrt{a} \arctan \sqrt{4 a - 1} - \arctan^{2} \sqrt{4 a - 1}}{- i \sqrt{4 a - 1}}

= \frac{{iln}^{2} \sqrt{a} - lna \arctan \sqrt{4 a - 1} - {iarctan}^{2} \sqrt{4 a - 1}}{\sqrt{4 a - 1}} \Rightarrow

Σ Res = \frac{1}{\sqrt{4 a - 1}} {- {iln}^{2} \sqrt{a} + lna \arctan \sqrt{4 a - 1} + i \arctan^{2} \sqrt{4 a - 1}

+ {iln}^{2} \sqrt{a} - lna \arctan \sqrt{4 a - 1} - iarctan \sqrt{4 a - 1}}

Re (Σ Res (\dots)) = \frac{1}{\sqrt{4 a - 1}} {o} = 0 \Rightarrow \int_{0}^{\infty} \frac{lnx}{x^{2} + x + a} dx = 0 \Rightarrow

\int_{0}^{\infty} \frac{lnx}{{(x^{2} + x + a)}^{2}} dx = 0