explicite-f-a-0-lnx-x-2-x-a-dx-with-a-gt-1-4-

Question Number 127777 by Bird last updated on 02/Jan/21

e x p l i c i t e f (a) = \int_{0}^{\infty} \frac{l n x}{x^{2} - x + a} d x

w i t h a > \frac{1}{4}

Answered by mathmax by abdo last updated on 03/Jan/21

let f (z) = \frac{\ln^{2} z}{z^{2} - z + a} we have f (a) = - \frac{1}{2} Re (Σ Res (f, z_{i}))

poles of f \to 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} we have ∣ 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}}

f (z) = \frac{\ln^{2} z}{(z - z_{1}) (z - z_{2})}

Res (f, z_{1}) = \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}}

Res (f, z_{2}) = \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}} \Rightarrow

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

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

= \frac{1}{i \sqrt{4 a - 1}} (2 iln (\sqrt{a}) \arctan \sqrt{4 a - 1}) = \frac{lna}{\sqrt{4 a - 1}} \arctan \sqrt{4 a - 1} \Rightarrow

f (a) = - \frac{lna}{2 \sqrt{4 a - 1}} \arctan \sqrt{4 a - 1}