explicite f(a)=∫_0 ^∞ ((lnx)/(x^2 −x+a))dx with a>(1/4)


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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
	$let f(z)=((ln^2 z)/(z^2 −z+a)) we have f(a)=−(1/2)Re(Σ Res(f,z_i )) poles of f →Δ=1−4a<0 ⇒z_1 =((1+i(√(4a−1)))/2) and z_2 =((1−i(√(4a−1)))/2) we have ∣z_1 ∣=(1/2)(√(1+4a−1)))=(√a) ⇒ z_1 =(√a)e^(iarctan(√(4a−1))) and z_2 =(√a)e^(−iarctan(√(4a−1))) f(z)=((ln^2 z)/((z−z_1 )(z−z_2 ))) Res(f,z_1 )=((ln^2 z_1 )/(z_1 −z_2 )) =(((ln(√a)+iarctan(√(4a))−1)^2 )/(i(√(4a−1)))) =((ln^2 ((√a))+2iln(√a)arctan(√(4a−1))−arctan^2 (√(4a−1)))/(i(√(4a−1)))) Res(f,z_2 )=((ln^2 z_2 )/(z_2 −z_1 )) =(((ln(√a)−iarctan(√(4a−1)))^2 )/(−i(√(4a−1)))) =((ln^2 (√a)−2iln(√a)arctan(√(4a−1))−arctan^2 (√(4a−1)))/(−i(√(4a−1)))) ⇒ Σ Res(f)=(1/(i(√(4a−1)))){ln^2 ((√a))+2iln(√a)arctan(√(4a−1))−arctan^2 (√(4a−1)) −ln^2 (√a)+2iln(√a)arctan(√(4a−1))+arctan^2 (√(4a−1))} =(1/(i(√(4a−1))))(2iln((√a))arctan(√(4a−1)))=((lna)/( (√(4a−1)))) arctan(√(4a−1)) ⇒ f(a) =−((lna)/(2(√(4a−1)))) arctan(√(4a−1))$
	$let f (z) = \frac{\ln^{2} z}{z^{2} - z + a} we have f (a) = - \frac{1}{2} Re (Σ Res (f, z_{i}))$ $poles of f \to Δ = 1 - 4 a < 0 \Rightarrow z_{1} = \frac{1 + i \sqrt{4 a - 1}}{2}$ $and z_{2} = \frac{1 - i \sqrt{4 a - 1}}{2} we have ∣ z_{1} ∣= \frac{1}{2} \sqrt{1 + 4 a - 1}) = \sqrt{a} \Rightarrow$ $z_{1} = \sqrt{a} e^{iarctan \sqrt{4 a - 1}} and z_{2} = \sqrt{a} e^{- iarctan \sqrt{4 a - 1}}$ $f (z) = \frac{\ln^{2} z}{(z - z_{1}) (z - z_{2})}$ $Res (f, z_{1}) = \frac{\ln^{2} z_{1}}{z_{1} - z_{2}} = \frac{{(\ln \sqrt{a} + iarctan \sqrt{4 a} - 1)}^{2}}{i \sqrt{4 a - 1}}$ $= \frac{\ln^{2} (\sqrt{a}) + 2 iln \sqrt{a} \arctan \sqrt{4 a - 1} - \arctan^{2} \sqrt{4 a - 1}}{i \sqrt{4 a - 1}}$ $Res (f, z_{2}) = \frac{\ln^{2} z_{2}}{z_{2} - z_{1}} = \frac{{(\ln \sqrt{a} - iarctan \sqrt{4 a - 1})}^{2}}{- i \sqrt{4 a - 1}}$ $= \frac{\ln^{2} \sqrt{a} - 2 iln \sqrt{a} \arctan \sqrt{4 a - 1} - \arctan^{2} \sqrt{4 a - 1}}{- i \sqrt{4 a - 1}} \Rightarrow$ $Σ Res (f) = \frac{1}{i \sqrt{4 a - 1}} {\ln^{2} (\sqrt{a}) + 2 iln \sqrt{a} \arctan \sqrt{4 a - 1} - \arctan^{2} \sqrt{4 a - 1}$ $- \ln^{2} \sqrt{a} + 2 iln \sqrt{a} \arctan \sqrt{4 a - 1} + \arctan^{2} \sqrt{4 a - 1}}$ $= \frac{1}{i \sqrt{4 a - 1}} (2 iln (\sqrt{a}) \arctan \sqrt{4 a - 1}) = \frac{lna}{\sqrt{4 a - 1}} \arctan \sqrt{4 a - 1} \Rightarrow$ $f (a) = - \frac{lna}{2 \sqrt{4 a - 1}} \arctan \sqrt{4 a - 1}$
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