calculate ∫_0 ^∞ ((lnx)/((x^(2 ) +x+2)^2 ))dx


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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
	$let f(a) =∫_0 ^∞ ((lnx)/(x^2 +x +a))dx with a>(1/4) we have f^′ (a) =−∫_0 ^∞ ((lnx)/((x^2 +x+a)^2 )) and ∫_0 ^∞ ((lnx)/((x^2 +x+a)^2 ))=−f^′ (2) let explicite f(a) by using ∫_0 ^∞ q(x)ln(x)dx=−(1/2)Re(Σ_i Res(q(z)ln^2 z ,a_i ) ϕ(z) =q(z)ln^2 z =((ln^2 z)/(z^2 +z+a)) poles of ϕ? Δ =1−4a<0 ⇒z_1 =((−1+i(√(4a−1)))/2) and z_2 =((−1−i(√(4a−1)))/2) ∣z_1 ∣ =(1/2)(√(1+4a−1))=(√a) ⇒z_1 =(√a)e^(−iarctan(√(4a−1))) and z_2 =(√a)e^(iarctan(√(4a−1))) we have ϕ(z) =((ln^2 z)/((z−z_1 )(z−z_2 ))) Res(ϕ,z_1 ) =lim_(z→z_1 ) (z−z_1 )ϕ(z) =((ln^2 z_1 )/(z_1 −z_2 )) =(((ln(√a)+iarctan(√(4a−1)))^2 )/(i(√(4a−1)))) =((ln^2 (√a)+2iln(√a)arctan(√(4a−1))−arctan^2 (√(4a−1)))/(i(√(4a−1)))) =−i((ln^2 (√a)+2iln((√a)) arctan(√(4a−1))−arctan^2 (√(4a−1)))/(√(4a−1))) =((−iln^2 (√a) +lna arctan(√(4a−1))+iarctan^2 (√(4a−1)))/(√(4a−1))) Res(ϕ,z_2 ) =lim_(z→z_2 ) (z−z_2 )ϕ(z) =((ln^2 z_2 )/(z_2 −z_1 )) =(((ln(√a)+iarctan(√(4a−1)))^2 )/(−i(√(4a−1)))) =((ln^2 (√a)+2iln(√a)arctan(√(4a−1))−arctan^2 (√(4a−1)))/(−i(√(4a−1)))) =((iln^2 (√a)−lna arctan(√(4a−1))−iarctan^2 (√(4a−1)))/(√(4a−1))) ⇒ Σ Res=(1/(√(4a−1))){−iln^2 (√a)+lna arctan(√(4a−1))+i arctan^2 (√(4a−1)) +iln^2 (√a)−lna arctan(√(4a−1))−iarctan(√(4a−1))} Re(Σ Res(...))=(1/(√(4a−1))){o} =0 ⇒∫_0 ^∞ ((lnx)/(x^2 +x+a))dx =0 ⇒ ∫_0 ^∞ ((lnx)/((x^2 +x+a)^2 ))dx =0$
	$let f (a) = \int_{0}^{\infty} \frac{lnx}{x^{2} + x + a} dx with a > \frac{1}{4}$ $we have f^{^{'}} (a) = - \int_{0}^{\infty} \frac{lnx}{{(x^{2} + x + a)}^{2}} and \int_{0}^{\infty} \frac{lnx}{{(x^{2} + x + a)}^{2}} = - f^{^{'}} (2)$ $let explicite f (a) by using \int_{0}^{\infty} q (x) \ln (x) dx = - \frac{1}{2} Re (\sum_{i} Res (q (z) \ln^{2} z, a_{i})$ $φ (z) = q (z) \ln^{2} z = \frac{\ln^{2} z}{z^{2} + z + a} poles of φ ?$ $Δ = 1 - 4 a < 0 \Rightarrow z_{1} = \frac{- 1 + i \sqrt{4 a - 1}}{2} and z_{2} = \frac{- 1 - i \sqrt{4 a - 1}}{2}$ $∣ z_{1} ∣ = \frac{1}{2} \sqrt{1 + 4 a - 1} = \sqrt{a} \Rightarrow z_{1} = \sqrt{a} e^{- iarctan \sqrt{4 a - 1}}$ $and z_{2} = \sqrt{a} e^{iarctan \sqrt{4 a - 1}} we have φ (z) = \frac{\ln^{2} z}{(z - z_{1}) (z - z_{2})}$ $Res (φ, z_{1}) = \lim_{z \to z_{1}} (z - z_{1}) φ (z) = \frac{\ln^{2} z_{1}}{z_{1} - z_{2}}$ $= \frac{{(\ln \sqrt{a} + iarctan \sqrt{4 a - 1})}^{2}}{i \sqrt{4 a - 1}} = \frac{\ln^{2} \sqrt{a} + 2 iln \sqrt{a} \arctan \sqrt{4 a - 1} - \arctan^{2} \sqrt{4 a - 1}}{i \sqrt{4 a - 1}}$ $= - i \frac{\ln^{2} \sqrt{a} + 2 iln (\sqrt{a}) \arctan \sqrt{4 a - 1} - \arctan^{2} \sqrt{4 a - 1}}{\sqrt{4 a - 1}}$ $= \frac{- {iln}^{2} \sqrt{a} + lna \arctan \sqrt{4 a - 1} + {iarctan}^{2} \sqrt{4 a - 1}}{\sqrt{4 a - 1}}$ $Res (φ, z_{2}) = \lim_{z \to z_{2}} (z - z_{2}) φ (z) = \frac{\ln^{2} z_{2}}{z_{2} - z_{1}} = \frac{{(\ln \sqrt{a} + iarctan \sqrt{4 a - 1})}^{2}}{- i \sqrt{4 a - 1}}$ $= \frac{\ln^{2} \sqrt{a} + 2 iln \sqrt{a} \arctan \sqrt{4 a - 1} - \arctan^{2} \sqrt{4 a - 1}}{- i \sqrt{4 a - 1}}$ $= \frac{{iln}^{2} \sqrt{a} - lna \arctan \sqrt{4 a - 1} - {iarctan}^{2} \sqrt{4 a - 1}}{\sqrt{4 a - 1}} \Rightarrow$ $Σ Res = \frac{1}{\sqrt{4 a - 1}} {- {iln}^{2} \sqrt{a} + lna \arctan \sqrt{4 a - 1} + i \arctan^{2} \sqrt{4 a - 1}$ $+ {iln}^{2} \sqrt{a} - lna \arctan \sqrt{4 a - 1} - iarctan \sqrt{4 a - 1}}$ $Re (Σ Res (. . .)) = \frac{1}{\sqrt{4 a - 1}} {o} = 0 \Rightarrow \int_{0}^{\infty} \frac{lnx}{x^{2} + x + a} dx = 0 \Rightarrow$ $\int_{0}^{\infty} \frac{lnx}{{(x^{2} + x + a)}^{2}} dx = 0$
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