calculste ∫_0 ^∞ ((ln(x))/(x^2 −x+1))dx


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Question Number 108750 by mathmax by abdo last updated on 19/Aug/20

$calculste \int_{0}^{\infty} \frac{\ln (x)}{x^{2} - x + 1} dx$
	Answered by mnjuly1970 last updated on 19/Aug/20

	$sol . . . . : put : x = \frac{1}{t} \Rightarrow Ω = \int_{0}^{\infty} \frac{l n (x)}{x^{2} - x + 1} d x = - Ω$ $2 Ω = 0 \Rightarrow Ω = \int_{0}^{\infty} \frac{l n (x)}{1 - x + x^{2}} d x = 0$ $. . . . . M . N . . . . .$
	Answered by mathmax by abdo last updated on 19/Aug/20
	$if q is a fraction with no real poles we have ∫_0 ^∞ q(x)lnx dx =−(1/2)Re(Σ_i Re( q(z)ln^2 z,a_i ) let use this we have q(z)ln^2 z =((ln^2 z)/(z^2 −z+1)) =w(z) poles? z^2 −z+1 =0 →Δ =−3 ⇒z_1 =((1+i(√3))/2) =e^((i2π)/3) and z_2 =((1−i(√3))/2) =e^(−((i2π)/3)) w(z) =((ln^2 z)/((z−z_1 )(z−z_2 ))) ⇒Res(w,z_1 ) =((ln^2 (z_1 ))/(z_1 −z_2 )) =((ln^2 (e^((i2π)/3) ))/(i(√3))) =(1/(i(√3)))(((i2π)/3))^2 =((−4π^2 )/(3i(√3))) also Res(w,z_2 ) =((ln^2 (z_2 ))/(z_2 −z_1 )) =((4π^2 )/(3i(√3))) ⇒ Σ Re(q(z)ln^2 (z),a_i )=0 ⇒∫_0 ^∞ ((ln(x))/(x^2 −x+1))dx =0$
	$if q is a fraction with no real poles we have$ $\int_{0}^{\infty} q (x) lnx dx = - \frac{1}{2} Re (\sum_{i} Re (q (z) \ln^{2} z, a_{i}) let use this$ $we have q (z) \ln^{2} z = \frac{\ln^{2} z}{z^{2} - z + 1} = w (z) poles ?$ $z^{2} - z + 1 = 0 \to Δ = - 3 \Rightarrow z_{1} = \frac{1 + i \sqrt{3}}{2} = e^{\frac{i 2 π}{3}} and z_{2} = \frac{1 - i \sqrt{3}}{2} = e^{- \frac{i 2 π}{3}}$ $w (z) = \frac{\ln^{2} z}{(z - z_{1}) (z - z_{2})} \Rightarrow Res (w, z_{1}) = \frac{\ln^{2} (z_{1})}{z_{1} - z_{2}} = \frac{\ln^{2} (e^{\frac{i 2 π}{3}})}{i \sqrt{3}}$ $= \frac{1}{i \sqrt{3}} {(\frac{i 2 π}{3})}^{2} = \frac{- 4 π^{2}}{3 i \sqrt{3}} also Res (w, z_{2}) = \frac{\ln^{2} (z_{2})}{z_{2} - z_{1}} = \frac{4 π^{2}}{3 i \sqrt{3}} \Rightarrow$ $Σ Re (q (z) \ln^{2} (z), a_{i}) = 0 \Rightarrow \int_{0}^{\infty} \frac{\ln (x)}{x^{2} - x + 1} dx = 0$
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