calculate ∫_0 ^∞ ((log^2 x)/(x^2 +x+1))dx


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Question Number 139402 by mathmax by abdo last updated on 26/Apr/21

$calculate \int_{0}^{\infty} \frac{\log^{2} x}{x^{2} + x + 1} dx$
	Answered by mathmax by abdo last updated on 29/Apr/21
	$let f(a)=∫_0 ^∞ ((x^a logx)/(x^2 +x+1))dx ⇒f(a)=∫_0 ^∞ ((e^(alogx) logx)/(x^2 +x+1))dx ⇒f^′ (a)=∫_0 ^∞ ((x^a log^2 x)/(x^2 +x+1))dx ⇒f^′ (o)=∫_0 ^∞ ((log^2 x)/(x^2 +x+1))dx f(a)=−(1/2)Re(Σ Res(ϕ a_i ) )withϕ(z)=(z^a /(z^2 +z+1))log^2 z we have ϕ(z)=((z^a log^2 z)/((z−e^((2iπ)/3) )(z−e^(−((2iπ)/3)) ))) Res(ϕ,e^((2iπ)/3) ) =((e^((2iπa)/3) (((2iπ)/3))^2 )/(2isin(((2π)/3)))) =−((4π^2 )/(9(i(√3))))e^((2iπa)/3) =−((4π^2 )/(9i(√3))) e^((2iπa)/3) Res(ϕ,e^(−((2iπ)/3)) ) =((e^(−((2iπa)/3)) (−((2iπ)/3))^2 )/((−2isin(((2π)/3)))))=−((4π^2 )/(9(−i(√3))))e^(−((2iπa)/3)) =((4π^2 )/(9i(√3)))e^(−((2iπa)/3)) ⇒Σ Res(ϕ z_i )=−((4π^2 )/(9i(√3))){ e^((2iπa)/3) −e^(−((2iπa)/3)) } =−((4π^2 )/(9i(√3)))(2i sin(((2πa)/3)) =−((8π^2 )/(9(√3)))sin(((2πa)/3)) ⇒ f(a) =((4π^2 )/(9(√3)))sin(((2πa)/3)) ⇒f^′ (a) =((4π^2 )/(9(√3)))×((2π)/3)cos(((2πa)/3)) =((8π^3 )/(27(√3))) cos(((2πa)/3)) ⇒f^′ (0)=((8π^3 )/(27(√3))) ⇒ ∫_0 ^∞ ((log^2 x)/(x^2 +x+1))dx =((8π^3 )/(27(√3)))$
	$let f (a) = \int_{0}^{\infty} \frac{x^{a} logx}{x^{2} + x + 1} dx \Rightarrow f (a) = \int_{0}^{\infty} \frac{e^{alogx} logx}{x^{2} + x + 1} dx$ $\Rightarrow f^{^{'}} (a) = \int_{0}^{\infty} \frac{x^{a} \log^{2} x}{x^{2} + x + 1} dx \Rightarrow f^{^{'}} (o) = \int_{0}^{\infty} \frac{\log^{2} x}{x^{2} + x + 1} dx$ $f (a) = - \frac{1}{2} Re (Σ Res (φ a_{i})) with φ (z) = \frac{z^{a}}{z^{2} + z + 1} \log^{2} z$ $we have φ (z) = \frac{z^{a} \log^{2} z}{(z - e^{\frac{2 i π}{3}}) (z - e^{- \frac{2 i π}{3}})}$ $Res (φ, e^{\frac{2 i π}{3}}) = \frac{e^{\frac{2 i π a}{3}} {(\frac{2 i π}{3})}^{2}}{2 isin (\frac{2 π}{3})} = - \frac{4 π^{2}}{9 (i \sqrt{3})} e^{\frac{2 i π a}{3}} = - \frac{4 π^{2}}{9 i \sqrt{3}} e^{\frac{2 i π a}{3}}$ $Res (φ, e^{- \frac{2 i π}{3}}) = \frac{e^{- \frac{2 i π a}{3}} {(- \frac{2 i π}{3})}^{2}}{(- 2 isin (\frac{2 π}{3}))} = - \frac{4 π^{2}}{9 (- i \sqrt{3})} e^{- \frac{2 i π a}{3}} = \frac{4 π^{2}}{9 i \sqrt{3}} e^{- \frac{2 i π a}{3}}$ $\Rightarrow Σ Res (φ z_{i}) = - \frac{4 π^{2}}{9 i \sqrt{3}} {e^{\frac{2 i π a}{3}} - e^{- \frac{2 i π a}{3}}}$ $= - \frac{4 π^{2}}{9 i \sqrt{3}} (2 i \sin (\frac{2 π a}{3}) = - \frac{8 π^{2}}{9 \sqrt{3}} \sin (\frac{2 π a}{3}) \Rightarrow$ $f (a) = \frac{4 π^{2}}{9 \sqrt{3}} \sin (\frac{2 π a}{3}) \Rightarrow f^{^{'}} (a) = \frac{4 π^{2}}{9 \sqrt{3}} \times \frac{2 π}{3} \cos (\frac{2 π a}{3})$ $= \frac{8 π^{3}}{27 \sqrt{3}} \cos (\frac{2 π a}{3}) \Rightarrow f^{^{'}} (0) = \frac{8 π^{3}}{27 \sqrt{3}} \Rightarrow$ $\int_{0}^{\infty} \frac{\log^{2} x}{x^{2} + x + 1} dx = \frac{8 π^{3}}{27 \sqrt{3}}$
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