calculate ∫_0 ^∞ ((x^2 −2cosx+1)/(x^4 +x^2 +1))dx


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Question Number 48720 by Abdo msup. last updated on 27/Nov/18

$c a l c u l a t e \int_{0}^{\infty} \frac{x^{2} - 2 c o s x + 1}{x^{4} + x^{2} + 1} d x$
Commented by Abdo msup. last updated on 28/Nov/18
$let I = ∫_0 ^∞ ((x^2 −2cosx +1)/(x^4 +x^2 +1))dx ⇒ 2I =∫_(−∞) ^(+∞) ((x^2 +1)/(x^4 +x^2 +1))dx −∫_(−∞) ^(+∞) ((2cosx)/(x^4 +x^2 +1))dx=H−K let find H let ϕ(z)=((z^2 +1)/(z^4 +z^2 +1)) poles of ϕ? z^4 +z^2 +1=0 ⇒t^2 +t+1=0 (t=z^2 ) Δ=−3=(i(√3))^2 ⇒z_1 =((−1+i(√3))/2) =e^((i2π)/3) and z_2 =e^(−((i2π)/3)) z^2 =e^((i2π)/3) ⇒ z =+^− e^((iπ)/3) z^2 =e^(−((i2π)/3)) ⇒z =+^− e^(−((iπ)/3)) ⇒ ϕ(z)=((z^2 +1)/((z−e^((iπ)/3) )(z+e^((iπ)/3) )(z−e^(−((iπ)/3)) )(z+e^(−((iπ)/3)) ))) ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ{ Res(ϕ,e^((iπ)/3) ) +Res(ϕ,−e^(−((iπ)/3)) ) Res(ϕ,e^((iπ)/3) )=((e^(2((iπ)/3)) +1)/(2e^((iπ)/3) (2isin((π/3))(2cos((π/3))))) =((e^((iπ)/3) +e^(−((iπ)/3)) )/(8i sin((π/3))cos((π/3)))) =((2cos((π/3)))/(8isin((π/3))cos((π/3)))) =(1/(4i((√3)/2))) =(1/(2i(√3))) Res(ϕ,−e^(−((iπ)/3)) ) =((e^(−((2iπ)/3)) +1)/(−2cos((π/3))(2isin((π/3)))(−2e^(−((iπ)/3)) ))) =((e^((iπ)/3) +e^((−iπ)/3) )/(8i cos((π/3))sin((π/3)))) =((2cos((π/3)))/(8i cos((π/3))sin((π/3)))) =(1/(4isin((π/3)))) = (1/(4i ((√3)/2))) =(1/(2i(√3))) ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ{ (1/(2i(√3))) +(1/(2i(√3)))} =((2π)/(√3)) =H$
$l e t I = \int_{0}^{\infty} \frac{x^{2} - 2 c o s x + 1}{x^{4} + x^{2} + 1} d x \Rightarrow$ $2 I = \int_{- \infty}^{+ \infty} \frac{x^{2} + 1}{x^{4} + x^{2} + 1} d x - \int_{- \infty}^{+ \infty} \frac{2 c o s x}{x^{4} + x^{2} + 1} d x = H - K$ $l e t f i n d H l e t φ (z) = \frac{z^{2} + 1}{z^{4} + z^{2} + 1} p o l e s o f φ ?$ $z^{4} + z^{2} + 1 = 0 \Rightarrow t^{2} + t + 1 = 0 (t = z^{2})$ $Δ = - 3 = {(i \sqrt{3})}^{2} \Rightarrow z_{1} = \frac{- 1 + i \sqrt{3}}{2} = e^{\frac{i 2 π}{3}} a n d z_{2} = e^{- \frac{i 2 π}{3}}$ $z^{2} = e^{\frac{i 2 π}{3}} \Rightarrow z = \bar{+} e^{\frac{i π}{3}}$ $z^{2} = e^{- \frac{i 2 π}{3}} \Rightarrow z = \bar{+} e^{- \frac{i π}{3}} \Rightarrow$ $φ (z) = \frac{z^{2} + 1}{(z - e^{\frac{i π}{3}}) (z + e^{\frac{i π}{3}}) (z - e^{- \frac{i π}{3}}) (z + e^{- \frac{i π}{3}})}$ $\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π {R e s (φ, e^{\frac{i π}{3}}) + R e s (φ, - e^{- \frac{i π}{3}})$ $R e s (φ, e^{\frac{i π}{3}}) = \frac{e^{2 \frac{i π}{3}} + 1}{2 e^{\frac{i π}{3}} (2 i s i n (\frac{π}{3}) (2 c o s (\frac{π}{3}))}$ $= \frac{e^{\frac{i π}{3}} + e^{- \frac{i π}{3}}}{8 i s i n (\frac{π}{3}) c o s (\frac{π}{3})} = \frac{2 c o s (\frac{π}{3})}{8 i s i n (\frac{π}{3}) c o s (\frac{π}{3})} = \frac{1}{4 i \frac{\sqrt{3}}{2}} = \frac{1}{2 i \sqrt{3}}$ $R e s (φ, - e^{- \frac{i π}{3}}) = \frac{e^{- \frac{2 i π}{3}} + 1}{- 2 c o s (\frac{π}{3}) (2 i s i n (\frac{π}{3})) (- 2 e^{- \frac{i π}{3}})}$ $= \frac{e^{\frac{i π}{3}} + e^{\frac{- i π}{3}}}{8 i c o s (\frac{π}{3}) s i n (\frac{π}{3})} = \frac{2 c o s (\frac{π}{3})}{8 i c o s (\frac{π}{3}) s i n (\frac{π}{3})}$ $= \frac{1}{4 i s i n (\frac{π}{3})} = \frac{1}{4 i \frac{\sqrt{3}}{2}} = \frac{1}{2 i \sqrt{3}} \Rightarrow$ $\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π {\frac{1}{2 i \sqrt{3}} + \frac{1}{2 i \sqrt{3}}} = \frac{2 π}{\sqrt{3}} = H$
Commented by Abdo msup. last updated on 28/Nov/18
$let find K K = ∫_(−∞) ^(+∞) ((2cosx)/(x^4 +x^2 +1))dx =Re(∫_(−∞) ^(+∞) ((2 e^(ix) )/(x^(4 ) +x^2 +1))dx) let W(z) =((2 e^(iz) )/(z^4 +z^2 +1)) ⇒ W(z)=((2 e^(iz) )/(z^4 +z^2 +1)) ⇒W(z) =((2 e^(iz) )/((z−e^((iπ)/3) )(z +e^((iπ)/3) )(z−e^(−((iπ)/3)) )(z+e^(−((iπ)/3)) ))) ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ { Res(ϕ,e^((iπ)/3) ) +Res(ϕ,−e^(−((iπ)/3)) )} Res(ϕ,e^((iπ)/3) ) = ((2 e^(i(cos(π/3)+isin((π/3)))) )/(2 e^((iπ)/3) (2i sin((π/3)))2cos((π/3)))) = (e^(i((1/2)+i((√3)/2))) /(4isin((π/3))cos((π/3)))) e^(−((iπ)/3)) =((e^(−((√3)/2)) e^(i((1/2)−(π/3))) )/(4i((√3)/2).(1/2))) =(e^(−((√3)/2)) /(i(√3))) e^(i((1/2)−(π/3))) Res(ϕ,−e^(−((iπ)/3)) ) = ((2 e^(i(cos(π/3)−isin((π/3)))) )/(−2cos((π/3))(2isin((π/3))(−2 e^((−iπ)/3) ))) =(e^(i((1/2)−i((√3)/2))) /(4i cos((π/3))sin((π/3)))) e^((iπ)/3) =((e^((√3)/2) e^(i((1/2)+(π/3))) )/(4i (1/2) ((√3)/2))) = ((e^((√3)/(2 )) e^(i((1/2)+(π/3))) )/(i(√3))) ⇒ ∫_(−∞) ^(+∞) W(z)dz =2iπ(1/(i(√3))){ e^(−((√3)/2)) e^(i((1/2)−(π/3))) + e^((√3)/2) e^(i((1/2)+(π/3){) } =((2π)/(√3)){ e^(−((√3)/2)) {cos((1/2)−(π/3))+isin((1/2)−(π/3))+ e^((√3)/2) { cos((1/2)+(π/3))+i sin((1/2)+(π/3))}} K =Re(∫_(−∞) ^(+∞) W(z)dz) =((2π)/(√3)){ e^(−((√3)/2)) cos((1/2)−(π/3)) +e^((√3)/2) cos((1/2) +(π/3))} so thevalue of I is determined.$
$l e t f i n d K$ $K = \int_{- \infty}^{+ \infty} \frac{2 c o s x}{x^{4} + x^{2} + 1} d x = R e (\int_{- \infty}^{+ \infty} \frac{2 e^{i x}}{x^{4} + x^{2} + 1} d x)$ $l e t W (z) = \frac{2 e^{i z}}{z^{4} + z^{2} + 1} \Rightarrow$ $W (z) = \frac{2 e^{i z}}{z^{4} + z^{2} + 1} \Rightarrow W (z) = \frac{2 e^{i z}}{(z - e^{\frac{i π}{3}}) (z + e^{\frac{i π}{3}}) (z - e^{- \frac{i π}{3}}) (z + e^{- \frac{i π}{3}})}$ $\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π {R e s (φ, e^{\frac{i π}{3}}) + R e s (φ, - e^{- \frac{i π}{3}})}$ $R e s (φ, e^{\frac{i π}{3}}) = \frac{2 e^{i (c o s \frac{π}{3} + i s i n (\frac{π}{3}))}}{2 e^{\frac{i π}{3}} (2 i s i n (\frac{π}{3})) 2 c o s (\frac{π}{3})}$ $= \frac{e^{i (\frac{1}{2} + i \frac{\sqrt{3}}{2})}}{4 i s i n (\frac{π}{3}) c o s (\frac{π}{3})} e^{- \frac{i π}{3}} = \frac{e^{- \frac{\sqrt{3}}{2}} e^{i (\frac{1}{2} - \frac{π}{3})}}{4 i \frac{\sqrt{3}}{2} . \frac{1}{2}}$ $= \frac{e^{- \frac{\sqrt{3}}{2}}}{i \sqrt{3}} e^{i (\frac{1}{2} - \frac{π}{3})}$ $R e s (φ, - e^{- \frac{i π}{3}}) = \frac{2 e^{i (c o s \frac{π}{3} - i s i n (\frac{π}{3}))}}{- 2 c o s (\frac{π}{3}) (2 i s i n (\frac{π}{3}) (- 2 e^{\frac{- i π}{3}})}$ $= \frac{e^{i (\frac{1}{2} - i \frac{\sqrt{3}}{2})}}{4 i c o s (\frac{π}{3}) s i n (\frac{π}{3})} e^{\frac{i π}{3}} = \frac{e^{\frac{\sqrt{3}}{2}} e^{i (\frac{1}{2} + \frac{π}{3})}}{4 i \frac{1}{2} \frac{\sqrt{3}}{2}} = \frac{e^{\frac{\sqrt{3}}{2}} e^{i (\frac{1}{2} + \frac{π}{3})}}{i \sqrt{3}} \Rightarrow$ $\int_{- \infty}^{+ \infty} W (z) d z = 2 i π \frac{1}{i \sqrt{3}} {e^{- \frac{\sqrt{3}}{2}} e^{i (\frac{1}{2} - \frac{π}{3})} + e^{\frac{\sqrt{3}}{2}} e^{i (\frac{1}{2} + \frac{π}{3} {}}$ $= \frac{2 π}{\sqrt{3}} {e^{- \frac{\sqrt{3}}{2}} {c o s (\frac{1}{2} - \frac{π}{3}) + i s i n (\frac{1}{2} - \frac{π}{3}) +$ $e^{\frac{\sqrt{3}}{2}} {c o s (\frac{1}{2} + \frac{π}{3}) + i s i n (\frac{1}{2} + \frac{π}{3})}}$ $K = R e (\int_{- \infty}^{+ \infty} W (z) d z)$ $= \frac{2 π}{\sqrt{3}} {e^{- \frac{\sqrt{3}}{2}} c o s (\frac{1}{2} - \frac{π}{3}) + e^{\frac{\sqrt{3}}{2}} c o s (\frac{1}{2} + \frac{π}{3})}$ $s o t h e v a l u e o f I i s d e t e r m i n e d .$
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