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


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Question Number 117249 by mathmax by abdo last updated on 10/Oct/20

$calculate \int_{0}^{\infty} \frac{{(- 1)}^{x^{2}}}{x^{4} + x^{2} + 1} dx$
	Answered by mathmax by abdo last updated on 11/Oct/20
	$A =∫_0 ^∞ (((−1)^x^2 )/(x^4 +x^2 +1))dx ⇒2A =∫_(−∞) ^(+∞) (e^(iπx^2 ) /(x^4 +x^2 +1))dx let ϕ(z) =(e^(iπz^2 ) /(z^4 +z^2 +1)) poles of ϕ? t^2 +t+1 =0 (t=z^2 )→Δ =−3 ⇒t_1 =((−1+i(√3))/2) =e^(i((2π)/3)) t_2 =((−1−i(√3))/2)=e^(−((i2π)/3)) ⇒ϕ(z) =(e^(iπz^2 ) /((z^2 −e^((i2π)/3) )(z^2 −e^(−((i2π)/3)) ))) =(e^(iπz^2 ) /((z−e^((iπ)/3) )(z+e^((iπ)/3) )(z−e^(−((iπ)/3)) )(z+e^(−((iπ)/3)) ))) so the poles are +^− e^((iπ)/3) and+^− e^(−((iπ)/3)) ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ{Res(ϕ,e^((iπ)/3) ) +Res(ϕ,−e^(−((iπ)/3)) )} Res(ϕ,e^((iπ)/3) ) =(e^(iπ(e^((2iπ)/3) )) /(2e^((iπ)/3) (2isin(((2π)/3))))) =(1/(4i(((√3)/2))))e^(−((iπ)/3)) e^(iπ(−(1/2)+((i(√3))/2))) =(1/(2i(√3))) e^(−((iπ)/3)) (−i) e^(−((π(√3))/2)) =−(1/(2(√3))) e^(−((iπ)/3)) e^(−((π(√3))/2)) Res(ϕ,−e^(−((iπ)/3)) ) =(e^(iπ(e^(−((2iπ)/3)) )) /((−2i sin(((2π)/3)))(−2e^(−((iπ)/3)) ))) =(1/(4i(((√3)/2))))e^((iπ)/3) e^(iπ(−(1/2)−((i(√3))/2))) =((−i)/(2i(√3))) e^((iπ)/3) .e^((π(√3))/2) =−(1/(2(√3))) e^((iπ)/3) e^((π(√3))/2) ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ(−(1/(2(√3)))){ e^(−((iπ)/3)) e^(−((π(√3))/2)) +e^((iπ)/3) e^((π(√3))/2) } =((−iπ)/(√3)){e^(−((π(√3))/2)) ((1/2)−((i(√3))/2)) +e^((π(√3))/2) ((1/2)+i((√3)/2))} =((−iπ)/(√3)){ (1/2)(e^((π(√2))/2) +e^(−((π(√2))/2)) )+i((√3)/2)(e^((π(√3))/2) −e^(−π((√3)/2)) )} =−((iπ)/(√3))ch(((π(√3))/2))+π sh(((π(√3))/2)) =2A ⇒ A =(π/2)sh(((π(√3))/2))−((iπ)/(2(√3)))ch(((π(√3))/2))$
	$A = \int_{0}^{\infty} \frac{{(- 1)}^{x^{2}}}{x^{4} + x^{2} + 1} dx \Rightarrow 2 A = \int_{- \infty}^{+ \infty} \frac{e^{i π x^{2}}}{x^{4} + x^{2} + 1} dx$ $let φ (z) = \frac{e^{i π z^{2}}}{z^{4} + z^{2} + 1} poles of φ ?$ $t^{2} + t + 1 = 0 (t = z^{2}) \to Δ = - 3 \Rightarrow t_{1} = \frac{- 1 + i \sqrt{3}}{2} = e^{i \frac{2 π}{3}}$ $t_{2} = \frac{- 1 - i \sqrt{3}}{2} = e^{- \frac{i 2 π}{3}} \Rightarrow φ (z) = \frac{e^{i π z^{2}}}{(z^{2} - e^{\frac{i 2 π}{3}}) (z^{2} - e^{- \frac{i 2 π}{3}})}$ $= \frac{e^{i π z^{2}}}{(z - e^{\frac{i π}{3}}) (z + e^{\frac{i π}{3}}) (z - e^{- \frac{i π}{3}}) (z + e^{- \frac{i π}{3}})} so the poles are \bar{+} e^{\frac{i π}{3}} and \bar{+} e^{- \frac{i π}{3}}$ $\int_{- \infty}^{+ \infty} φ (z) dz = 2 i π {Res (φ, e^{\frac{i π}{3}}) + Res (φ, - e^{- \frac{i π}{3}})}$ $Res (φ, e^{\frac{i π}{3}}) = \frac{e^{i π (e^{\frac{2 i π}{3}})}}{{2 e}^{\frac{i π}{3}} (2 isin (\frac{2 π}{3}))} = \frac{1}{4 i (\frac{\sqrt{3}}{2})} e^{- \frac{i π}{3}} e^{i π (- \frac{1}{2} + \frac{i \sqrt{3}}{2})}$ $= \frac{1}{2 i \sqrt{3}} e^{- \frac{i π}{3}} (- i) e^{- \frac{π \sqrt{3}}{2}} = - \frac{1}{2 \sqrt{3}} e^{- \frac{i π}{3}} e^{- \frac{π \sqrt{3}}{2}}$ $Res (φ, - e^{- \frac{i π}{3}}) = \frac{e^{i π (e^{- \frac{2 i π}{3}})}}{(- 2 i \sin (\frac{2 π}{3})) (- {2 e}^{- \frac{i π}{3}})} = \frac{1}{4 i (\frac{\sqrt{3}}{2})} e^{\frac{i π}{3}} e^{i π (- \frac{1}{2} - \frac{i \sqrt{3}}{2})}$ $= \frac{- i}{2 i \sqrt{3}} e^{\frac{i π}{3}} . e^{\frac{π \sqrt{3}}{2}} = - \frac{1}{2 \sqrt{3}} e^{\frac{i π}{3}} e^{\frac{π \sqrt{3}}{2}} \Rightarrow$ $\int_{- \infty}^{+ \infty} φ (z) dz = 2 i π (- \frac{1}{2 \sqrt{3}}) {e^{- \frac{i π}{3}} e^{- \frac{π \sqrt{3}}{2}} + e^{\frac{i π}{3}} e^{\frac{π \sqrt{3}}{2}}}$ $= \frac{- i π}{\sqrt{3}} {e^{- \frac{π \sqrt{3}}{2}} (\frac{1}{2} - \frac{i \sqrt{3}}{2}) + e^{\frac{π \sqrt{3}}{2}} (\frac{1}{2} + i \frac{\sqrt{3}}{2})}$ $= \frac{- i π}{\sqrt{3}} {\frac{1}{2} (e^{\frac{π \sqrt{2}}{2}} + e^{- \frac{π \sqrt{2}}{2}}) + i \frac{\sqrt{3}}{2} (e^{\frac{π \sqrt{3}}{2}} - e^{- π \frac{\sqrt{3}}{2}})}$ $= - \frac{i π}{\sqrt{3}} ch (\frac{π \sqrt{3}}{2}) + π sh (\frac{π \sqrt{3}}{2}) = 2 A \Rightarrow$ $A = \frac{π}{2} sh (\frac{π \sqrt{3}}{2}) - \frac{i π}{2 \sqrt{3}} ch (\frac{π \sqrt{3}}{2})$
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