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


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Question Number 97620 by mathmax by abdo last updated on 08/Jun/20

$calculate \int_{0}^{\infty} \frac{\sin (π x^{2})}{x^{4} - x^{2} + 1} dx$
	Answered by mathmax by abdo last updated on 10/Jun/20
	$let I =∫_0 ^∞ ((sin(πx^2 ))/(x^4 −x^2 +1))dx ⇒2I =∫_(−∞) ^(+∞) ((sin(πx^2 ))/(x^4 −x^2 +1))dx =Im(∫_(−∞) ^(+∞) (e^(iπx^2 ) /(x^4 −x^2 +1))dx) let ϕ(z) =(e^(iπz^2 ) /(z^4 −x^2 +1)) poles of ϕ? z^4 −z^2 +1 =0⇒t^2 −t +1 =0 (t=z^2 ) Δ =−3 ⇒t_1 =((1+i(√3))/2) =e^((iπ)/3) and t_2 =((1−i(√3))/2) =e^(−((iπ)/3)) ⇒ ϕ(z) =(e^(iπz^2 ) /((z^2 −e^((iπ)/3) )(z^2 −e^(−((iπ)/3)) ))) =(e^(iπz^2 ) /((z−e^((iπ)/6) )(z+e^((iπ)/6) )(z−e^(−((iπ)/6)) )(z+e^(−((iπ)/6)) ))) residus theorem give ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ{ Res(ϕ,e^((iπ)/6) ) +Res(ϕ,−e^(−((iπ)/6)) )} Res(ϕ,e^((iπ)/6) ) =lim_(z→e^((iπ)/6) ) (z−e^((iπ)/6) )ϕ(z) =(e^(iπe^((iπ)/3) ) /(2e^((iπ)/6) (2isin((π/3))))) =(e^(iπ((1/2)+((i(√3))/2))) /(4i×((√3)/2))) e^(−((iπ)/6)) =(1/(2i(√3))) e^(−((iπ)/6)) i . e^(−((π(√3))/4)) =(1/(2(√3))) e^(−((π(√3))/4)) e^(−((iπ)/6)) Res(ϕ,−e^(−((iπ)/6)) ) =(e^(iπe^(−((iπ)/3)) ) /(−2e^(−((iπ)/6)) (−2i sin((π/3))))) =(1/(4i×((√3)/2))) e^((iπ)/6) e^(iπ((1/2)−((i(√3))/2))) =(1/(2i(√3))) .i .e^((π(√3))/4) e^((iπ)/6) =(1/(2(√3))) e^((π(√3))/4) e^((iπ)/6) ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ{ (1/(2(√3))) e^(−((π(√3))/4)) e^(−((iπ)/6)) +(1/(2(√3))) e^((π(√3))/4) e^((iπ)/6) } =((iπ)/(√3)){ e^(−((π(√3))/2)) (((√3)/2)−(i/2)) +e^((π(√3))/4) (((√3)/2)+(i/2))} =((iπ)/(√3)){ ((√3)/2) e^(−((π(√3))/4)) −(i/2) e^(−((π(√3))/4)) +((√3)/2) e^((π(√3))/4) +(i/2) e^((π(√3))/4) } =((iπ)/2) e^(−((π(√3))/4)) +((iπ)/2) e^((π(√3))/4) +(...) ⇒Im(∫_(−∞) ^(+∞) ϕ(z)dz =(π/2) (e^((π(√3))/4) +e^(−((π(√3))/4)) ) =2I ⇒ ★I =(π/4)( e^((π(√3))/4) +e^(−((π(√3))/4)) )★$
	$let I = \int_{0}^{\infty} \frac{\sin (π x^{2})}{x^{4} - x^{2} + 1} dx \Rightarrow 2 I = \int_{- \infty}^{+ \infty} \frac{\sin (π x^{2})}{x^{4} - x^{2} + 1} dx = Im (\int_{- \infty}^{+ \infty} \frac{e^{i π x^{2}}}{x^{4} - x^{2} + 1} dx)$ $let φ (z) = \frac{e^{i π z^{2}}}{z^{4} - x^{2} + 1} poles of φ ?$ $z^{4} - z^{2} + 1 = 0 \Rightarrow t^{2} - t + 1 = 0 (t = z^{2})$ $Δ = - 3 \Rightarrow t_{1} = \frac{1 + i \sqrt{3}}{2} = e^{\frac{i π}{3}} and t_{2} = \frac{1 - i \sqrt{3}}{2} = e^{- \frac{i π}{3}} \Rightarrow$ $φ (z) = \frac{e^{i π z^{2}}}{(z^{2} - e^{\frac{i π}{3}}) (z^{2} - e^{- \frac{i π}{3}})} = \frac{e^{i π z^{2}}}{(z - e^{\frac{i π}{6}}) (z + e^{\frac{i π}{6}}) (z - e^{- \frac{i π}{6}}) (z + e^{- \frac{i π}{6}})}$ $residus theorem give$ $\int_{- \infty}^{+ \infty} φ (z) dz = 2 i π {Res (φ, e^{\frac{i π}{6}}) + Res (φ, - e^{- \frac{i π}{6}})}$ $Res (φ, e^{\frac{i π}{6}}) = \lim_{z \to e^{\frac{i π}{6}}} (z - e^{\frac{i π}{6}}) φ (z) = \frac{e^{i π e^{\frac{i π}{3}}}}{{2 e}^{\frac{i π}{6}} (2 isin (\frac{π}{3}))} = \frac{e^{i π (\frac{1}{2} + \frac{i \sqrt{3}}{2})}}{4 i \times \frac{\sqrt{3}}{2}} e^{- \frac{i π}{6}}$ $= \frac{1}{2 i \sqrt{3}} e^{- \frac{i π}{6}} i . e^{- \frac{π \sqrt{3}}{4}} = \frac{1}{2 \sqrt{3}} e^{- \frac{π \sqrt{3}}{4}} e^{- \frac{i π}{6}}$ $Res (φ, - e^{- \frac{i π}{6}}) = \frac{e^{i π e^{- \frac{i π}{3}}}}{- {2 e}^{- \frac{i π}{6}} (- 2 i \sin (\frac{π}{3}))} = \frac{1}{4 i \times \frac{\sqrt{3}}{2}} e^{\frac{i π}{6}} e^{i π (\frac{1}{2} - \frac{i \sqrt{3}}{2})}$ $= \frac{1}{2 i \sqrt{3}} . i . e^{\frac{π \sqrt{3}}{4}} e^{\frac{i π}{6}} = \frac{1}{2 \sqrt{3}} e^{\frac{π \sqrt{3}}{4}} e^{\frac{i π}{6}} \Rightarrow$ $\int_{- \infty}^{+ \infty} φ (z) dz = 2 i π {\frac{1}{2 \sqrt{3}} e^{- \frac{π \sqrt{3}}{4}} e^{- \frac{i π}{6}} + \frac{1}{2 \sqrt{3}} e^{\frac{π \sqrt{3}}{4}} e^{\frac{i π}{6}}}$ $= \frac{i π}{\sqrt{3}} {e^{- \frac{π \sqrt{3}}{2}} (\frac{\sqrt{3}}{2} - \frac{i}{2}) + e^{\frac{π \sqrt{3}}{4}} (\frac{\sqrt{3}}{2} + \frac{i}{2})}$ $= \frac{i π}{\sqrt{3}} {\frac{\sqrt{3}}{2} e^{- \frac{π \sqrt{3}}{4}} - \frac{i}{2} e^{- \frac{π \sqrt{3}}{4}} + \frac{\sqrt{3}}{2} e^{\frac{π \sqrt{3}}{4}} + \frac{i}{2} e^{\frac{π \sqrt{3}}{4}}}$ $= \frac{i π}{2} e^{- \frac{π \sqrt{3}}{4}} + \frac{i π}{2} e^{\frac{π \sqrt{3}}{4}} + (. . .) \Rightarrow Im (\int_{- \infty}^{+ \infty} φ (z) dz = \frac{π}{2} (e^{\frac{π \sqrt{3}}{4}} + e^{- \frac{π \sqrt{3}}{4}}) = 2 I \Rightarrow$ $★ I = \frac{π}{4} (e^{\frac{π \sqrt{3}}{4}} + e^{- \frac{π \sqrt{3}}{4}}) ★$
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