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


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Question Number 76194 by abdomathmax last updated on 25/Dec/19

$c a l c u l a t e \int_{0}^{\infty} \frac{c o s (x^{2})}{x^{4} - x^{2} + 1} d x$
Commented by abdomathmax last updated on 25/Dec/19
$let A =∫_0 ^∞ ((cos(x^2 ))/(x^4 −x^2 +1))dx ⇒A =(1/2) ∫_(−∞) ^(+∞) ((cos(x^2 ))/(x^4 −x^2 +1))dx ⇒2A =Re (∫_(−∞) ^(+∞) (e^(ix^2 ) /(x^4 −x^2 +1))dx) let W(z) =(e^(iz^2 ) /(z^4 −z^2 +1)) poles of W? z^4 −z^2 +1 =0 ⇒t^2 −t +1 =0 with t=z^2 Δ=1−4=−3 ⇒t_1 =((1+(√3))/2) and t_2 =((1−(√3))/2) t_1 =e^((iπ)/3) and t_2 =e^((iπ)/3) ⇒W(z) =(e^(iz^2 ) /((z^2 −e^((iπ)/3) )(z^2 −e^(−((iπ)/3)) ))) =(e^(iz^2 ) /((z−e^((iπ)/6) )(z+e^((iπ)/6) )(z−e^(−((iπ)/6)) )(z+e^(−((iπ)/6)) ))) ∫_(−∞) ^(+∞) W(z)dz =2iπ { Res(W,e^((iπ)/6) ) +Res(W,−e^(−((iπ)/6)) )} we have Res(W, e^((iπ)/6) ) =(e^(i(((iπ)/3))) /(2e^((iπ)/6) (e^((iπ)/3) −e^(−((iπ)/3)) ))) =((e^(−(π/3)) e^(−((iπ)/6)) )/(4 sin((π/3)))) =((e^(−(π/3)) e^(−((iπ)/6)) )/(4((√3)/2))) =(e^(−(π/3)) /(2(√3))) e^(−((iπ)/6)) Res(W,−e^(−((iπ)/6)) ) =(e^(i(−((iπ)/3))) /((−2 e^(−((iπ)/6)) )( e^(−((iπ)/3)) −e^((iπ)/3) ))) =(e^(π/3) /(4 sin((π/3))))e^((iπ)/6) =((e^(π/3) e^((iπ)/3) )/(4 ((√3)/2))) =((e^(π/3) e^((iπ)/6) )/(2(√3))) ⇒ ∫_(−∞) ^(+∞) W(z)dz =((2iπ)/(2(√3))){ e^(−(π/3)) e^(−((iπ)/6)) +e^(π/3) e^((iπ)/6) } =((iπ)/(√3)){ e^(−(π/3)) ( ((√3)/2)−(i/2))+e^(π/3) (((√3)/2)+(i/2))} =(π/(√3)){i((√3)/2) e^(−(π/3)) +(1/2)e^(−(π/3)) +i((√3)/2)e^(π/3) −(1/2)e^(π/3) } ⇒ 2A =−(π/(2(√3)))( e^(π/3) −e^(−(π/3)) ) ⇒ A =−(π/(4(√3)))( e^(π/3) −e^(−(π/3)) )=(π/(4(√3)))( e^(−(π/3)) −e^(π/3) ) × ×$
$l e t A = \int_{0}^{\infty} \frac{c o s (x^{2})}{x^{4} - x^{2} + 1} d x \Rightarrow A = \frac{1}{2} \int_{- \infty}^{+ \infty} \frac{c o s (x^{2})}{x^{4} - x^{2} + 1} d x$ $\Rightarrow 2 A = R e (\int_{- \infty}^{+ \infty} \frac{e^{{i x}^{2}}}{x^{4} - x^{2} + 1} d x) l e t$ $W (z) = \frac{e^{{i z}^{2}}}{z^{4} - z^{2} + 1} p o l e s o f W ?$ $z^{4} - z^{2} + 1 = 0 \Rightarrow t^{2} - t + 1 = 0 w i t h t = z^{2}$ $Δ = 1 - 4 = - 3 \Rightarrow t_{1} = \frac{1 + \sqrt{3}}{2} a n d t_{2} = \frac{1 - \sqrt{3}}{2}$ $t_{1} = e^{\frac{i π}{3}} a n d t_{2} = e^{\frac{i π}{3}} \Rightarrow W (z) = \frac{e^{{i z}^{2}}}{(z^{2} - e^{\frac{i π}{3}}) (z^{2} - e^{- \frac{i π}{3}})}$ $= \frac{e^{{i z}^{2}}}{(z - e^{\frac{i π}{6}}) (z + e^{\frac{i π}{6}}) (z - e^{- \frac{i π}{6}}) (z + e^{- \frac{i π}{6}})}$ $\int_{- \infty}^{+ \infty} W (z) d z = 2 i π {R e s (W, e^{\frac{i π}{6}}) + R e s (W, - e^{- \frac{i π}{6}})}$ $w e h a v e R e s (W, e^{\frac{i π}{6}}) = \frac{e^{i (\frac{i π}{3})}}{2 e^{\frac{i π}{6}} (e^{\frac{i π}{3}} - e^{- \frac{i π}{3}})}$ $= \frac{e^{- \frac{π}{3}} e^{- \frac{i π}{6}}}{4 s i n (\frac{π}{3})} = \frac{e^{- \frac{π}{3}} e^{- \frac{i π}{6}}}{4 \frac{\sqrt{3}}{2}} = \frac{e^{- \frac{π}{3}}}{2 \sqrt{3}} e^{- \frac{i π}{6}}$ $R e s (W, - e^{- \frac{i π}{6}}) = \frac{e^{i (- \frac{i π}{3})}}{(- 2 e^{- \frac{i π}{6}}) (e^{- \frac{i π}{3}} - e^{\frac{i π}{3}})}$ $= \frac{e^{\frac{π}{3}}}{4 s i n (\frac{π}{3})} e^{\frac{i π}{6}} = \frac{e^{\frac{π}{3}} e^{\frac{i π}{3}}}{4 \frac{\sqrt{3}}{2}} = \frac{e^{\frac{π}{3}} e^{\frac{i π}{6}}}{2 \sqrt{3}} \Rightarrow$ $\int_{- \infty}^{+ \infty} W (z) d z = \frac{2 i π}{2 \sqrt{3}} {e^{- \frac{π}{3}} e^{- \frac{i π}{6}} + e^{\frac{π}{3}} e^{\frac{i π}{6}}}$ $= \frac{i π}{\sqrt{3}} {e^{- \frac{π}{3}} (\frac{\sqrt{3}}{2} - \frac{i}{2}) + e^{\frac{π}{3}} (\frac{\sqrt{3}}{2} + \frac{i}{2})}$ $= \frac{π}{\sqrt{3}} {i \frac{\sqrt{3}}{2} e^{- \frac{π}{3}} + \frac{1}{2} e^{- \frac{π}{3}} + i \frac{\sqrt{3}}{2} e^{\frac{π}{3}} - \frac{1}{2} e^{\frac{π}{3}}} \Rightarrow$ $2 A = - \frac{π}{2 \sqrt{3}} (e^{\frac{π}{3}} - e^{- \frac{π}{3}}) \Rightarrow$ $A = - \frac{π}{4 \sqrt{3}} (e^{\frac{π}{3}} - e^{- \frac{π}{3}}) = \frac{π}{4 \sqrt{3}} (e^{- \frac{π}{3}} - e^{\frac{π}{3}})$ $\times$ $\times$
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