calculate-0-cos-x-2-x-4-x-2-1-dx-

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

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