calculate-0-dx-x-4-x-2-2-

Question Number 138973 by mathmax by abdo last updated on 20/Apr/21

calculate \int_{0}^{\infty} \frac{dx}{x^{4} - x^{2} + 2}

Answered by mathmax by abdo last updated on 22/Apr/21

Φ = \int_{0}^{\infty} \frac{dx}{x^{4} - x^{2} + 2}

let f (x) = \frac{1}{x^{4} - x^{2} + 2}

x^{4} - x^{2} + 2 = 0 \to t^{2} - t + 2 = 0 \to Δ = 1 - 8 = - 7 \Rightarrow t_{1} = \frac{1 + i \sqrt{7}}{2}

ant t_{2} = \frac{1 - i \sqrt{7}}{2} \Rightarrow f (x) = \frac{1}{(x^{2} - t_{1}) (x^{2} - t_{2})} = \frac{1}{t_{1} - t_{2}} (\frac{1}{x^{2} - t_{1}} - \frac{1}{x^{2} - t_{2}})

= \frac{1}{i \sqrt{7}} (\frac{1}{x^{2} - t_{1}} - \frac{1}{x^{2} - t_{2}}) \Rightarrow Φ = \frac{1}{i \sqrt{7}} \int_{0}^{\infty} \frac{dx}{x^{2} - t_{1}} - \frac{1}{i \sqrt{7}} \int_{0}^{\infty} \frac{dx}{x^{2} - t_{2}}

= Φ_{1} - Φ_{2}

Φ_{1} = \frac{1}{2 i \sqrt{7}} \int_{- \infty}^{+ \infty} \frac{dx}{x^{2} - t_{1}} let w (z) = \frac{1}{z^{2} - \frac{1 + i \sqrt{7}}{2}}

we have ∣ \frac{1 + i \sqrt{7}}{2} ∣= \frac{1}{2} \sqrt{1 + 7} = \frac{2 \sqrt{2}}{2} = \sqrt{2} \Rightarrow \frac{1 + i \sqrt{7}}{2} = \sqrt{2} e^{iarctan (\sqrt{7})}

\Rightarrow w (z) = \frac{1}{(z - (^{4} \sqrt{2}) e^{\frac{i}{2} \arctan (\sqrt{7})}) (z + (^{4} \sqrt{2}) e^{\frac{i}{2} \arctan (\sqrt{7})})}

\int_{- \infty}^{+ \infty} w (z) dx = 2 i π Res (w, (^{4} \sqrt{2} e^{\frac{i}{2} \arctan (\sqrt{7})})

= 2 i π . \frac{1}{2 (^{4} \sqrt{2}) e^{\frac{i}{2} \arctan (\sqrt{7})}} = \frac{i π}{(^{4} \sqrt{2})} e^{- \frac{i}{2} \arctan (\sqrt{7})} \Rightarrow

Φ_{1} = \frac{1}{2 i \sqrt{7}} \times \frac{i π}{(^{4} \sqrt{2})} e^{- \frac{i}{2} \arctan (\sqrt{7})} = \frac{π}{2 \sqrt{7} (^{4} \sqrt{2})} e^{- \frac{i}{2} \arctan (\sqrt{7})}

Φ_{2} = \frac{1}{i \sqrt{7}} \int_{0}^{\infty} \frac{dx}{x - t_{2}} = \frac{1}{2 i \sqrt{7}} \times conj (\int_{- \infty}^{\infty} \frac{dx}{x^{2} - t_{1}})

= \frac{1}{2 i \sqrt{7}} \times (- \frac{i π}{(^{4} \sqrt{2})} e^{\frac{i}{2} \arctan (\sqrt{7})}) = - \frac{π}{2 \sqrt{7} (^{4} \sqrt{2})} e^{\frac{i}{2} \arctan (\sqrt{7})} \Rightarrow

Φ = Φ_{1} - Φ_{2} = \frac{π}{2 \sqrt{7} (^{4} \sqrt{2})} (e^{\frac{i}{2} \arctan (\sqrt{7})} + e^{- \frac{i}{2} \arctan (\sqrt{7})})

= \frac{π}{2 \sqrt{7} (^{4} \sqrt{2})} (2 \cos (\arctan (\sqrt{7})) = \frac{π}{\sqrt{7} (^{4} \sqrt{2})} \cos (\arctan (\sqrt{7}))

we have \cos (arctanx) = \frac{1}{\sqrt{1 + x^{2}}} \Rightarrow \cos (\arctan (\sqrt{7}))

= \frac{1}{\sqrt{1 + 7}} = \frac{1}{2 \sqrt{2}} \Rightarrow Φ = \frac{π}{\sqrt{7} (^{4} \sqrt{2})} \times \frac{1}{2 \sqrt{2}}

= \frac{π}{2 \sqrt{14} (^{4} \sqrt{2})}

Commented by mathmax by abdo last updated on 23/Apr/21

sorry Φ = Φ_{1} - Φ_{2} = \frac{π}{2 \sqrt{7} (^{4} \sqrt{2})} (2 \cos (\frac{1}{2} \arctan (\sqrt{7}))

= \frac{π}{\sqrt{7} (^{4} \sqrt{2})} . \cos (\frac{1}{2} \arctan (\sqrt{7}))