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Question Number 98721 by mathmax by abdo last updated on 15/Jun/20

calculate \int_{0}^{\infty} \frac{dx}{x^{4} + x^{2} + 1}

1) by using residue theorem

2) by using complex decomposition

Answered by mathmax by abdo last updated on 16/Jun/20

let I = \int_{0}^{\infty} \frac{dx}{x^{4} + x^{2} + 1} \Rightarrow 2 I = \int_{- \infty}^{+ \infty} \frac{dx}{x^{4} + x^{2} + 1} let φ (z) = \frac{1}{z^{4} + z^{2} + 1} poles of φ ?

z^{4} + z^{2} + 1 = 0 \Rightarrow u^{2} + u + 1 = 0 (u = z^{2})

Δ = - 3 \Rightarrow u_{1} = \frac{- 1 + i \sqrt{3}}{2} = e^{i \frac{2 π}{3}}, u_{2} = \frac{- 1 - i \sqrt{3}}{2} = e^{- \frac{i 2 π}{3}} \Rightarrow

φ (z) = \frac{1}{(z^{2} - e^{\frac{i 2 π}{3}}) (z^{2} - e^{- \frac{i 2 π}{3}})} = \frac{1}{(z - e^{\frac{i π}{3}}) (z + e^{\frac{i π}{3}}) (z - e^{- \frac{i π}{3}}) (z + 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{1}{{2 e}^{\frac{i π}{3}} (2 isin (\frac{2 π}{3}))} = \frac{e^{- \frac{i π}{3}}}{4 i (\frac{\sqrt{3}}{2})} = \frac{e^{- \frac{i π}{3}}}{2 i \sqrt{3}}

Res (φ, - e^{- \frac{i π}{3}}) = \frac{1}{- {2 e}^{- \frac{i π}{3}} (- 2 isin (\frac{2 π}{3}))} = \frac{e^{\frac{i π}{3}}}{4 i (\frac{\sqrt{3}}{2})} = \frac{e^{\frac{i π}{3}}}{2 i \sqrt{3}} \Rightarrow

\int_{- \infty}^{+ \infty} φ (z) dz = 2 i π {\frac{e^{- \frac{i π}{3}}}{2 i \sqrt{3}} + \frac{e^{\frac{i π}{3}}}{2 i \sqrt{3}}} = \frac{π}{\sqrt{3}} (2 \cos (\frac{π}{3})) = \frac{π}{\sqrt{3}} (2 \times \frac{1}{2}) = \frac{π}{\sqrt{3}}

\Rightarrow I = \frac{π}{2 \sqrt{3}}

Answered by mathmax by abdo last updated on 16/Jun/20

2) let use complex decompositon 2 I = \int_{- \infty}^{+ \infty} \frac{dx}{x^{4} + x^{2} + 1}

x^{4} + x^{2} + 1 = 0 \Rightarrow t^{2} + t + 1 = 0 (t = x^{2})

Δ = - 3 \Rightarrow t_{1} = \frac{- 1 + i \sqrt{3}}{2} = e^{\frac{i 2 π}{3}} and t_{2} = \frac{- 1 - i \sqrt{3}}{2} = e^{- \frac{i 2 π}{3}}

\Rightarrow F (x) = \frac{1}{x^{4} + x^{2} + 1} = \frac{1}{(x^{2} - e^{\frac{i 2 π}{3}}) (x^{2} - e^{- \frac{i 2 π}{3}})} = \frac{1}{(x - e^{\frac{i π}{3}}) (x + e^{\frac{i π}{3}}) (x - e^{- \frac{i π}{3}}) (x + e^{- \frac{i π}{3}})}

= \frac{a}{x - e^{\frac{i π}{3}}} + \frac{b}{x + e^{\frac{i π}{3}}} + \frac{c}{x - e^{- \frac{i π}{3}}} + \frac{d}{x + e^{- \frac{i π}{3}}}

a = \frac{1}{{2 e}^{\frac{i π}{3}} (2 isin (\frac{2 π}{3}))} = \frac{e^{- \frac{i π}{3}}}{2 i \sqrt{3}}, b = \frac{1}{- {2 e}^{\frac{i π}{3}} (2 isin (\frac{2 π}{3}))} = - \frac{e^{- \frac{i π}{3}}}{2 i \sqrt{3}}

c = \frac{1}{{2 e}^{- \frac{i π}{3}} (- 2 isin (\frac{2 π}{3}))} = - \frac{e^{\frac{i π}{3}}}{2 i \sqrt{3}}, d = \frac{1}{- {2 e}^{- \frac{i π}{3}} (- 2 isin (\frac{2 π}{3}))} = \frac{e^{\frac{i π}{3}}}{2 i \sqrt{3}}

\Rightarrow \int_{- \infty}^{+ \infty} F (x) dx = \frac{1}{2 i \sqrt{3}} {e^{- \frac{i π}{3}} \int_{- \infty}^{+ \infty} \frac{dx}{x - e^{\frac{i π}{3}}} - e^{- \frac{i π}{3}} \int_{- \infty}^{+ \infty} \frac{dx}{x + e^{\frac{i π}{3}}} - e^{\frac{i π}{3}} \int_{- \infty}^{+ \infty} \frac{dx}{x - e^{- \frac{i π}{3}}}

+ e^{\frac{i π}{3}} \int_{- \infty}^{+ \infty} \frac{dx}{x + e^{- \frac{i π}{3}}}} we have proved thst \int_{- \infty}^{+ \infty} \frac{dx}{x - a} = i π if im (a) > 0 and

- i π if Im (a) < 0 \Rightarrow

\int_{- \infty}^{+ \infty} F (x) dx = \frac{1}{2 i \sqrt{3}} {(i π) e^{- \frac{i π}{3}} - (- i π) e^{- \frac{i π}{3}} - (- i π) e^{\frac{i π}{3}} + (i π) e^{\frac{i π}{3}}}

= \frac{π}{2 \sqrt{3}} {e^{- \frac{i π}{3}} + e^{- \frac{i π}{3}} + e^{\frac{i π}{3}} + e^{\frac{i π}{3}}} = \frac{π}{\sqrt{3}} {e^{\frac{i π}{3}} + e^{- \frac{i π}{3}}} = \frac{π}{\sqrt{3}} {2 \cos (\frac{π}{3}))

= \frac{π}{\sqrt{3}} = 2 I \Rightarrow I = \frac{π}{2 \sqrt{3}}