calculate-0-dx-x-8-x-4-1-

Question Number 98883 by mathmax by abdo last updated on 16/Jun/20

calculate \int_{0}^{\infty} \frac{dx}{x^{8} + x^{4} + 1}

Answered by maths mind last updated on 18/Jun/20

X^{8} + X^{4} + 1

X^{4} = y

y^{2} + y + 1 = 0 \Rightarrow y = e^{i \frac{2 k π}{3}}, k \in {1, 2}

x_{k} = e^{i \frac{π}{6} + \frac{i 2 k π}{4}}, k \in {0, 1, 2, 3}

x_{k} = e^{i \frac{π}{3} + \frac{2 i k π}{4}}

a_{1} = e^{i \frac{π}{6}}, a_{2} = e^{i \frac{π}{3}}, a_{3} = e^{i (\frac{π}{6} + \frac{π}{2})}, a_{4} = e^{i (\frac{π}{3} + \frac{π}{2})},

\int_{0}^{+ \infty} \frac{d x}{x^{8} + x^{4} + 1} = \frac{1}{2} \int_{- \infty}^{+ \infty} \frac{d x}{x^{8} + x^{4} + 1} = \frac{1}{2} . 2 i π \sum_{a k} R e s (f (z), a_{k})

= i π . \underset{k = 1}{\sum^{4}} \frac{1}{8 a_{k}^{7} + 4 a_{k}^{3}}