calculate ∫_0 ^∞ (dx/(x^8 +x^4 +1))


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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
	$X^8 +X^4 +1 X^4 =y y^2 +y+1=0⇒y=e^(i((2kπ)/3)) ,k∈{1,2} x_k =e^(i(π/6)+((i2kπ)/4)) ,k∈{0,1,2,3} x_k =e^(i(π/3)+((2ikπ)/4)) a_1 =e^(i(π/6)) ,a_2 =e^(i(π/3)) ,a_3 =e^(i((π/6)+(π/2))) ,a_4 =e^(i((π/3)+(π/2))) , ∫_0 ^(+∞) (dx/(x^8 +x^4 +1))=(1/2)∫_(−∞) ^(+∞) (dx/(x^8 +x^4 +1))=(1/2).2iπ Σ_(ak) Res(f(z),a_k ) =iπ.Σ_(k=1) ^4 (1/(8a_k ^7 +4a_k ^3 ))$
	$X^{8} + X^{4} + 1$ $X^{4} = y$ $y^{2} + y + 1 = 0 \Rightarrow y = e^{i \frac{2 k π}{3}}, k \in {1, 2}$ $x_{k} = e^{i \frac{π}{6} + \frac{i 2 k π}{4}}, k \in {0, 1, 2, 3}$ $x_{k} = e^{i \frac{π}{3} + \frac{2 i k π}{4}}$ $a_{1} = e^{i \frac{π}{6}}, a_{2} = e^{i \frac{π}{3}}, a_{3} = e^{i (\frac{π}{6} + \frac{π}{2})}, a_{4} = e^{i (\frac{π}{3} + \frac{π}{2})},$ $\int_{0}^{+ \infty} \frac{d x}{x^{8} + x^{4} + 1} = \frac{1}{2} \int_{- \infty}^{+ \infty} \frac{d x}{x^{8} + x^{4} + 1} = \frac{1}{2} . 2 i π \sum_{a k} R e s (f (z), a_{k})$ $= i π . \underset{k = 1}{\sum^{4}} \frac{1}{8 a_{k}^{7} + 4 a_{k}^{3}}$
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