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

Question Number 113562 by mathmax by abdo last updated on 14/Sep/20

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

Answered by Olaf last updated on 14/Sep/20

x^{4} + {i x}^{2} + 2 = 0

x_{1} = - 1 + i

x_{2} = 1 - i

x_{3} = - \frac{1}{\sqrt{2}} (1 + i)

x_{4} = \frac{1}{\sqrt{2}} (1 + i)

I = \int_{0}^{\infty} \frac{d x}{x^{4} + {i x}^{2} + 2}

I = \int_{0}^{\infty} (\underset{k = 1}{\sum^{4}} \frac{A_{k}}{x - x_{k}}) d x

A_{j} = \underset{\underset{k \neq j}{k = 1}}{\prod^{4}} \frac{1}{x_{j} - x_{k}}

I = {[\underset{k = 1}{\sum^{4}} A_{k} \ln ∣ x - x_{k} ∣]}_{0}^{\infty} \dots .

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