Question Number 113562 by mathmax by abdo last updated on 14/Sep/20
$$\mathrm{calculate}\:\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{dx}}{\mathrm{x}^{\mathrm{4}} +\mathrm{ix}^{\mathrm{2}} \:+\mathrm{2}} \\ $$
Answered by Olaf last updated on 14/Sep/20
![x^4 +ix^2 +2 = 0 x_1 = −1+i x_2 = 1−i x_3 = −(1/( (√2)))(1+i) x_4 = (1/( (√2)))(1+i) I = ∫_0 ^∞ (dx/(x^4 +ix^2 +2)) I = ∫_0 ^∞ (Σ_(k=1) ^4 (A_k /(x−x_k )))dx A_j = Π_(k=1_(k≠j) ) ^4 (1/(x_j −x_k )) I = [Σ_(k=1) ^4 A_k ln∣x−x_k ∣]_0 ^∞ ....](https://www.tinkutara.com/question/Q113609.png)
$${x}^{\mathrm{4}} +{ix}^{\mathrm{2}} +\mathrm{2}\:=\:\mathrm{0} \\ $$$${x}_{\mathrm{1}} \:=\:−\mathrm{1}+{i} \\ $$$${x}_{\mathrm{2}} \:=\:\mathrm{1}−{i} \\ $$$${x}_{\mathrm{3}} \:=\:−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\left(\mathrm{1}+{i}\right) \\ $$$${x}_{\mathrm{4}} \:=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\left(\mathrm{1}+{i}\right) \\ $$$$\mathrm{I}\:=\:\int_{\mathrm{0}} ^{\infty} \frac{{dx}}{{x}^{\mathrm{4}} +{ix}^{\mathrm{2}} +\mathrm{2}} \\ $$$$\mathrm{I}\:=\:\int_{\mathrm{0}} ^{\infty} \left(\underset{{k}=\mathrm{1}} {\overset{\mathrm{4}} {\sum}}\frac{\mathrm{A}_{{k}} }{{x}−{x}_{{k}} }\right){dx} \\ $$$$\mathrm{A}_{{j}} \:=\:\underset{\underset{{k}\neq{j}} {{k}=\mathrm{1}}} {\overset{\mathrm{4}} {\prod}}\frac{\mathrm{1}}{{x}_{{j}} −{x}_{{k}} } \\ $$$$ \\ $$$$\mathrm{I}\:=\:\left[\underset{{k}=\mathrm{1}} {\overset{\mathrm{4}} {\sum}}\mathrm{A}_{{k}} \mathrm{ln}\mid{x}−{x}_{{k}} \mid\right]_{\mathrm{0}} ^{\infty} …. \\ $$$$ \\ $$