Math Question and Answers


Question and Answers Forum

All Questions Topic List
Algebra Questions
Previous in All Question Next in All Question
Previous in Algebra Next in Algebra
Question Number 210573 by Spillover last updated on 12/Aug/24

	Answered by mathmax last updated on 13/Aug/24
	$2I=∫_(−∞) ^(+∞) (dx/(x^4 +ix^2 +2)) (fonction paire) roots x^2 =t→t^2 +it+2 Δ=−1−8 =−9 ⇒t_1 =((−i+3i)/2)=i and t_2 =((−i−3i)/2)=−2i ⇒2I=∫_(−∞) ^(+∞) (dx/((x^2 −i)(x^2 +2i))) =(1/(3i))∫_(−∞) ^(+∞) ((1/(x^2 −i))−(1/(x^2 +2i)))dx ⇒ 6i I =∫_(−∞) ^(+∞) (dx/(x^2 −i))−∫_(−∞) ^(+∞) (dx/(x^2 −(−2i))) =I_1 −I_2 I_1 =(1/(2(√i)))∫_(−∞) ^(+∞) ((1/(x−(√i)))−(1/(x+(√i))))dx on rappelle que ∫_(−∞) ^(+∞) (dx/(x−z))=iπ si Im(z)>0 et −iπ si Im(z)<0 ⇒ I_1 =(1/(2(√i))){iπ−(−iπ)=((2iπ)/(2(√i)))=π e^((iπ)/4) I_2 =∫_(−∞) ^(+∞) (dx/((x−(√(−2i)))(x+(√(−2i))))) =(1/( 2(√(−2i))))∫_(−∞) ^(+∞) ((1/(x−(√(−2i))))−(1/(x+(√(−2i)))))dx =(1/( 2(√2)))e^((iπ)/4) ∫_(−∞) ^(+∞) ((1/(x−(√2)e^(−((iπ)/4)) ))−(1/(x+(√2)e^(−((iπ)/4)) )))dx =(1/( 2(√2)))e^((iπ)/4) (−iπ−(iπ))=((−2iπ)/( 2(√2))) e^((iπ)/4) =−((iπ)/( (√2))) e^((iπ)/4) 6iI=πe^((iπ)/4) −((iπ)/( (√2))) e^((iπ)/4) =π(1−(i/( (√2))))e^((iπ)/4) ⇒I=(π/(6i))(1−(i/( (√2))))e^((iπ)/4) =(π/6)(−i−(1/( (√2))))((1/( (√2)))+(1/( (√2)))i) =(π/(12))(−1−i(√2))(1+i)$
	$2 I = \int_{- \infty}^{+ \infty} \frac{d x}{x^{4} + {i x}^{2} + 2} (f o n c t i o n p a i r e)$ $r o o t s x^{2} = t \to t^{2} + i t + 2$ $Δ = - 1 - 8 = - 9 \Rightarrow t_{1} = \frac{- i + 3 i}{2} = i$ $a n d t_{2} = \frac{- i - 3 i}{2} = - 2 i \Rightarrow 2 I = \int_{- \infty}^{+ \infty} \frac{d x}{(x^{2} - i) (x^{2} + 2 i)}$ $= \frac{1}{3 i} \int_{- \infty}^{+ \infty} (\frac{1}{x^{2} - i} - \frac{1}{x^{2} + 2 i}) d x \Rightarrow$ $6 i I = \int_{- \infty}^{+ \infty} \frac{d x}{x^{2} - i} - \int_{- \infty}^{+ \infty} \frac{d x}{x^{2} - (- 2 i)}$ $= I_{1} - I_{2}$ $I_{1} = \frac{1}{2 \sqrt{i}} \int_{- \infty}^{+ \infty} (\frac{1}{x - \sqrt{i}} - \frac{1}{x + \sqrt{i}}) d x$ $o n r a p p e l l e q u e \int_{- \infty}^{+ \infty} \frac{d x}{x - z} = i π s i I m (z) > 0$ $e t - i π s i I m (z) < 0 \Rightarrow$ $I_{1} = \frac{1}{2 \sqrt{i}} {i π - (- i π) = \frac{2 i π}{2 \sqrt{i}} = π e^{\frac{i π}{4}}$ $I_{2} = \int_{- \infty}^{+ \infty} \frac{d x}{(x - \sqrt{- 2 i}) (x + \sqrt{- 2 i})}$ $= \frac{1}{2 \sqrt{- 2 i}} \int_{- \infty}^{+ \infty} (\frac{1}{x - \sqrt{- 2 i}} - \frac{1}{x + \sqrt{- 2 i}}) d x$ $= \frac{1}{2 \sqrt{2}} e^{\frac{i π}{4}} \int_{- \infty}^{+ \infty} (\frac{1}{x - \sqrt{2} e^{- \frac{i π}{4}}} - \frac{1}{x + \sqrt{2} e^{- \frac{i π}{4}}}) d x$ $= \frac{1}{2 \sqrt{2}} e^{\frac{i π}{4}} (- i π - (i π)) = \frac{- 2 i π}{2 \sqrt{2}} e^{\frac{i π}{4}}$ $= - \frac{i π}{\sqrt{2}} e^{\frac{i π}{4}}$ $6 i I = π e^{\frac{i π}{4}} - \frac{i π}{\sqrt{2}} e^{\frac{i π}{4}}$ $= π (1 - \frac{i}{\sqrt{2}}) e^{\frac{i π}{4}} \Rightarrow I = \frac{π}{6 i} (1 - \frac{i}{\sqrt{2}}) e^{\frac{i π}{4}}$ $= \frac{π}{6} (- i - \frac{1}{\sqrt{2}}) (\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} i)$ $= \frac{π}{12} (- 1 - i \sqrt{2}) (1 + i)$
	Commented by Frix last updated on 13/Aug/24

	$Error of sign somewhere, it must be$ $\frac{π}{12} (+ 1 - i \sqrt{2}) (1 + i)$
	Commented by mathmax last updated on 13/Aug/24

	$t h a n k s$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum