Question-210573

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)

$$\mathrm{2}{I}=\int_{−\infty} ^{+\infty} \frac{{dx}}{{x}^{\mathrm{4}} +{ix}^{\mathrm{2}} +\mathrm{2}}\:\left({fonction}\:{paire}\right) \\ $$$${roots}\:\:\:\:\:\:{x}^{\mathrm{2}} ={t}\rightarrow{t}^{\mathrm{2}} +{it}+\mathrm{2} \\ $$$$\Delta=−\mathrm{1}−\mathrm{8}\:=−\mathrm{9}\:\Rightarrow{t}_{\mathrm{1}} =\frac{−{i}+\mathrm{3}{i}}{\mathrm{2}}={i} \\ $$$${and}\:{t}_{\mathrm{2}} =\frac{−{i}−\mathrm{3}{i}}{\mathrm{2}}=−\mathrm{2}{i}\:\Rightarrow\mathrm{2}{I}=\int_{−\infty} ^{+\infty} \frac{{dx}}{\left({x}^{\mathrm{2}} −{i}\right)\left({x}^{\mathrm{2}} +\mathrm{2}{i}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}{i}}\int_{−\infty} ^{+\infty} \left(\frac{\mathrm{1}}{{x}^{\mathrm{2}} −{i}}−\frac{\mathrm{1}}{{x}^{\mathrm{2}} +\mathrm{2}{i}}\right){dx}\:\Rightarrow \\ $$$$\mathrm{6}{i}\:{I}\:=\int_{−\infty} ^{+\infty} \frac{{dx}}{{x}^{\mathrm{2}} −{i}}−\int_{−\infty} ^{+\infty} \frac{{dx}}{{x}^{\mathrm{2}} −\left(−\mathrm{2}{i}\right)} \\ $$$$={I}_{\mathrm{1}} −{I}_{\mathrm{2}} \\ $$$${I}_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}\sqrt{{i}}}\int_{−\infty} ^{+\infty} \left(\frac{\mathrm{1}}{{x}−\sqrt{{i}}}−\frac{\mathrm{1}}{{x}+\sqrt{{i}}}\right){dx} \\ $$$${on}\:{rappelle}\:{que}\:\int_{−\infty} ^{+\infty} \frac{{dx}}{{x}−{z}}={i}\pi\:{si}\:{Im}\left({z}\right)>\mathrm{0} \\ $$$${et}\:−{i}\pi\:{si}\:{Im}\left({z}\right)<\mathrm{0}\:\Rightarrow \\ $$$${I}_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}\sqrt{{i}}}\left\{{i}\pi−\left(−{i}\pi\right)=\frac{\mathrm{2}{i}\pi}{\mathrm{2}\sqrt{{i}}}=\pi\:{e}^{\frac{{i}\pi}{\mathrm{4}}} \right. \\ $$$${I}_{\mathrm{2}} =\int_{−\infty} ^{+\infty} \frac{{dx}}{\left({x}−\sqrt{−\mathrm{2}{i}}\right)\left({x}+\sqrt{−\mathrm{2}{i}}\right)} \\ $$$$=\frac{\mathrm{1}}{\:\mathrm{2}\sqrt{−\mathrm{2}{i}}}\int_{−\infty} ^{+\infty} \left(\frac{\mathrm{1}}{{x}−\sqrt{−\mathrm{2}{i}}}−\frac{\mathrm{1}}{{x}+\sqrt{−\mathrm{2}{i}}}\right){dx} \\ $$$$=\frac{\mathrm{1}}{\:\mathrm{2}\sqrt{\mathrm{2}}}{e}^{\frac{{i}\pi}{\mathrm{4}}} \int_{−\infty} ^{+\infty} \left(\frac{\mathrm{1}}{{x}−\sqrt{\mathrm{2}}{e}^{−\frac{{i}\pi}{\mathrm{4}}} }−\frac{\mathrm{1}}{{x}+\sqrt{\mathrm{2}}{e}^{−\frac{{i}\pi}{\mathrm{4}}} }\right){dx} \\ $$$$=\frac{\mathrm{1}}{\:\mathrm{2}\sqrt{\mathrm{2}}}{e}^{\frac{{i}\pi}{\mathrm{4}}} \left(−{i}\pi−\left({i}\pi\right)\right)=\frac{−\mathrm{2}{i}\pi}{\:\mathrm{2}\sqrt{\mathrm{2}}}\:{e}^{\frac{{i}\pi}{\mathrm{4}}} \\ $$$$=−\frac{{i}\pi}{\:\sqrt{\mathrm{2}}}\:{e}^{\frac{{i}\pi}{\mathrm{4}}} \\ $$$$\mathrm{6}{iI}=\pi{e}^{\frac{{i}\pi}{\mathrm{4}}} −\frac{{i}\pi}{\:\sqrt{\mathrm{2}}}\:{e}^{\frac{{i}\pi}{\mathrm{4}}} \\ $$$$=\pi\left(\mathrm{1}−\frac{{i}}{\:\sqrt{\mathrm{2}}}\right){e}^{\frac{{i}\pi}{\mathrm{4}}} \:\Rightarrow{I}=\frac{\pi}{\mathrm{6}{i}}\left(\mathrm{1}−\frac{{i}}{\:\sqrt{\mathrm{2}}}\right){e}^{\frac{{i}\pi}{\mathrm{4}}} \\ $$$$=\frac{\pi}{\mathrm{6}}\left(−{i}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}{i}\right) \\ $$$$=\frac{\pi}{\mathrm{12}}\left(−\mathrm{1}−{i}\sqrt{\mathrm{2}}\right)\left(\mathrm{1}+{i}\right) \\ $$$$ \\ $$$$ \\ $$

Commented by Frix last updated on 13/Aug/24

Error of sign somewhere, it must be (π/(12))(+1−i(√2))(1+i)

$$\mathrm{Error}\:\mathrm{of}\:\mathrm{sign}\:\mathrm{somewhere},\:\mathrm{it}\:\mathrm{must}\:\mathrm{be} \\ $$$$\frac{\pi}{\mathrm{12}}\left(+\mathrm{1}−\mathrm{i}\sqrt{\mathrm{2}}\right)\left(\mathrm{1}+\mathrm{i}\right) \\ $$

Commented by mathmax last updated on 13/Aug/24

$${thanks} \\ $$

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