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Question Number 36167 by abdo mathsup 649 cc last updated on 29/May/18
let give I = ∫_0 ^∞ (dx/((x^2 +i)^2 )) 1) extract Re(I) and Im(I) 2) find the value of I 3) calculate Re(I) and Im(I) .
$${let}\:{give}\:\:{I}\:\:=\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{dx}}{\left({x}^{\mathrm{2}} \:+{i}\right)^{\mathrm{2}} } \\ $$$$\left.\mathrm{1}\right)\:{extract}\:{Re}\left({I}\right)\:{and}\:{Im}\left({I}\right) \\ $$$$\left.\mathrm{2}\right)\:{find}\:{the}\:{value}\:{of}\:{I} \\ $$$$\left.\mathrm{3}\right)\:{calculate}\:{Re}\left({I}\right)\:{and}\:{Im}\left({I}\right)\:. \\ $$
Commented by maxmathsup by imad last updated on 22/Aug/18
1) we have I =∫_0 ^∞ (((x^2 −i)^2 )/((x^2 +i)^2 (x^2 −i)^2 ))dx =∫_0 ^∞ ((x^4 −2ix^2 −1)/((x^4 +1)^2 )) dx = ∫_0 ^∞ ((x^4 −1)/((x^4 +1)^2 )) dx +i ∫_0 ^∞ ((−2x^2 )/((x^4 +1)^2 )) dx ⇒ Re(I) =∫_0 ^∞ ((x^4 −1)/((x^4 +1)^2 )) dx and Im(I) =∫_0 ^∞ ((−2x^2 )/((x^4 +1)^2 ))dx
$$\left.\mathrm{1}\right)\:{we}\:{have}\:\:{I}\:=\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\left({x}^{\mathrm{2}} −{i}\right)^{\mathrm{2}} }{\left({x}^{\mathrm{2}} \:+{i}\right)^{\mathrm{2}} \left({x}^{\mathrm{2}} −{i}\right)^{\mathrm{2}} }{dx} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{x}^{\mathrm{4}} \:−\mathrm{2}{ix}^{\mathrm{2}} \:−\mathrm{1}}{\left({x}^{\mathrm{4}} \:+\mathrm{1}\right)^{\mathrm{2}} }\:{dx}\:\:=\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{x}^{\mathrm{4}} −\mathrm{1}}{\left({x}^{\mathrm{4}} \:+\mathrm{1}\right)^{\mathrm{2}} }\:{dx}\:+{i}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{−\mathrm{2}{x}^{\mathrm{2}} }{\left({x}^{\mathrm{4}} \:+\mathrm{1}\right)^{\mathrm{2}} }\:{dx}\:\Rightarrow \\ $$$${Re}\left({I}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{x}^{\mathrm{4}} −\mathrm{1}}{\left({x}^{\mathrm{4}} \:+\mathrm{1}\right)^{\mathrm{2}} }\:{dx}\:\:{and}\:\:{Im}\left({I}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{−\mathrm{2}{x}^{\mathrm{2}} }{\left({x}^{\mathrm{4}} \:+\mathrm{1}\right)^{\mathrm{2}} }{dx} \\ $$
Commented by maxmathsup by imad last updated on 22/Aug/18
2) we have 2I =∫_(−∞) ^(+∞) (dx/((x^2 +i)^2 )) let consider the complex function ϕ(z) = (1/((z^2 +i)^2 )) ⇒ϕ(z) = (1/((z^2 −((√(−i)))^2 )^2 )) =(1/((z−(√(−i)))^2 (z+(√(−i)))^2 )) = (1/((z−e^(−((iπ)/4)) )^2 (z +e^(−((iπ)/4)) )^2 )) so the poles of ϕ are +^− e^(−((iπ)/4)) (doubles) ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ Res(ϕ,−e^(−((iπ)/4)) ) but Res(ϕ,e^(−((iπ)/4)) ) =lim_(z→−e^(−((iπ)/4)) ) (1/((2−1)!)){(z+e^(−((iπ)/4)) )^2 ϕ(z)}^((1)) =lim_(z→−e^(−((iπ)/4)) ) {(z−e^(−((iπ)/4)) )^(−2) }^((1)) =lim_(z→−e^(−((iπ)/4)) ) −2 (z−e^(−((iπ)/4)) )^(−3) =−2(−2 e^(−((iπ)/4)) )^(−3) =((−2)/((−2)^3 )) e^(i((3π)/4)) =(1/4) e^((i3π)/4) =(1/4){cos(((3π)/4)) +isin(((3π)/4))} =(1/4){−(1/( (√2))) +(i/( (√2)))} ⇒∫_(−∞) ^(+∞) ϕ(z)dz =((2iπ)/4){−(1/( (√2))) +(i/( (√2)))}=((iπ)/2){−(1/( (√2))) +(i/( (√2)))} =2I ⇒ I =((iπ)/4){−(1/( (√2))) +(i/( (√2)))} =−(π/(4(√2))) −(π/(4(√2))) i 3) we have I =Re(I) +iIm(I) ⇒Re(I) =−(π/(4(√2))) and Im(I) =−(π/(4(√2)))
$$\left.\mathrm{2}\right)\:{we}\:{have}\:\mathrm{2}{I}\:=\int_{−\infty} ^{+\infty} \:\:\:\frac{{dx}}{\left({x}^{\mathrm{2}} +{i}\right)^{\mathrm{2}} }\:\:{let}\:{consider}\:{the}\:{complex}\:{function} \\ $$$$\varphi\left({z}\right)\:=\:\frac{\mathrm{1}}{\left({z}^{\mathrm{2}} +{i}\right)^{\mathrm{2}} }\:\:\Rightarrow\varphi\left({z}\right)\:=\:\frac{\mathrm{1}}{\left({z}^{\mathrm{2}} −\left(\sqrt{−{i}}\right)^{\mathrm{2}} \right)^{\mathrm{2}} }\:=\frac{\mathrm{1}}{\left({z}−\sqrt{−{i}}\right)^{\mathrm{2}} \left({z}+\sqrt{−{i}}\right)^{\mathrm{2}} } \\ $$$$=\:\frac{\mathrm{1}}{\left({z}−{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)^{\mathrm{2}} \left({z}\:+{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)^{\mathrm{2}} }\:\:{so}\:{the}\:{poles}\:{of}\:\varphi\:{are}\:\overset{−} {+}\:{e}^{−\frac{{i}\pi}{\mathrm{4}}} \:\:\left({doubles}\right)\:\Rightarrow \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{Res}\left(\varphi,−{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)\:\:{but} \\ $$$${Res}\left(\varphi,{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)\:={lim}_{{z}\rightarrow−{e}^{−\frac{{i}\pi}{\mathrm{4}}} } \:\:\:\:\:\frac{\mathrm{1}}{\left(\mathrm{2}−\mathrm{1}\right)!}\left\{\left({z}+{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)^{\mathrm{2}} \varphi\left({z}\right)\right\}^{\left(\mathrm{1}\right)} \\ $$$$={lim}_{{z}\rightarrow−{e}^{−\frac{{i}\pi}{\mathrm{4}}} } \left\{\left({z}−{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)^{−\mathrm{2}} \right\}^{\left(\mathrm{1}\right)} \:={lim}_{{z}\rightarrow−{e}^{−\frac{{i}\pi}{\mathrm{4}}} } \:\:\:−\mathrm{2}\:\left({z}−{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)^{−\mathrm{3}} \\ $$$$=−\mathrm{2}\left(−\mathrm{2}\:{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)^{−\mathrm{3}} \:=\frac{−\mathrm{2}}{\left(−\mathrm{2}\right)^{\mathrm{3}} }\:{e}^{{i}\frac{\mathrm{3}\pi}{\mathrm{4}}} \:\:=\frac{\mathrm{1}}{\mathrm{4}}\:{e}^{\frac{{i}\mathrm{3}\pi}{\mathrm{4}}} \:=\frac{\mathrm{1}}{\mathrm{4}}\left\{{cos}\left(\frac{\mathrm{3}\pi}{\mathrm{4}}\right)\:+{isin}\left(\frac{\mathrm{3}\pi}{\mathrm{4}}\right)\right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\left\{−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:+\frac{{i}}{\:\sqrt{\mathrm{2}}}\right\}\:\Rightarrow\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\frac{\mathrm{2}{i}\pi}{\mathrm{4}}\left\{−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:+\frac{{i}}{\:\sqrt{\mathrm{2}}}\right\}=\frac{{i}\pi}{\mathrm{2}}\left\{−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:+\frac{{i}}{\:\sqrt{\mathrm{2}}}\right\}\:=\mathrm{2}{I}\:\Rightarrow \\ $$$${I}\:=\frac{{i}\pi}{\mathrm{4}}\left\{−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:+\frac{{i}}{\:\sqrt{\mathrm{2}}}\right\}\:=−\frac{\pi}{\mathrm{4}\sqrt{\mathrm{2}}}\:−\frac{\pi}{\mathrm{4}\sqrt{\mathrm{2}}}\:{i} \\ $$$$\left.\mathrm{3}\right)\:{we}\:{have}\:{I}\:={Re}\left({I}\right)\:+{iIm}\left({I}\right)\:\Rightarrow{Re}\left({I}\right)\:=−\frac{\pi}{\mathrm{4}\sqrt{\mathrm{2}}}\:{and}\:{Im}\left({I}\right)\:=−\frac{\pi}{\mathrm{4}\sqrt{\mathrm{2}}} \\ $$$$ \\ $$

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