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let-A-0-dx-x-2-i-2-i-2-1-1-calculate-A-2-let-R-Re-A-and-I-Im-A-find-the-value-of-R-and-I-

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Question Number 62141 by maxmathsup by imad last updated on 15/Jun/19
let A =∫_0 ^(+∞) (dx/((x^2 −i)^2 )) ( i^2 =−1) 1) calculate A 2) let R =Re(A) and I =Im(A) find the value of R and I .
$${let}\:{A}\:=\int_{\mathrm{0}} ^{+\infty} \:\:\frac{{dx}}{\left({x}^{\mathrm{2}} \:−{i}\right)^{\mathrm{2}} }\:\:\:\:\:\left(\:{i}^{\mathrm{2}} =−\mathrm{1}\right) \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:{A} \\ $$$$\left.\mathrm{2}\right)\:{let}\:{R}\:={Re}\left({A}\right)\:{and}\:{I}\:={Im}\left({A}\right) \\ $$$${find}\:\:{the}\:{value}\:{of}\:{R}\:{and}\:{I}\:. \\ $$
Commented by maxmathsup by imad last updated on 16/Jun/19
1) we have 2A =∫_(−∞) ^(+∞) (dx/((x^2 −i)^2 )) let w(z) =(1/((z^2 −i)^2 )) ⇒ w(z) =(1/((z−(√i))^2 (z+(√i))^2 )) =(1/((z−e^((iπ)/4) )^2 (z +e^((iπ)/4) )^2 )) so the poles of w are +^− e^((iπ)/4) (doubles) residus theorem give ∫_(−∞) ^(+∞) w(z)dz =2iπ Res(w,e^((iπ)/4) ) Res(w,e^((iπ)/4) ) =lim_(z→e^((iπ)/4) ) (1/((2−1)!)){(z−e^((iπ)/4) )^2 w(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^((−3iπ)/4) =−(1/4){ cos(((3π)/4))−isin(((3π)/4))} =−(1/4){−cos((π/4))−isin((π/4))} =(1/4)((√2)/2) +i(1/4)((√2)/2) =((√2)/8) +i((√2)/8) ⇒ ∫_(−∞) ^(+∞) w(z)dz =2iπ {((√2)/8) +i((√2)/8)} =((iπ(√2))/4) −((π(√2))/4) =2A ⇒ A =−((π(√2))/8) +((iπ(√2))/8) 2) we have A = ∫_0 ^∞ (dx/((x^2 −i)^2 )) =∫_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 +2ix^2 )/(x^8 +2x^4 +1))dx = ∫_0 ^∞ ((x^4 −1)/(x^8 +2x^4 +1)) dx +i ∫_0 ^∞ ((2x^2 )/(x^8 +2x^4 +1)) dx ⇒ R =∫_0 ^∞ ((x^4 −1)/(x^8 +2x^4 +1)) and I = ∫_0 ^∞ ((2x^2 )/(x^8 +2x^4 +1)) dx ⇒ R =−((π(√2))/8) and I =((π(√2))/8) .
$$\left.\mathrm{1}\right)\:{we}\:{have}\:\mathrm{2}{A}\:=\int_{−\infty} ^{+\infty} \:\:\frac{{dx}}{\left({x}^{\mathrm{2}} −{i}\right)^{\mathrm{2}} }\:\:{let}\:{w}\left({z}\right)\:=\frac{\mathrm{1}}{\left({z}^{\mathrm{2}} −{i}\right)^{\mathrm{2}} }\:\Rightarrow \\ $$$${w}\left({z}\right)\:=\frac{\mathrm{1}}{\left({z}−\sqrt{{i}}\right)^{\mathrm{2}} \left({z}+\sqrt{{i}}\right)^{\mathrm{2}} }\:=\frac{\mathrm{1}}{\left({z}−{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)^{\mathrm{2}} \left({z}\:+{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)^{\mathrm{2}} }\:\:{so}\:{the}\:{poles}\:{of}\:{w}\:{are}\:\overset{−} {+}{e}^{\frac{{i}\pi}{\mathrm{4}}} \left({doubles}\right) \\ $$$${residus}\:{theorem}\:{give}\:\int_{−\infty} ^{+\infty} \:{w}\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{Res}\left({w},{e}^{\frac{{i}\pi}{\mathrm{4}}} \right) \\ $$$${Res}\left({w},{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)\:={lim}_{{z}\rightarrow{e}^{\frac{{i}\pi}{\mathrm{4}}} } \:\:\:\:\frac{\mathrm{1}}{\left(\mathrm{2}−\mathrm{1}\right)!}\left\{\left({z}−{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)^{\mathrm{2}} {w}\left({z}\right)\right\}^{\left(\mathrm{1}\right)} \\ $$$$={lim}_{{z}\rightarrow{e}^{\frac{{i}\pi}{\mathrm{4}}} } \:\:\:\:\:\left\{\:\:\:\left({z}+{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)^{−\mathrm{2}} \right\}^{\left(\mathrm{1}\right)} \:={lim}_{{z}\rightarrow{e}^{\frac{{i}\pi}{\mathrm{4}}} } \:\:\:\:\:−\mathrm{2}\left({z}+{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)^{−\mathrm{3}} \\ $$$$=−\mathrm{2}\:\left(\mathrm{2}\:{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)^{−\mathrm{3}} \:=−\mathrm{2}.\mathrm{2}^{−\mathrm{3}} \:.{e}^{\frac{−\mathrm{3}{i}\pi}{\mathrm{4}}} \:=−\frac{\mathrm{1}}{\mathrm{4}}\left\{\:{cos}\left(\frac{\mathrm{3}\pi}{\mathrm{4}}\right)−{isin}\left(\frac{\mathrm{3}\pi}{\mathrm{4}}\right)\right\} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{4}}\left\{−{cos}\left(\frac{\pi}{\mathrm{4}}\right)−{isin}\left(\frac{\pi}{\mathrm{4}}\right)\right\}\:=\frac{\mathrm{1}}{\mathrm{4}}\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\:+{i}\frac{\mathrm{1}}{\mathrm{4}}\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\:=\frac{\sqrt{\mathrm{2}}}{\mathrm{8}}\:+{i}\frac{\sqrt{\mathrm{2}}}{\mathrm{8}}\:\:\Rightarrow \\ $$$$\int_{−\infty} ^{+\infty} \:{w}\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:\left\{\frac{\sqrt{\mathrm{2}}}{\mathrm{8}}\:+{i}\frac{\sqrt{\mathrm{2}}}{\mathrm{8}}\right\}\:=\frac{{i}\pi\sqrt{\mathrm{2}}}{\mathrm{4}}\:−\frac{\pi\sqrt{\mathrm{2}}}{\mathrm{4}}\:=\mathrm{2}{A}\:\Rightarrow \\ $$$${A}\:=−\frac{\pi\sqrt{\mathrm{2}}}{\mathrm{8}}\:+\frac{{i}\pi\sqrt{\mathrm{2}}}{\mathrm{8}} \\ $$$$\left.\mathrm{2}\right)\:{we}\:{have}\:{A}\:=\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dx}}{\left({x}^{\mathrm{2}} −{i}\right)^{\mathrm{2}} }\:=\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\left({x}^{\mathrm{2}} +{i}\right)^{\mathrm{2}} }{\left({x}^{\mathrm{2}} −{i}\right)^{\mathrm{2}} \left({x}^{\mathrm{2}} +{i}\right)^{\mathrm{2}} }\:{dx} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{x}^{\mathrm{4}} \:+\mathrm{2}{ix}^{\mathrm{2}} \:−\mathrm{1}}{\left({x}^{\mathrm{4}} \:+\mathrm{1}\right)^{\mathrm{2}} }\:{dx}\:=\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{x}^{\mathrm{4}} −\mathrm{1}\:+\mathrm{2}{ix}^{\mathrm{2}} }{{x}^{\mathrm{8}} \:+\mathrm{2}{x}^{\mathrm{4}} \:+\mathrm{1}}{dx} \\ $$$$=\:\int_{\mathrm{0}} ^{\infty} \:\frac{{x}^{\mathrm{4}} −\mathrm{1}}{{x}^{\mathrm{8}} \:+\mathrm{2}{x}^{\mathrm{4}} \:+\mathrm{1}}\:{dx}\:+{i}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\mathrm{2}{x}^{\mathrm{2}} }{{x}^{\mathrm{8}} \:+\mathrm{2}{x}^{\mathrm{4}} \:+\mathrm{1}}\:{dx}\:\Rightarrow \\ $$$${R}\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{x}^{\mathrm{4}} −\mathrm{1}}{{x}^{\mathrm{8}} \:+\mathrm{2}{x}^{\mathrm{4}} \:+\mathrm{1}}\:\:{and}\:\:{I}\:=\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{2}{x}^{\mathrm{2}} }{{x}^{\mathrm{8}} \:+\mathrm{2}{x}^{\mathrm{4}} \:+\mathrm{1}}\:{dx}\:\Rightarrow \\ $$$${R}\:=−\frac{\pi\sqrt{\mathrm{2}}}{\mathrm{8}}\:\:{and}\:\:{I}\:=\frac{\pi\sqrt{\mathrm{2}}}{\mathrm{8}}\:. \\ $$$$ \\ $$

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