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1-calculate-I-dx-x-2-i-and-J-dx-x-2-i-2-find-the-value-of-dx-x-4-1-

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Question Number 56629 by maxmathsup by imad last updated on 19/Mar/19
1) calculate I =∫_(−∞) ^(+∞) (dx/(x^2 −i)) and J =∫_(−∞) ^(+∞) (dx/(x^2 −i)) 2) find the value of ∫_(−∞) ^(+∞) (dx/(x^4 +1))
$$\left.\mathrm{1}\right)\:{calculate}\:{I}\:=\int_{−\infty} ^{+\infty} \:\:\frac{{dx}}{{x}^{\mathrm{2}} −{i}}\:\:\:{and}\:{J}\:=\int_{−\infty} ^{+\infty} \:\:\frac{{dx}}{{x}^{\mathrm{2}} −{i}} \\ $$$$\left.\mathrm{2}\right)\:{find}\:{the}\:{value}\:{of}\:\int_{−\infty} ^{+\infty} \:\:\frac{{dx}}{{x}^{\mathrm{4}} \:+\mathrm{1}} \\ $$
Commented by maxmathsup by imad last updated on 19/Mar/19
J =∫_(−∞) ^(+∞) (dx/(x^2 +i))
$$\:{J}\:=\int_{−\infty} ^{+\infty} \:\:\frac{{dx}}{{x}^{\mathrm{2}} \:+{i}} \\ $$
Commented by maxmathsup by imad last updated on 20/Mar/19
1) let ϕ(z)=(1/(z^2 −i)) ⇒ ϕ(z)= (1/((z−(√i))(z+(√i)))) =(1/((z−e^((iπ)/4) )(z+e^((iπ)/4) ))) so the poles of ϕ are +^− e^((iπ)/4) residus theorem give ∫_(−∞) ^(+∞) ϕ(z)dz=2iπRes(ϕ,e^((iπ)/4) ) Res(ϕ, e^((iπ)/4) )=lim_(z→e^((iπ)/4) ) (z−e^((iπ)/4) )ϕ(z)=(1/(2 e^((iπ)/4) )) =(1/2) e^(−((iπ)/4)) ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz=2iπ(1/2) e^(−((iπ)/4)) =iπ e^(−((iπ)/4)) ⇒ I =iπ e^(−((iπ)/4)) let calculate J =∫_(−∞) ^(+∞) (dx/(x^2 +i)) let W(z)=(1/(z^2 +i)) ⇒W(z)=(1/((z−e^(−((iπ)/4)) )(z +e^(−((iπ)/4)) ))) so the poles of ϕ are +^− e^(−((iπ)/4)) residus theorem give ∫_(−∞) ^(+∞) W(z)dz =2iπ Res(W,−e^(−((iπ)/4)) ) Res(W,−e^(−((iπ)/4)) ) =lim_(z→−e^(−((iπ)/4)) ) (z+e^(−((iπ)/4)) )W(z) =(1/(−2 e^(−((iπ)/4)) )) =−(1/2) e^((iπ)/4) ⇒ ∫_(−∞) ^(+∞) W(z)dz =2iπ (−(1/2) e^((iπ)/4) ) =−iπ e^((iπ)/4) =J also we can use that J =conj(I) =conj(iπ e^(−((iπ)/4)) )=−iπ e^((iπ)/4) 2) ∫_(−∞) ^(+∞) (dx/(x^4 +1)) =∫_(−∞) ^(+∞) (dx/((x^2 −i)(x^2 +i))) =(1/(2i))∫_(−∞) ^(+∞) {(1/(x^2 −i)) −(1/(x^2 +i))}dx =(1/(2i)){ I−J} =(1/(2i)){iπ e^(−((iπ)/4)) +i π e^((iπ)/4) } =(π/2) (2cos((π/4)))=π (1/( (√2))) =(π/( (√2))) .
$$\left.\mathrm{1}\right)\:{let}\:\varphi\left({z}\right)=\frac{\mathrm{1}}{{z}^{\mathrm{2}} −{i}}\:\Rightarrow\:\varphi\left({z}\right)=\:\frac{\mathrm{1}}{\left({z}−\sqrt{{i}}\right)\left({z}+\sqrt{{i}}\right)}\:=\frac{\mathrm{1}}{\left({z}−{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)\left({z}+{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)}\:{so}\:{the}\:{poles}\:{of} \\ $$$$\varphi\:{are}\:\overset{−} {+}{e}^{\frac{{i}\pi}{\mathrm{4}}} \:\:{residus}\:{theorem}\:{give}\:\int_{−\infty} ^{+\infty} \varphi\left({z}\right){dz}=\mathrm{2}{i}\pi{Res}\left(\varphi,{e}^{\frac{{i}\pi}{\mathrm{4}}} \right) \\ $$$${Res}\left(\varphi,\:{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)={lim}_{{z}\rightarrow{e}^{\frac{{i}\pi}{\mathrm{4}}} } \:\:\:\left({z}−{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)\varphi\left({z}\right)=\frac{\mathrm{1}}{\mathrm{2}\:{e}^{\frac{{i}\pi}{\mathrm{4}}} }\:=\frac{\mathrm{1}}{\mathrm{2}}\:{e}^{−\frac{{i}\pi}{\mathrm{4}}} \:\Rightarrow\:\int_{−\infty} ^{+\infty} \varphi\left({z}\right){dz}=\mathrm{2}{i}\pi\frac{\mathrm{1}}{\mathrm{2}}\:{e}^{−\frac{{i}\pi}{\mathrm{4}}} \\ $$$$={i}\pi\:{e}^{−\frac{{i}\pi}{\mathrm{4}}} \:\Rightarrow\:{I}\:={i}\pi\:{e}^{−\frac{{i}\pi}{\mathrm{4}}} \:\:\:\:{let}\:{calculate}\:{J}\:=\int_{−\infty} ^{+\infty} \:\frac{{dx}}{{x}^{\mathrm{2}} \:+{i}} \\ $$$${let}\:{W}\left({z}\right)=\frac{\mathrm{1}}{{z}^{\mathrm{2}} \:+{i}}\:\Rightarrow{W}\left({z}\right)=\frac{\mathrm{1}}{\left({z}−{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)\left({z}\:+{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)}\:\:{so}\:{the}\:{poles}\:{of}\:\varphi\:{are}\:\overset{−} {+}{e}^{−\frac{{i}\pi}{\mathrm{4}}} \\ $$$${residus}\:{theorem}\:{give}\:\int_{−\infty} ^{+\infty} \:{W}\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{Res}\left({W},−{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right) \\ $$$${Res}\left({W},−{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)\:={lim}_{{z}\rightarrow−{e}^{−\frac{{i}\pi}{\mathrm{4}}} } \:\:\:\left({z}+{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right){W}\left({z}\right)\:=\frac{\mathrm{1}}{−\mathrm{2}\:{e}^{−\frac{{i}\pi}{\mathrm{4}}} }\:=−\frac{\mathrm{1}}{\mathrm{2}}\:{e}^{\frac{{i}\pi}{\mathrm{4}}} \:\Rightarrow \\ $$$$\int_{−\infty} ^{+\infty} \:{W}\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:\left(−\frac{\mathrm{1}}{\mathrm{2}}\:{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)\:=−{i}\pi\:{e}^{\frac{{i}\pi}{\mathrm{4}}} \:\:\:={J}\:\:\:{also}\:{we}\:{can}\:{use}\:{that} \\ $$$${J}\:={conj}\left({I}\right)\:={conj}\left({i}\pi\:{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)=−{i}\pi\:{e}^{\frac{{i}\pi}{\mathrm{4}}} \: \\ $$$$\left.\mathrm{2}\right)\:\:\int_{−\infty} ^{+\infty} \:\:\frac{{dx}}{{x}^{\mathrm{4}} \:+\mathrm{1}}\:=\int_{−\infty} ^{+\infty} \:\:\frac{{dx}}{\left({x}^{\mathrm{2}} −{i}\right)\left({x}^{\mathrm{2}} \:+{i}\right)}\:=\frac{\mathrm{1}}{\mathrm{2}{i}}\int_{−\infty} ^{+\infty} \left\{\frac{\mathrm{1}}{{x}^{\mathrm{2}} −{i}}\:−\frac{\mathrm{1}}{{x}^{\mathrm{2}} \:+{i}}\right\}{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{i}}\left\{\:{I}−{J}\right\}\:=\frac{\mathrm{1}}{\mathrm{2}{i}}\left\{{i}\pi\:{e}^{−\frac{{i}\pi}{\mathrm{4}}} \:+{i}\:\pi\:{e}^{\frac{{i}\pi}{\mathrm{4}}} \right\}\:=\frac{\pi}{\mathrm{2}}\:\left(\mathrm{2}{cos}\left(\frac{\pi}{\mathrm{4}}\right)\right)=\pi\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:=\frac{\pi}{\:\sqrt{\mathrm{2}}}\:. \\ $$$$ \\ $$

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