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let-A-dx-x-2-j-with-j-e-i-2pi-3-extract-ReA-and-Im-A-and-calculste-its-values-

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Question Number 34421 by abdo mathsup 649 cc last updated on 06/May/18
let A = ∫_(−∞) ^(+∞) (dx/(x^2 −j)) with j=e^(i((2π)/3)) extract ReA and Im(A) and calculste its values.
$${let}\:{A}\:\:=\:\int_{−\infty} ^{+\infty} \:\:\:\:\frac{{dx}}{{x}^{\mathrm{2}} \:−{j}}\:\:\:\:{with}\:{j}={e}^{{i}\frac{\mathrm{2}\pi}{\mathrm{3}}} \\ $$$${extract}\:\:{ReA}\:{and}\:{Im}\left({A}\right)\:{and}\:{calculste}\:{its}\:{values}. \\ $$
Commented by abdo mathsup 649 cc last updated on 07/May/18
we have A =∫_(−∞) ^(+∞) (dx/(x^2 −(−(1/2) +i((√3)/2)))) = ∫_(−∞) ^(+∞) (dx/(x^2 +(1/2) −i((√3)/2))) = ∫_(−∞) ^(+∞) ((x^2 +(1/2) +i((√3)/2))/((x^2 +(1/2))^2 +(3/4)))dx ⇒ Re(A) = ∫_(−∞) ^(+∞) ((x^2 +(1/2))/((x^2 +(1/2))^2 +(3/4)))dx and Im(A) = ((√3)/2) ∫_(−∞) ^(+∞) (dx/((x^2 +(1/2))^2 +(3/4))) let introduce the complex function ϕ(z) = (1/(z^2 −j)) ϕ(z) = (1/((z −(√j))(z +(√j)))) = (1/((z − e^(i(π/3)) )(z + e^(i(π/3)) ))) the poles of ϕ are e^(i(π/3)) , −e^(i(π/3)) ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ Res(ϕ,e^(i(π/3)) ) Res(ϕ,e^(i(π/3)) )= (1/(2 e^(i(π/3)) )) =(1/2) e^(−i(π/3)) =(1/2)( cos(−(π/3)) +isin(−(π/3))) =(1/2)( (1/2) −i((√3)/2)) ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ (1/2)( (1/2) −i((√3)/2)) =i(π/2) +π((√3)/2) ⇒Re(A) = π((√3)/2) and Im(A) = (π/2) .
$${we}\:{have}\:{A}\:=\int_{−\infty} ^{+\infty} \:\:\:\frac{{dx}}{{x}^{\mathrm{2}} \:−\left(−\frac{\mathrm{1}}{\mathrm{2}}\:+{i}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)} \\ $$$$=\:\int_{−\infty} ^{+\infty} \:\:\:\:\frac{{dx}}{{x}^{\mathrm{2}} \:+\frac{\mathrm{1}}{\mathrm{2}}\:−{i}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}} \\ $$$$=\:\int_{−\infty} ^{+\infty} \:\:\:\frac{{x}^{\mathrm{2}} \:+\frac{\mathrm{1}}{\mathrm{2}}\:+{i}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}{\left({x}^{\mathrm{2}} \:+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} \:+\frac{\mathrm{3}}{\mathrm{4}}}{dx}\:\Rightarrow \\ $$$${Re}\left({A}\right)\:=\:\int_{−\infty} ^{+\infty} \:\:\frac{{x}^{\mathrm{2}} \:+\frac{\mathrm{1}}{\mathrm{2}}}{\left({x}^{\mathrm{2}} \:+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{3}}{\mathrm{4}}}{dx}\:{and} \\ $$$${Im}\left({A}\right)\:=\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:\int_{−\infty} ^{+\infty} \:\:\:\:\frac{{dx}}{\left({x}^{\mathrm{2}} \:+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} \:+\frac{\mathrm{3}}{\mathrm{4}}}\:\:{let}\:{introduce} \\ $$$${the}\:{complex}\:{function}\:\varphi\left({z}\right)\:=\:\:\frac{\mathrm{1}}{{z}^{\mathrm{2}} \:−{j}} \\ $$$$\varphi\left({z}\right)\:=\:\frac{\mathrm{1}}{\left({z}\:−\sqrt{{j}}\right)\left({z}\:+\sqrt{{j}}\right)}\:=\:\:\frac{\mathrm{1}}{\left({z}\:−\:{e}^{{i}\frac{\pi}{\mathrm{3}}} \right)\left({z}\:\:+\:\mathrm{e}^{\mathrm{i}\frac{\pi}{\mathrm{3}}} \right)} \\ $$$${the}\:{poles}\:{of}\:\varphi\:{are}\:{e}^{{i}\frac{\pi}{\mathrm{3}}} \:\:,\:−{e}^{{i}\frac{\pi}{\mathrm{3}}} \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{Res}\left(\varphi,{e}^{{i}\frac{\pi}{\mathrm{3}}} \right) \\ $$$${Res}\left(\varphi,{e}^{{i}\frac{\pi}{\mathrm{3}}} \right)=\:\:\frac{\mathrm{1}}{\mathrm{2}\:{e}^{{i}\frac{\pi}{\mathrm{3}}} }\:=\frac{\mathrm{1}}{\mathrm{2}}\:{e}^{−{i}\frac{\pi}{\mathrm{3}}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left(\:{cos}\left(−\frac{\pi}{\mathrm{3}}\right)\:+{isin}\left(−\frac{\pi}{\mathrm{3}}\right)\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left(\:\frac{\mathrm{1}}{\mathrm{2}}\:−{i}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right) \\ $$$$\int_{−\infty} ^{+\infty} \:\:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:\frac{\mathrm{1}}{\mathrm{2}}\left(\:\frac{\mathrm{1}}{\mathrm{2}}\:−{i}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right) \\ $$$$={i}\frac{\pi}{\mathrm{2}}\:\:+\pi\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:\:\Rightarrow{Re}\left({A}\right)\:=\:\pi\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$${and}\:{Im}\left({A}\right)\:=\:\frac{\pi}{\mathrm{2}}\:. \\ $$

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