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Question Number 34320 by abdo mathsup 649 cc last updated on 04/May/18

calculate ∫_(−∞) ^(+∞) (dx/(x^2 +1 −i))

$${calculate}\:\:\:\int_{−\infty} ^{+\infty} \:\:\:\:\:\frac{{dx}}{{x}^{\mathrm{2}} \:+\mathrm{1}\:−{i}} \\ $$

Commented by math khazana by abdo last updated on 05/May/18

∫_(−∞) ^(+∞) (dx/(x^2 +1−i)) =I we consider the complex function ϕ(z) = (1/(z^2 +1−i)) poles of ϕ? ϕ(z) = (1/(z^2 −(−1 +i))) = (1/(z^2 − (√2) e^(i((3π)/4)) )) = (1/((z− 2^(1/4) e^(i((3π)/8)) )(z +2^(1/4) e^(i((3π)/8)) ))) so the poles of ϕ are z_0 =2^(1/4) e^(i((3π)/8)) and z_1 = −2^(1/4) e^(i((3π)/8)) ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ Res (ϕ,z_0 ) Res(ϕ,z_0 ) =(1/(2z_0 )) = (1/(2^(5/4) e^(i((3π)/8)) )) = 2^(−(5/4)) e^(−i((3π)/8)) ∫_(−∞) ^(+∞) ϕ(z)dz = 2iπ . 2^(−(5/4)) e^(−i((3π)/8)) = 2^(−(1/4)) iπ( cos(((3π)/8)) −isin(((3π)/8))) =I .

$$\int_{−\infty} ^{+\infty} \:\:\:\frac{{dx}}{{x}^{\mathrm{2}} \:+\mathrm{1}−{i}}\:\:={I}\:\:{we}\:{consider}\:{the}\:{complex} \\ $$$${function}\:\varphi\left({z}\right)\:=\:\frac{\mathrm{1}}{{z}^{\mathrm{2}} \:+\mathrm{1}−{i}}\:{poles}\:{of}\:\varphi? \\ $$$$\varphi\left({z}\right)\:=\:\frac{\mathrm{1}}{{z}^{\mathrm{2}} \:−\left(−\mathrm{1}\:+{i}\right)}\:=\:\frac{\mathrm{1}}{{z}^{\mathrm{2}} \:−\:\sqrt{\mathrm{2}}\:{e}^{{i}\frac{\mathrm{3}\pi}{\mathrm{4}}} } \\ $$$$=\:\frac{\mathrm{1}}{\left({z}−\:\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{4}}} \:{e}^{{i}\frac{\mathrm{3}\pi}{\mathrm{8}}} \right)\left({z}\:+\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{4}}} \:{e}^{{i}\frac{\mathrm{3}\pi}{\mathrm{8}}} \:\right)}\:{so}\:{the}\:{poles}\:{of}\:\varphi\:{are} \\ $$$${z}_{\mathrm{0}} =\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{4}}} \:{e}^{{i}\frac{\mathrm{3}\pi}{\mathrm{8}}} \:\:{and}\:\:{z}_{\mathrm{1}} =\:−\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{4}}} \:\:{e}^{{i}\frac{\mathrm{3}\pi}{\mathrm{8}}} \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{Res}\:\left(\varphi,{z}_{\mathrm{0}} \right) \\ $$$${Res}\left(\varphi,{z}_{\mathrm{0}} \right)\:\:=\frac{\mathrm{1}}{\mathrm{2}{z}_{\mathrm{0}} }\:=\:\:\frac{\mathrm{1}}{\mathrm{2}^{\frac{\mathrm{5}}{\mathrm{4}}} \:{e}^{{i}\frac{\mathrm{3}\pi}{\mathrm{8}}} }\:=\:\mathrm{2}^{−\frac{\mathrm{5}}{\mathrm{4}}} \:{e}^{−{i}\frac{\mathrm{3}\pi}{\mathrm{8}}} \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\:\mathrm{2}{i}\pi\:.\:\mathrm{2}^{−\frac{\mathrm{5}}{\mathrm{4}}} \:{e}^{−{i}\frac{\mathrm{3}\pi}{\mathrm{8}}} \\ $$$$=\:\mathrm{2}^{−\frac{\mathrm{1}}{\mathrm{4}}} \:{i}\pi\left(\:{cos}\left(\frac{\mathrm{3}\pi}{\mathrm{8}}\right)\:−{isin}\left(\frac{\mathrm{3}\pi}{\mathrm{8}}\right)\right)\:={I}\:. \\ $$$$ \\ $$

Commented by math khazana by abdo last updated on 05/May/18

remark we have ∫_(−∞) ^(+∞) (dx/(x^2 +1 −i)) = ∫_(−∞) ^(+∞) ((x^2 +1 +i)/((x^2 +1)^2 +1))dx = ∫_(−∞) ^(+∞) ((x^2 +1)/((x^2 +1)^2 +1))dx +i ∫_(−∞) ^(+∞) (dx/((x^2 +1)^2 +1)) ⇒ ∫_(−∞) ^(+∞) ((x^2 +1)/((x^2 +1)^2 +1))dx = π 2^(−(1/4)) sin(((3π)/8)) ∫_(−∞) ^(+∞) (dx/((x^2 +1)^2 +1)) = π 2^(−(1/4)) cos(((3π)/8)) .

$${remark}\:{we}\:{have} \\ $$$$\int_{−\infty} ^{+\infty} \:\:\:\frac{{dx}}{{x}^{\mathrm{2}} +\mathrm{1}\:−{i}}\:=\:\int_{−\infty} ^{+\infty} \:\:\:\frac{{x}^{\mathrm{2}} +\mathrm{1}\:+{i}}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} \:+\mathrm{1}}{dx} \\ $$$$=\:\int_{−\infty} ^{+\infty} \:\frac{{x}^{\mathrm{2}} \:+\mathrm{1}}{\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} \:+\mathrm{1}}{dx}\:\:+{i}\:\int_{−\infty} ^{+\infty} \:\:\:\:\:\frac{{dx}}{\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} +\mathrm{1}}\:\Rightarrow \\ $$$$\int_{−\infty} ^{+\infty} \:\:\:\:\frac{{x}^{\mathrm{2}} +\mathrm{1}}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} +\mathrm{1}}{dx}\:=\:\pi\:\mathrm{2}^{−\frac{\mathrm{1}}{\mathrm{4}}} \:\:{sin}\left(\frac{\mathrm{3}\pi}{\mathrm{8}}\right) \\ $$$$\int_{−\infty} ^{+\infty} \:\:\:\:\:\:\frac{{dx}}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} \:+\mathrm{1}}\:=\:\pi\:\mathrm{2}^{−\frac{\mathrm{1}}{\mathrm{4}}} \:{cos}\left(\frac{\mathrm{3}\pi}{\mathrm{8}}\right)\:. \\ $$

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