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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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