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Question Number 29853 by abdo imad last updated on 13/Feb/18

$$find{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{dx}{x^2+2ix+2-4i}.$$

Commented by abdo imad last updated on 18/Feb/18

$$letputf\left(z\right)=\frac{1}{z^2+2iz+2-4i}polesoff?letfindtherootsof$$$$of{z}^2+2iz+2-4i=0\Rightarrow {\mathrm{\Delta }}^{{}^{\prime }}=i^2-(2-4i)=-1-2+4i$$$$=-3+4i={(1+2i)}^2\Rightarrow z_1=-i+1+2i=1+i$$$$z_2=-i-(1+2i)=-1-3i$$$$\mid z_1\mid -1=\sqrt{2}-1<1and\mid z_2\mid -1=\sqrt{10}-1>1(toeliminate$$$$fromresidus)$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}f\left(z\right)dz=2i\pi Res(f,z_1)but$$$$f\left(z\right)=\frac{1}{(z-z_1)(z-z_2)}\Rightarrow Res(f,z_1)={lim}_{z\to z_1}(z-z_1)f\left(z\right)$$$$=\frac{1}{z_1-z_2}=\frac{1}{2+4i}=\frac{2-4i}{4-16}=\frac{2-4i}{-12}=-\frac{1}{6}+\frac{1}{3}i$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}f\left(z\right)dz=2i\pi (-\frac{1}{6}+\frac{1}{3}i)=\frac{-i\pi }{3}-\frac{2\pi }{3}\Rightarrow$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{dx}{x^2+2ix+2-4i}=-\frac{2\pi }{3}-\frac{i\pi }{3}.$$$$$$

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