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Question Number 110448 by mathmax by abdo last updated on 29/Aug/20
calculate ∫_(−∞) ^(+∞) ((cos(2x))/((x^2 −4i)^3 ))dx (i=(√(−1)))
$$\mathrm{calculate}\:\:\int_{−\infty} ^{+\infty} \:\frac{\mathrm{cos}\left(\mathrm{2x}\right)}{\left(\mathrm{x}^{\mathrm{2}} −\mathrm{4i}\right)^{\mathrm{3}} }\mathrm{dx}\:\:\:\:\:\left(\mathrm{i}=\sqrt{−\mathrm{1}}\right) \\ $$
Answered by mathmax by abdo last updated on 30/Aug/20
A =∫_(−∞) ^(+∞) ((cos(2x))/((x^2 −4i)^3 ))dx ⇒ A =Re(∫_(−∞) ^(+∞) (e^(2ix) /((x^2 −4i)^3 ))dx) let ϕ(z) =(e^(2iz) /((z^2 −4i)^3 )) poles of ϕ we have ϕ(z)=(e^(2iz) /((z^2 −(2e^((iπ)/4) )^2 )^3 )) =(e^(2iz) /((z−2e^((iπ)/4) )^3 (z+2e^((iπ)/4) )^3 )) so the poles are +^− 2e^((iπ)/4) (triples) residus theorem ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ Res(ϕ,2e^((iπ)/4) ) and Res(ϕ,2e^((iπ)/4) ) =lim_(z→2 e^((iπ)/4) ) (1/((3−1)!)){(z−2e^((iπ)/4) )^3 ϕ(z)}^((2)) =lim_(z→2e^((iπ)/4) ) (1/2){ (e^(2iz) /((z+2e^((iπ)/4) )^3 ))}^((2)) =lim_(z→2e^((iπ)/4) ) (1/2){((2iz (z+2e^((iπ)/4) )^3 e^(2iz) −3(z+2e^((iπ)/4) )^2 e^(2iz) )/((z+2e^((iπ)/4) )^6 ))}^((1)) =lim_(z→2e^((iπ)/4) ) (1/2){((2iz(z +2e^((iπ)/4) )e^(2iz) −3e^(2iz) )/((z+2e^((iπ)/4) )^4 ))}^((1)) =lim_(z→2e^((iπ)/4) ) (1/2){ (((2iz^2 +4iz e^((iπ)/4) −3)e^(2iz) )/((z+2e^((iπ)/4) )^4 ))}^((1)) =lim_(z→2e^((iπ)/4) ) (1/2)(({(4iz +4ie^((iπ)/4) )e^(2iz) +2ie^(2iz) (2iz^2 +4ize^((iπ)/4) −3)}(z+2e^((iπ)/4) )^4 −4(z +2e^((iπ)/4) )^3 (2iz^2 +4iz e^((iπ)/4) −3)e^(2iz) )/((z+2e^((iπ)/4) )^8 )) ...be continued....
$$\mathrm{A}\:=\int_{−\infty} ^{+\infty} \:\frac{\mathrm{cos}\left(\mathrm{2x}\right)}{\left(\mathrm{x}^{\mathrm{2}} −\mathrm{4i}\right)^{\mathrm{3}} }\mathrm{dx}\:\Rightarrow\:\mathrm{A}\:=\mathrm{Re}\left(\int_{−\infty} ^{+\infty} \:\frac{\mathrm{e}^{\mathrm{2ix}} }{\left(\mathrm{x}^{\mathrm{2}} −\mathrm{4i}\right)^{\mathrm{3}} }\mathrm{dx}\right)\:\mathrm{let}\: \\ $$$$\varphi\left(\mathrm{z}\right)\:=\frac{\mathrm{e}^{\mathrm{2iz}} }{\left(\mathrm{z}^{\mathrm{2}} −\mathrm{4i}\right)^{\mathrm{3}} }\:\:\mathrm{poles}\:\mathrm{of}\:\varphi\:\:\mathrm{we}\:\mathrm{have}\:\varphi\left(\mathrm{z}\right)=\frac{\mathrm{e}^{\mathrm{2iz}} }{\left(\mathrm{z}^{\mathrm{2}} −\left(\mathrm{2e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)^{\mathrm{2}} \right)^{\mathrm{3}} } \\ $$$$=\frac{\mathrm{e}^{\mathrm{2iz}} }{\left(\mathrm{z}−\mathrm{2e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)^{\mathrm{3}} \left(\mathrm{z}+\mathrm{2e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)^{\mathrm{3}} }\:\mathrm{so}\:\mathrm{the}\:\mathrm{poles}\:\mathrm{are}\:\overset{−} {+}\:\mathrm{2e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} \left(\mathrm{triples}\right) \\ $$$$\mathrm{residus}\:\mathrm{theorem}\:\Rightarrow\:\int_{−\infty} ^{+\infty} \:\varphi\left(\mathrm{z}\right)\mathrm{dz}\:=\mathrm{2i}\pi\:\mathrm{Res}\left(\varphi,\mathrm{2e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} \right) \\ $$$$\mathrm{and}\:\mathrm{Res}\left(\varphi,\mathrm{2e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)\:=\mathrm{lim}_{\mathrm{z}\rightarrow\mathrm{2}\:\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} } \:\:\:\frac{\mathrm{1}}{\left(\mathrm{3}−\mathrm{1}\right)!}\left\{\left(\mathrm{z}−\mathrm{2e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)^{\mathrm{3}} \varphi\left(\mathrm{z}\right)\right\}^{\left(\mathrm{2}\right)} \\ $$$$=\mathrm{lim}_{\mathrm{z}\rightarrow\mathrm{2e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} } \:\:\:\frac{\mathrm{1}}{\mathrm{2}}\left\{\:\frac{\mathrm{e}^{\mathrm{2iz}} }{\left(\mathrm{z}+\mathrm{2e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)^{\mathrm{3}} }\right\}^{\left(\mathrm{2}\right)} \\ $$$$=\mathrm{lim}_{\mathrm{z}\rightarrow\mathrm{2e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} } \:\:\:\frac{\mathrm{1}}{\mathrm{2}}\left\{\frac{\mathrm{2iz}\:\left(\mathrm{z}+\mathrm{2e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)^{\mathrm{3}} \mathrm{e}^{\mathrm{2iz}} −\mathrm{3}\left(\mathrm{z}+\mathrm{2e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)^{\mathrm{2}} \:\mathrm{e}^{\mathrm{2iz}} }{\left(\mathrm{z}+\mathrm{2e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)^{\mathrm{6}} }\right\}^{\left(\mathrm{1}\right)} \\ $$$$=\mathrm{lim}_{\mathrm{z}\rightarrow\mathrm{2e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} } \:\:\frac{\mathrm{1}}{\mathrm{2}}\left\{\frac{\mathrm{2iz}\left(\mathrm{z}\:+\mathrm{2e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)\mathrm{e}^{\mathrm{2iz}} −\mathrm{3e}^{\mathrm{2iz}} }{\left(\mathrm{z}+\mathrm{2e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)^{\mathrm{4}} }\right\}^{\left(\mathrm{1}\right)} \\ $$$$=\mathrm{lim}_{\mathrm{z}\rightarrow\mathrm{2e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} } \:\:\:\frac{\mathrm{1}}{\mathrm{2}}\left\{\:\:\:\frac{\left(\mathrm{2iz}^{\mathrm{2}} +\mathrm{4iz}\:\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} −\mathrm{3}\right)\mathrm{e}^{\mathrm{2iz}} }{\left(\mathrm{z}+\mathrm{2e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)^{\mathrm{4}} }\right\}^{\left(\mathrm{1}\right)} \\ $$$$=\mathrm{lim}_{\mathrm{z}\rightarrow\mathrm{2e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} } \:\:\frac{\mathrm{1}}{\mathrm{2}}\frac{\left\{\left(\mathrm{4iz}\:+\mathrm{4ie}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)\mathrm{e}^{\mathrm{2iz}} \:+\mathrm{2ie}^{\mathrm{2iz}} \left(\mathrm{2iz}^{\mathrm{2}} +\mathrm{4ize}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} −\mathrm{3}\right)\right\}\left(\mathrm{z}+\mathrm{2e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)^{\mathrm{4}} −\mathrm{4}\left(\mathrm{z}\:+\mathrm{2e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)^{\mathrm{3}} \left(\mathrm{2iz}^{\mathrm{2}} +\mathrm{4iz}\:\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} −\mathrm{3}\right)\mathrm{e}^{\mathrm{2iz}} }{\left(\mathrm{z}+\mathrm{2e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)^{\mathrm{8}} } \\ $$$$…\mathrm{be}\:\mathrm{continued}…. \\ $$

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