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Question Number 66465 by mathmax by abdo last updated on 15/Aug/19

calculate ∫_0 ^∞ (dx/((x^2 +2i)( x^2 +4j))) with i=e^((iπ)/2) and j=e^(i((2π)/3))

$${calculate}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{dx}}{\left({x}^{\mathrm{2}} +\mathrm{2}{i}\right)\left(\:{x}^{\mathrm{2}} \:+\mathrm{4}{j}\right)}\:\:\:{with}\:{i}={e}^{\frac{{i}\pi}{\mathrm{2}}} \:{and}\:{j}={e}^{{i}\frac{\mathrm{2}\pi}{\mathrm{3}}} \\ $$

Commented by mathmax by abdo last updated on 16/Aug/19

let A =∫_0 ^∞ (dx/((x^2 +2i)(x^2 +4j))) ⇒2A =∫_(−∞) ^(+∞) (dx/((x^2 +2i)(x^2 +4j))) let ϕ(z) =(1/((z^2 +2i)(z^2 +4j))) poles of ϕ? we have ϕ(z) =(1/((z^2 −(−2i)(z^2 −(−4j)))) =(1/((z−(√(−2i)))(z+(√(−2i)))(z−2(√(−j)))(z+2(√(−j))))) −j =e^(iπ) .e^(i((2π)/3)) =e^(i(π+((2π)/3))) =e^(i(((5π)/3))) =e^(−((iπ)/3)) ⇒(√(−j))=e^(−((iπ)/6)) ⇒ ϕ(z) =(1/((z−(√2)e^(−((iπ)/4)) )(z+(√2)e^(−((iπ)/4)) )(z−2e^(−((iπ)/6)) )(z+2 e^(−((iπ)/6)) ))) the poles of ϕ are +^− (√2)e^(−((iπ)/4)) and +^− 2e^(−((iπ)/6)) residus tbeorem give ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ{ Res(ϕ,−(√2)e^(−((iπ)/4)) )+Res(ϕ,−2e^(−((iπ)/6)) )} Res(ϕ,−(√2)e^(−((iπ)/4)) ) =(1/(−2(√2)e^(−((iπ)/4)) (2 e^(−((iπ)/2)) +4j))) =−(e^((iπ)/4) /(4(√2)(−i +4j))) Res(ϕ,−2e^(−((iπ)/6)) ) =(1/(−4e^(−((iπ)/6)) (e^(−((iπ)/3)) +2i))) =−(e^(i(π/6)) /(4(2i +e^(−((iπ)/3)) ))) ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz =−((iπ)/2){ (e^((iπ)/4) /((√2)(−i+4j))) +(e^((iπ)/6) /(2i+e^(−((iπ)/3)) ))} =2A

$${let}\:{A}\:=\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{dx}}{\left({x}^{\mathrm{2}} \:+\mathrm{2}{i}\right)\left({x}^{\mathrm{2}} \:+\mathrm{4}{j}\right)}\:\Rightarrow\mathrm{2}{A}\:=\int_{−\infty} ^{+\infty} \:\frac{{dx}}{\left({x}^{\mathrm{2}} \:+\mathrm{2}{i}\right)\left({x}^{\mathrm{2}} \:+\mathrm{4}{j}\right)} \\ $$$${let}\:\varphi\left({z}\right)\:=\frac{\mathrm{1}}{\left({z}^{\mathrm{2}} \:+\mathrm{2}{i}\right)\left({z}^{\mathrm{2}} \:+\mathrm{4}{j}\right)}\:\:{poles}\:{of}\:\varphi? \\ $$$${we}\:{have}\:\varphi\left({z}\right)\:=\frac{\mathrm{1}}{\left({z}^{\mathrm{2}} −\left(−\mathrm{2}{i}\right)\left({z}^{\mathrm{2}} −\left(−\mathrm{4}{j}\right)\right)\right.}\: \\ $$$$=\frac{\mathrm{1}}{\left({z}−\sqrt{−\mathrm{2}{i}}\right)\left({z}+\sqrt{−\mathrm{2}{i}}\right)\left({z}−\mathrm{2}\sqrt{−{j}}\right)\left({z}+\mathrm{2}\sqrt{−{j}}\right)} \\ $$$$−{j}\:={e}^{{i}\pi} .{e}^{{i}\frac{\mathrm{2}\pi}{\mathrm{3}}} \:={e}^{{i}\left(\pi+\frac{\mathrm{2}\pi}{\mathrm{3}}\right)} \:={e}^{{i}\left(\frac{\mathrm{5}\pi}{\mathrm{3}}\right)} \:={e}^{−\frac{{i}\pi}{\mathrm{3}}} \:\Rightarrow\sqrt{−{j}}={e}^{−\frac{{i}\pi}{\mathrm{6}}} \:\Rightarrow \\ $$$$\varphi\left({z}\right)\:=\frac{\mathrm{1}}{\left({z}−\sqrt{\mathrm{2}}{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)\left({z}+\sqrt{\mathrm{2}}{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)\left({z}−\mathrm{2}{e}^{−\frac{{i}\pi}{\mathrm{6}}} \right)\left({z}+\mathrm{2}\:{e}^{−\frac{{i}\pi}{\mathrm{6}}} \right)} \\ $$$${the}\:{poles}\:{of}\:\varphi\:{are}\:\overset{−} {+}\sqrt{\mathrm{2}}{e}^{−\frac{{i}\pi}{\mathrm{4}}} \:{and}\:\overset{−} {+}\mathrm{2}{e}^{−\frac{{i}\pi}{\mathrm{6}}} \:{residus}\:{tbeorem}\:{give} \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\left\{\:{Res}\left(\varphi,−\sqrt{\mathrm{2}}{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)+{Res}\left(\varphi,−\mathrm{2}{e}^{−\frac{{i}\pi}{\mathrm{6}}} \right)\right\} \\ $$$${Res}\left(\varphi,−\sqrt{\mathrm{2}}{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)\:=\frac{\mathrm{1}}{−\mathrm{2}\sqrt{\mathrm{2}}{e}^{−\frac{{i}\pi}{\mathrm{4}}} \left(\mathrm{2}\:{e}^{−\frac{{i}\pi}{\mathrm{2}}} \:+\mathrm{4}{j}\right)}\:=−\frac{{e}^{\frac{{i}\pi}{\mathrm{4}}} }{\mathrm{4}\sqrt{\mathrm{2}}\left(−{i}\:+\mathrm{4}{j}\right)} \\ $$$${Res}\left(\varphi,−\mathrm{2}{e}^{−\frac{{i}\pi}{\mathrm{6}}} \right)\:=\frac{\mathrm{1}}{−\mathrm{4}{e}^{−\frac{{i}\pi}{\mathrm{6}}} \left({e}^{−\frac{{i}\pi}{\mathrm{3}}} \:+\mathrm{2}{i}\right)}\:=−\frac{{e}^{{i}\frac{\pi}{\mathrm{6}}} }{\mathrm{4}\left(\mathrm{2}{i}\:+{e}^{−\frac{{i}\pi}{\mathrm{3}}} \right)}\:\Rightarrow \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=−\frac{{i}\pi}{\mathrm{2}}\left\{\:\:\frac{{e}^{\frac{{i}\pi}{\mathrm{4}}} }{\sqrt{\mathrm{2}}\left(−{i}+\mathrm{4}{j}\right)}\:+\frac{{e}^{\frac{{i}\pi}{\mathrm{6}}} }{\mathrm{2}{i}+{e}^{−\frac{{i}\pi}{\mathrm{3}}} }\right\}\:=\mathrm{2}{A} \\ $$

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