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Question Number 65131 by turbo msup by abdo last updated on 25/Jul/19

calculate ∫_0 ^∞ (dx/((x^4 −4i)^3 ))

$${calculate}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dx}}{\left({x}^{\mathrm{4}} −\mathrm{4}{i}\right)^{\mathrm{3}} } \\ $$

Commented by mathmax by abdo last updated on 26/Jul/19

let I =∫_0 ^∞ (dx/((x^4 −4i)^3 )) ⇒2I =∫_(−∞) ^(+∞) (dx/((x^4 −4i)^3 )) let ϕ(z) =(1/((z^4 −4i)^3 )) poles of ϕ? first let solve z^4 =4i z =r e^(iθ) ⇒ r^4 e^(i4θ) =4 e^((iπ)/2) ⇒r^4 =4 and 4θ =(π/2) +2kπ ⇒r =4^(1/4) =2^(1/2) =(√2) and θ =(π/8) +((kπ)/2) ⇒4i =((√2)e^((iπ)/8) )^4 ⇒ z^4 −4i =z^4 −((√2)e^((iπ)/8) )^4 =(z^2 −((√2)e^((iπ)/8) )^2 )(z^2 +((√2)e^((iπ)/8) )^2 ) =(z−(√2)e^((iπ)/8) )(z+(√2)e^((iπ)/8) )(z^2 −((√2)i e^((iπ)/8) )^2 ) =(z−(√2)e^((iπ)/8) )(z+(√2)e^((iπ)/8) )(z^2 −((√2)e^((i5π)/8) )^2 ) (z−(√2)e^((iπ)/8) )(z+(√2)e^((iπ)/8) )(z−(√2)e^((i5π)/8) )(z+(√2)e^((i5π)/8) ) ⇒ ϕ(z) =(1/((z−(√2)e^((iπ)/8) )^3 (z+(√2)e^((iπ)/8) )^3 (z−(√2)e^((i5π)/8) )^3 (z+(√2)e^((i5π)/8) )^3 )) the poles of ϕ are +^− (√2)e^((iπ)/8) and +^− e^((i5π)/8) (ordre 3) residus theorem give ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ {Res(ϕ,(√2)e^((iπ)/8) )+Res(ϕ,(√2)e^((i5π)/8) )} Res(ϕ,(√2)e^((iπ)/8) ) =lim_(z→(√2)e^((iπ)/8) ) (1/((3−1)!)){ (z−(√2)e^((iπ)/8) )^3 ϕ(z)}^((2)) =lim_(z→(√2)e^((iπ)/8) ) (1/2) {(1/((z+(√2)e^((iπ)/8) )^3 (z−(√2)e^((i5π)/8) )^3 (z+(√2)e^((i5π)/8) )^3 ))}^((2)) ...be continued....

$${let}\:{I}\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dx}}{\left({x}^{\mathrm{4}} −\mathrm{4}{i}\right)^{\mathrm{3}} }\:\Rightarrow\mathrm{2}{I}\:=\int_{−\infty} ^{+\infty} \:\:\frac{{dx}}{\left({x}^{\mathrm{4}} −\mathrm{4}{i}\right)^{\mathrm{3}} } \\ $$$$\:{let}\:\varphi\left({z}\right)\:=\frac{\mathrm{1}}{\left({z}^{\mathrm{4}} −\mathrm{4}{i}\right)^{\mathrm{3}} }\:\:{poles}\:{of}\:\varphi?\:\:{first}\:{let}\:{solve}\:{z}^{\mathrm{4}} \:=\mathrm{4}{i} \\ $$$${z}\:={r}\:{e}^{{i}\theta} \:\Rightarrow\:{r}^{\mathrm{4}} \:{e}^{{i}\mathrm{4}\theta} \:=\mathrm{4}\:{e}^{\frac{{i}\pi}{\mathrm{2}}} \:\Rightarrow{r}^{\mathrm{4}} \:=\mathrm{4}\:\:{and}\:\mathrm{4}\theta\:=\frac{\pi}{\mathrm{2}}\:+\mathrm{2}{k}\pi\:\Rightarrow{r}\:=\mathrm{4}^{\frac{\mathrm{1}}{\mathrm{4}}} \\ $$$$=\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{2}}} \:=\sqrt{\mathrm{2}}\:{and}\:\:\theta\:=\frac{\pi}{\mathrm{8}}\:+\frac{{k}\pi}{\mathrm{2}}\:\Rightarrow\mathrm{4}{i}\:=\left(\sqrt{\mathrm{2}}{e}^{\frac{{i}\pi}{\mathrm{8}}} \right)^{\mathrm{4}} \:\Rightarrow \\ $$$${z}^{\mathrm{4}} −\mathrm{4}{i}\:={z}^{\mathrm{4}} −\left(\sqrt{\mathrm{2}}{e}^{\frac{{i}\pi}{\mathrm{8}}} \right)^{\mathrm{4}} \:=\left({z}^{\mathrm{2}} \:−\left(\sqrt{\mathrm{2}}{e}^{\frac{{i}\pi}{\mathrm{8}}} \right)^{\mathrm{2}} \right)\left({z}^{\mathrm{2}} \:+\left(\sqrt{\mathrm{2}}{e}^{\frac{{i}\pi}{\mathrm{8}}} \right)^{\mathrm{2}} \right) \\ $$$$=\left({z}−\sqrt{\mathrm{2}}{e}^{\frac{{i}\pi}{\mathrm{8}}} \right)\left({z}+\sqrt{\mathrm{2}}{e}^{\frac{{i}\pi}{\mathrm{8}}} \right)\left({z}^{\mathrm{2}} −\left(\sqrt{\mathrm{2}}{i}\:{e}^{\frac{{i}\pi}{\mathrm{8}}} \right)^{\mathrm{2}} \right) \\ $$$$=\left({z}−\sqrt{\mathrm{2}}{e}^{\frac{{i}\pi}{\mathrm{8}}} \right)\left({z}+\sqrt{\mathrm{2}}{e}^{\frac{{i}\pi}{\mathrm{8}}} \right)\left({z}^{\mathrm{2}} −\left(\sqrt{\mathrm{2}}{e}^{\frac{{i}\mathrm{5}\pi}{\mathrm{8}}} \right)^{\mathrm{2}} \right) \\ $$$$\left({z}−\sqrt{\mathrm{2}}{e}^{\frac{{i}\pi}{\mathrm{8}}} \right)\left({z}+\sqrt{\mathrm{2}}{e}^{\frac{{i}\pi}{\mathrm{8}}} \right)\left({z}−\sqrt{\mathrm{2}}{e}^{\frac{{i}\mathrm{5}\pi}{\mathrm{8}}} \right)\left({z}+\sqrt{\mathrm{2}}{e}^{\frac{{i}\mathrm{5}\pi}{\mathrm{8}}} \right)\:\Rightarrow \\ $$$$\varphi\left({z}\right)\:=\frac{\mathrm{1}}{\left({z}−\sqrt{\mathrm{2}}{e}^{\frac{{i}\pi}{\mathrm{8}}} \right)^{\mathrm{3}} \left({z}+\sqrt{\mathrm{2}}{e}^{\frac{{i}\pi}{\mathrm{8}}} \right)^{\mathrm{3}} \left({z}−\sqrt{\mathrm{2}}{e}^{\frac{{i}\mathrm{5}\pi}{\mathrm{8}}} \right)^{\mathrm{3}} \left({z}+\sqrt{\mathrm{2}}{e}^{\frac{{i}\mathrm{5}\pi}{\mathrm{8}}} \right)^{\mathrm{3}} } \\ $$$${the}\:{poles}\:{of}\:\varphi\:{are}\:\overset{−} {+}\sqrt{\mathrm{2}}{e}^{\frac{{i}\pi}{\mathrm{8}}} \:\:{and}\:\overset{−} {+}\:{e}^{\frac{{i}\mathrm{5}\pi}{\mathrm{8}}} \:\:\:\left({ordre}\:\mathrm{3}\right) \\ $$$${residus}\:{theorem}\:{give}\:\int_{−\infty} ^{+\infty} \:\:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:\left\{{Res}\left(\varphi,\sqrt{\mathrm{2}}{e}^{\frac{{i}\pi}{\mathrm{8}}} \right)+{Res}\left(\varphi,\sqrt{\mathrm{2}}{e}^{\frac{{i}\mathrm{5}\pi}{\mathrm{8}}} \right)\right\} \\ $$$${Res}\left(\varphi,\sqrt{\mathrm{2}}{e}^{\frac{{i}\pi}{\mathrm{8}}} \right)\:={lim}_{{z}\rightarrow\sqrt{\mathrm{2}}{e}^{\frac{{i}\pi}{\mathrm{8}}} } \:\:\:\frac{\mathrm{1}}{\left(\mathrm{3}−\mathrm{1}\right)!}\left\{\:\left({z}−\sqrt{\mathrm{2}}{e}^{\frac{{i}\pi}{\mathrm{8}}} \right)^{\mathrm{3}} \varphi\left({z}\right)\right\}^{\left(\mathrm{2}\right)} \\ $$$$={lim}_{{z}\rightarrow\sqrt{\mathrm{2}}{e}^{\frac{{i}\pi}{\mathrm{8}}} } \:\:\:\frac{\mathrm{1}}{\mathrm{2}}\:\left\{\frac{\mathrm{1}}{\left({z}+\sqrt{\mathrm{2}}{e}^{\frac{{i}\pi}{\mathrm{8}}} \right)^{\mathrm{3}} \left({z}−\sqrt{\mathrm{2}}{e}^{\frac{{i}\mathrm{5}\pi}{\mathrm{8}}} \right)^{\mathrm{3}} \left({z}+\sqrt{\mathrm{2}}{e}^{\frac{{i}\mathrm{5}\pi}{\mathrm{8}}} \right)^{\mathrm{3}} }\right\}^{\left(\mathrm{2}\right)} \\ $$$$...{be}\:{continued}.... \\ $$

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