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Question Number 69563 by mathmax by abdo last updated on 25/Sep/19
calculate ∫_0 ^∞ (dx/(x^4 −x^2 +1))
$${calculate}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dx}}{{x}^{\mathrm{4}} −{x}^{\mathrm{2}} \:+\mathrm{1}} \\ $$
Commented by mathmax by abdo last updated on 25/Sep/19
let I =∫_0 ^∞ (dx/(x^4 −x^2 +1)) ⇒ 2I =∫_(−∞) ^(+∞) (dx/(x^4 −x^2 +1)) let ϕ(z) =(1/(z^4 −z^2 +1)) poles of ϕ? z^4 −z^2 +1 =0 →t^2 −t +1 =0 with t=z^2 Δ =1−4=−3 =(i(√3))^2 ⇒t_1 =((1+i(√3))/2) and t_2 =((1−i(√3))/2) ∣t_1 ∣ =1 ⇒t_1 =e^(iarctan((√3))) =e^((iπ)/3) and t_2 =e^(−((iπ)/3)) ⇒ ϕ(z) =(1/((z^2 −e^((iπ)/3) )(z^2 −e^(−((iπ)/3)) ))) =(1/((z−e^((iπ)/6) )(z+e^((iπ)/6) )(z−e^(−((iπ)/6)) )(z+e^(−((iπ)/6)) ))) so the poles of ϕ are +^− e^((iπ)/6) and +^− e^(−((iπ)/6)) ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ { Res(ϕ,e^((iπ)/6) )+Res(ϕ,−e^(−((iπ)/6)) )} Res(ϕ,e^((iπ)/6) ) =lim_(z→e^((iπ)/6) ) (z−e^((iπ)/6) )ϕ(z) =(1/(2e^((iπ)/6) (e^((iπ)/3) −e^(−((iπ)/3)) ))) =(e^(−((iπ)/6)) /(2(2isin((π/3))))) =(e^((−iπ)/6) /(4i((√3)/2))) = (e^(−((iπ)/6)) /(2i(√3))) Res(ϕ,−e^(−((iπ)/6)) ) =lim_(z→−e^(−((iπ)/6)) ) (z+e^((iπ)/6) )ϕ(z) =(1/((−2i e^(−((iπ)/6)) )(e^(−i(π/3)) −e^((iπ)/3) ))) =(e^((iπ)/6) /(4i sin((π/3)))) =(e^((iπ)/6) /(4i((√3)/2))) =(e^((iπ)/6) /(2i(√3))) ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ{(e^(−((iπ)/6)) /(2i(√3))) +(e^((iπ)/6) /(2i(√3)))} =(π/( (√3))){2cos((π/6))} =((2π)/( (√3)))×((√3)/2) =π =2I ⇒ I =(π/2)
$${let}\:{I}\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dx}}{{x}^{\mathrm{4}} −{x}^{\mathrm{2}} \:+\mathrm{1}}\:\Rightarrow\:\mathrm{2}{I}\:=\int_{−\infty} ^{+\infty} \:\frac{{dx}}{{x}^{\mathrm{4}} −{x}^{\mathrm{2}} \:+\mathrm{1}} \\ $$$${let}\:\varphi\left({z}\right)\:=\frac{\mathrm{1}}{{z}^{\mathrm{4}} −{z}^{\mathrm{2}} \:+\mathrm{1}}\:\:{poles}\:{of}\:\varphi? \\ $$$${z}^{\mathrm{4}} −{z}^{\mathrm{2}} \:+\mathrm{1}\:=\mathrm{0}\:\rightarrow{t}^{\mathrm{2}} −{t}\:+\mathrm{1}\:=\mathrm{0}\:\:{with}\:{t}={z}^{\mathrm{2}} \\ $$$$\Delta\:=\mathrm{1}−\mathrm{4}=−\mathrm{3}\:=\left({i}\sqrt{\mathrm{3}}\right)^{\mathrm{2}} \:\Rightarrow{t}_{\mathrm{1}} =\frac{\mathrm{1}+{i}\sqrt{\mathrm{3}}}{\mathrm{2}}\:\:{and}\:{t}_{\mathrm{2}} =\frac{\mathrm{1}−{i}\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$\mid{t}_{\mathrm{1}} \mid\:=\mathrm{1}\:\Rightarrow{t}_{\mathrm{1}} ={e}^{{iarctan}\left(\sqrt{\mathrm{3}}\right)} \:={e}^{\frac{{i}\pi}{\mathrm{3}}} \:{and}\:{t}_{\mathrm{2}} ={e}^{−\frac{{i}\pi}{\mathrm{3}}} \:\Rightarrow \\ $$$$\varphi\left({z}\right)\:=\frac{\mathrm{1}}{\left({z}^{\mathrm{2}} −{e}^{\frac{{i}\pi}{\mathrm{3}}} \right)\left({z}^{\mathrm{2}} −{e}^{−\frac{{i}\pi}{\mathrm{3}}} \right)}\:=\frac{\mathrm{1}}{\left({z}−{e}^{\frac{{i}\pi}{\mathrm{6}}} \right)\left({z}+{e}^{\frac{{i}\pi}{\mathrm{6}}} \right)\left({z}−{e}^{−\frac{{i}\pi}{\mathrm{6}}} \right)\left({z}+{e}^{−\frac{{i}\pi}{\mathrm{6}}} \right)} \\ $$$${so}\:{the}\:{poles}\:{of}\:\varphi\:{are}\:\overset{−} {+}{e}^{\frac{{i}\pi}{\mathrm{6}}} \:{and}\:\overset{−} {+}{e}^{−\frac{{i}\pi}{\mathrm{6}}} \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:\left\{\:{Res}\left(\varphi,{e}^{\frac{{i}\pi}{\mathrm{6}}} \right)+{Res}\left(\varphi,−{e}^{−\frac{{i}\pi}{\mathrm{6}}} \right)\right\} \\ $$$${Res}\left(\varphi,{e}^{\frac{{i}\pi}{\mathrm{6}}} \right)\:={lim}_{{z}\rightarrow{e}^{\frac{{i}\pi}{\mathrm{6}}} } \:\:\left({z}−{e}^{\frac{{i}\pi}{\mathrm{6}}} \right)\varphi\left({z}\right)\:=\frac{\mathrm{1}}{\mathrm{2}{e}^{\frac{{i}\pi}{\mathrm{6}}} \left({e}^{\frac{{i}\pi}{\mathrm{3}}} \:−{e}^{−\frac{{i}\pi}{\mathrm{3}}} \right)} \\ $$$$=\frac{{e}^{−\frac{{i}\pi}{\mathrm{6}}} }{\mathrm{2}\left(\mathrm{2}{isin}\left(\frac{\pi}{\mathrm{3}}\right)\right)}\:=\frac{{e}^{\frac{−{i}\pi}{\mathrm{6}}} }{\mathrm{4}{i}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}\:=\:\frac{{e}^{−\frac{{i}\pi}{\mathrm{6}}} }{\mathrm{2}{i}\sqrt{\mathrm{3}}} \\ $$$${Res}\left(\varphi,−{e}^{−\frac{{i}\pi}{\mathrm{6}}} \right)\:={lim}_{{z}\rightarrow−{e}^{−\frac{{i}\pi}{\mathrm{6}}} } \:\:\:\left({z}+{e}^{\frac{{i}\pi}{\mathrm{6}}} \right)\varphi\left({z}\right) \\ $$$$=\frac{\mathrm{1}}{\left(−\mathrm{2}{i}\:{e}^{−\frac{{i}\pi}{\mathrm{6}}} \right)\left({e}^{−{i}\frac{\pi}{\mathrm{3}}} −{e}^{\frac{{i}\pi}{\mathrm{3}}} \right)}\:=\frac{{e}^{\frac{{i}\pi}{\mathrm{6}}} }{\mathrm{4}{i}\:{sin}\left(\frac{\pi}{\mathrm{3}}\right)}\:=\frac{{e}^{\frac{{i}\pi}{\mathrm{6}}} }{\mathrm{4}{i}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}\:=\frac{{e}^{\frac{{i}\pi}{\mathrm{6}}} }{\mathrm{2}{i}\sqrt{\mathrm{3}}}\:\Rightarrow \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\left\{\frac{{e}^{−\frac{{i}\pi}{\mathrm{6}}} }{\mathrm{2}{i}\sqrt{\mathrm{3}}}\:+\frac{{e}^{\frac{{i}\pi}{\mathrm{6}}} }{\mathrm{2}{i}\sqrt{\mathrm{3}}}\right\} \\ $$$$=\frac{\pi}{\:\sqrt{\mathrm{3}}}\left\{\mathrm{2}{cos}\left(\frac{\pi}{\mathrm{6}}\right)\right\}\:=\frac{\mathrm{2}\pi}{\:\sqrt{\mathrm{3}}}×\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:=\pi\:=\mathrm{2}{I}\:\Rightarrow\:{I}\:=\frac{\pi}{\mathrm{2}} \\ $$

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