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Question Number 63023 by mathmax by abdo last updated on 27/Jun/19

calculate ∫_(−∞) ^(+∞) ((x^2 −3)/(x^4 +x^2 +1))dx .

$${calculate}\:\int_{−\infty} ^{+\infty} \:\:\:\frac{{x}^{\mathrm{2}} −\mathrm{3}}{{x}^{\mathrm{4}} \:+{x}^{\mathrm{2}} \:+\mathrm{1}}{dx}\:. \\ $$

Commented by mathmax by abdo last updated on 28/Jun/19

let A =∫_(−∞) ^(+∞) ((x^2 −3)/(x^4 +x^2 +1))dx let W(z) =((z^2 −3)/(z^4 +z^2 +1)) poles of W? z^4 +z^2 +1 =0 →t^2 +t +1 =0( with z^2 =t) → Δ =1−4 =(i(√3))^2 ⇒t_1 =((−1+i(√3))/2) and t_2 =((−1−i(√3))/2) ⇒ t_1 =e^((i2π)/3) and t_2 =e^(−((i2π)/3)) ⇒z^4 +z^2 +1 =(t−t_1 )(t−t_2 ) =(z^2 −e^((i2π)/3) )(z^2 −e^(−((i2π)/3)) ) =(z−e^(i(π/3)) )(z+e^((iπ)/3) )(z−e^(−((iπ)/3)) )(z +e^((iπ)/3) )⇒W(z) =((z^2 −3)/((z−e^((iπ)/3) )(z+e^((iπ)/3) )(z−e^(−((iπ)/3)) )(z+e^(−((iπ)/3)) ))) =((z^2 −3)/((z^2 −e^((i2π)/3) )(z^2 −e^(−((i2π)/3)) ))) the poles of W are +^− e^((iπ)/3) and +^− e^(−((iπ)/3)) residus theorem give ∫_(−∞) ^(+∞) W(z)dz =2iπ { Res(W,e^((iπ)/3) ) +Res(W,−e^(−((iπ)/3)) )} Res(W,e^((iπ)/3) ) =lim_(z→e^((iπ)/3) ) (z−e^((iπ)/3) )W(z) =((e^(i((2π)/3)) −3)/(2e^((iπ)/3) (e^((2iπ)/3) −e^(−((i2π)/3)) ))) =(1/2) e^(−((iπ)/3)) ((e^(i((2π)/3)) −3)/(2isin(((2π)/3)))) =(1/(4i((√3)/2))) (e^((iπ)/3) −3 e^(−((iπ)/3)) ) =(1/(2i(√3))){ e^((iπ)/3) −3 e^(−((iπ)/3)) } Res(W,−e^(−((iπ)/3)) ) =lim_(z→−e^(−((iπ)/3)) ) (z+e^(−((iπ)/3)) )W(z) =((e^(−((2iπ)/3)) −3)/((e^(−((2iπ)/3)) −e^((i2π)/3) )(−2e^(−((iπ)/3)) ))) =(e^((iπ)/3) /(2(2i sin(((2π)/3))))){ e^(−((2iπ)/3)) −3} =(1/(4i ((√3)/2))){ e^(−((iπ)/3)) −3 e^((iπ)/3) } =(1/(2i(√3))){ e^(−((iπ)/3)) −3 e^((iπ)/3) } ⇒ ∫_(−∞) ^(+∞) W(z)dz =2iπ (1/(2i(√3))){ e^((iπ)/3) −3 e^(−((iπ)/3)) +e^(−((iπ)/3)) −3 e^((iπ)/3) } =(π/(√3)){ −2 e^((iπ)/3) −2 e^(−((iπ)/3)) } =((−2π)/(√3)){ e^((iπ)/3) +e^(−((iπ)/3)) } =((−2π)/(√3)){ 2cos((π/3))} =((−2π)/(√3))2(1/2) =((−2π)/(√3)) ⇒ ∫_(−∞) ^(+∞) ((x^2 −3)/(x^4 +x^2 +1))dx =((−2π)/(√3)) .

$${let}\:{A}\:=\int_{−\infty} ^{+\infty} \:\frac{{x}^{\mathrm{2}} −\mathrm{3}}{{x}^{\mathrm{4}} \:+{x}^{\mathrm{2}} \:+\mathrm{1}}{dx}\:{let}\:{W}\left({z}\right)\:=\frac{{z}^{\mathrm{2}} −\mathrm{3}}{{z}^{\mathrm{4}} \:+{z}^{\mathrm{2}} \:+\mathrm{1}}\:\:{poles}\:{of}\:{W}? \\ $$$${z}^{\mathrm{4}} \:+{z}^{\mathrm{2}} \:+\mathrm{1}\:=\mathrm{0}\:\rightarrow{t}^{\mathrm{2}} \:+{t}\:+\mathrm{1}\:=\mathrm{0}\left(\:{with}\:{z}^{\mathrm{2}} \:={t}\right)\:\rightarrow \\ $$$$\Delta\:=\mathrm{1}−\mathrm{4}\:=\left({i}\sqrt{\mathrm{3}}\right)^{\mathrm{2}} \:\Rightarrow{t}_{\mathrm{1}} =\frac{−\mathrm{1}+{i}\sqrt{\mathrm{3}}}{\mathrm{2}}\:\:{and}\:{t}_{\mathrm{2}} =\frac{−\mathrm{1}−{i}\sqrt{\mathrm{3}}}{\mathrm{2}}\:\Rightarrow \\ $$$${t}_{\mathrm{1}} ={e}^{\frac{{i}\mathrm{2}\pi}{\mathrm{3}}} \:{and}\:{t}_{\mathrm{2}} ={e}^{−\frac{{i}\mathrm{2}\pi}{\mathrm{3}}} \:\Rightarrow{z}^{\mathrm{4}} \:+{z}^{\mathrm{2}} \:+\mathrm{1}\:=\left({t}−{t}_{\mathrm{1}} \right)\left({t}−{t}_{\mathrm{2}} \right)\:=\left({z}^{\mathrm{2}} −{e}^{\frac{{i}\mathrm{2}\pi}{\mathrm{3}}} \right)\left({z}^{\mathrm{2}} −{e}^{−\frac{{i}\mathrm{2}\pi}{\mathrm{3}}} \right) \\ $$$$=\left({z}−{e}^{{i}\frac{\pi}{\mathrm{3}}} \right)\left({z}+{e}^{\frac{{i}\pi}{\mathrm{3}}} \right)\left({z}−{e}^{−\frac{{i}\pi}{\mathrm{3}}} \right)\left({z}\:+{e}^{\frac{{i}\pi}{\mathrm{3}}} \right)\Rightarrow{W}\left({z}\right)\:=\frac{{z}^{\mathrm{2}} −\mathrm{3}}{\left({z}−{e}^{\frac{{i}\pi}{\mathrm{3}}} \right)\left({z}+{e}^{\frac{{i}\pi}{\mathrm{3}}} \right)\left({z}−{e}^{−\frac{{i}\pi}{\mathrm{3}}} \right)\left({z}+{e}^{−\frac{{i}\pi}{\mathrm{3}}} \right)} \\ $$$$=\frac{{z}^{\mathrm{2}} −\mathrm{3}}{\left({z}^{\mathrm{2}} −{e}^{\frac{{i}\mathrm{2}\pi}{\mathrm{3}}} \right)\left({z}^{\mathrm{2}} −{e}^{−\frac{{i}\mathrm{2}\pi}{\mathrm{3}}} \right)} \\ $$$${the}\:{poles}\:{of}\:\:{W}\:{are}\:\overset{−} {+}\:{e}^{\frac{{i}\pi}{\mathrm{3}}} \:{and}\:\:\overset{−} {+}\:{e}^{−\frac{{i}\pi}{\mathrm{3}}} \:\:\:\:{residus}\:{theorem}\:{give} \\ $$$$\int_{−\infty} ^{+\infty} \:{W}\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:\left\{\:{Res}\left({W},{e}^{\frac{{i}\pi}{\mathrm{3}}} \right)\:+{Res}\left({W},−{e}^{−\frac{{i}\pi}{\mathrm{3}}} \right)\right\} \\ $$$${Res}\left({W},{e}^{\frac{{i}\pi}{\mathrm{3}}} \right)\:={lim}_{{z}\rightarrow{e}^{\frac{{i}\pi}{\mathrm{3}}} } \:\:\left({z}−{e}^{\frac{{i}\pi}{\mathrm{3}}} \right){W}\left({z}\right)\:=\frac{{e}^{{i}\frac{\mathrm{2}\pi}{\mathrm{3}}} −\mathrm{3}}{\mathrm{2}{e}^{\frac{{i}\pi}{\mathrm{3}}} \left({e}^{\frac{\mathrm{2}{i}\pi}{\mathrm{3}}} \:−{e}^{−\frac{{i}\mathrm{2}\pi}{\mathrm{3}}} \right)}\:=\frac{\mathrm{1}}{\mathrm{2}}\:{e}^{−\frac{{i}\pi}{\mathrm{3}}} \:\:\frac{{e}^{{i}\frac{\mathrm{2}\pi}{\mathrm{3}}} −\mathrm{3}}{\mathrm{2}{isin}\left(\frac{\mathrm{2}\pi}{\mathrm{3}}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}{i}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}\:\left({e}^{\frac{{i}\pi}{\mathrm{3}}} \:−\mathrm{3}\:{e}^{−\frac{{i}\pi}{\mathrm{3}}} \right)\:=\frac{\mathrm{1}}{\mathrm{2}{i}\sqrt{\mathrm{3}}}\left\{\:{e}^{\frac{{i}\pi}{\mathrm{3}}} \:−\mathrm{3}\:{e}^{−\frac{{i}\pi}{\mathrm{3}}} \right\} \\ $$$${Res}\left({W},−{e}^{−\frac{{i}\pi}{\mathrm{3}}} \right)\:={lim}_{{z}\rightarrow−{e}^{−\frac{{i}\pi}{\mathrm{3}}} } \:\:\:\:\left({z}+{e}^{−\frac{{i}\pi}{\mathrm{3}}} \right){W}\left({z}\right)\:=\frac{{e}^{−\frac{\mathrm{2}{i}\pi}{\mathrm{3}}} −\mathrm{3}}{\left({e}^{−\frac{\mathrm{2}{i}\pi}{\mathrm{3}}} \:−{e}^{\frac{{i}\mathrm{2}\pi}{\mathrm{3}}} \right)\left(−\mathrm{2}{e}^{−\frac{{i}\pi}{\mathrm{3}}} \right)} \\ $$$$=\frac{{e}^{\frac{{i}\pi}{\mathrm{3}}} }{\mathrm{2}\left(\mathrm{2}{i}\:{sin}\left(\frac{\mathrm{2}\pi}{\mathrm{3}}\right)\right)}\left\{\:{e}^{−\frac{\mathrm{2}{i}\pi}{\mathrm{3}}} −\mathrm{3}\right\}\:=\frac{\mathrm{1}}{\mathrm{4}{i}\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}\left\{\:{e}^{−\frac{{i}\pi}{\mathrm{3}}} \:−\mathrm{3}\:{e}^{\frac{{i}\pi}{\mathrm{3}}} \right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{i}\sqrt{\mathrm{3}}}\left\{\:\:{e}^{−\frac{{i}\pi}{\mathrm{3}}} \:−\mathrm{3}\:{e}^{\frac{{i}\pi}{\mathrm{3}}} \right\}\:\Rightarrow \\ $$$$\int_{−\infty} ^{+\infty} \:{W}\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:\frac{\mathrm{1}}{\mathrm{2}{i}\sqrt{\mathrm{3}}}\left\{\:\:{e}^{\frac{{i}\pi}{\mathrm{3}}} \:−\mathrm{3}\:{e}^{−\frac{{i}\pi}{\mathrm{3}}} \:+{e}^{−\frac{{i}\pi}{\mathrm{3}}} \:−\mathrm{3}\:{e}^{\frac{{i}\pi}{\mathrm{3}}} \right\} \\ $$$$=\frac{\pi}{\sqrt{\mathrm{3}}}\left\{\:−\mathrm{2}\:{e}^{\frac{{i}\pi}{\mathrm{3}}} −\mathrm{2}\:{e}^{−\frac{{i}\pi}{\mathrm{3}}} \right\}\:=\frac{−\mathrm{2}\pi}{\sqrt{\mathrm{3}}}\left\{\:\:{e}^{\frac{{i}\pi}{\mathrm{3}}} \:+{e}^{−\frac{{i}\pi}{\mathrm{3}}} \right\}\:=\frac{−\mathrm{2}\pi}{\sqrt{\mathrm{3}}}\left\{\:\mathrm{2}{cos}\left(\frac{\pi}{\mathrm{3}}\right)\right\}\:=\frac{−\mathrm{2}\pi}{\sqrt{\mathrm{3}}}\mathrm{2}\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$=\frac{−\mathrm{2}\pi}{\sqrt{\mathrm{3}}}\:\Rightarrow\:\int_{−\infty} ^{+\infty} \:\:\frac{{x}^{\mathrm{2}} −\mathrm{3}}{{x}^{\mathrm{4}} \:+{x}^{\mathrm{2}} \:+\mathrm{1}}{dx}\:=\frac{−\mathrm{2}\pi}{\sqrt{\mathrm{3}}}\:. \\ $$$$ \\ $$

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