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Question Number 57233 by maxmathsup by imad last updated on 31/Mar/19

find the value of ∫_0 ^(+∞) (x^4 /((1+x^2 +x^4 )^2 ))dx

$${find}\:{the}\:{value}\:{of}\:\int_{\mathrm{0}} ^{+\infty} \:\:\frac{{x}^{\mathrm{4}} }{\left(\mathrm{1}+{x}^{\mathrm{2}} \:+{x}^{\mathrm{4}} \right)^{\mathrm{2}} }{dx} \\ $$

Commented by maxmathsup by imad last updated on 04/Apr/19

let I =∫_0 ^∞ (x^4 /((1+x^2 +x^4 )^2 )) dx ⇒2I =∫_(−∞) ^(+∞) (x^4 /((x^4 +x^2 +1)^2 )) dx let ϕ(z) =(z^4 /((z^4 +z^2 +1)^2 )) .poles of ϕ? z^4 +z^2 +1 =0 ⇒t^2 +t +1 =0 (witht=z^2 ) →Δ=1−4=−3 ⇒ t_1 =((−1+i(√3))/2) =e^((i2π)/3) and t_2 =((−1−i(√3))/2) =e^(−((i2π)/3)) z^2 =t_1 ⇒z =+^− e^((iπ)/3) z^2 =t_2 ⇒z =+^− e^(−((iπ)/3)) ⇒the poles of ϕ are +^− e^((iπ)/3) and +^− e^(−((iπ)/3)) (doubles) and ϕ(z) =(z^4 /((z−e^((iπ)/3) )^2 (z+e^((iπ)/3) )^2 (z−e^(−((iπ)/3)) )^2 (z+e^(−((iπ)/3)) )^2 )) residus theorem give ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ { Res(ϕ,e^((iπ)/3) ) +Res(ϕ,−e^(−((iπ)/3)) )} Res(ϕ,e^((iπ)/3) ) =lim_(z→e^((iπ)/3) ) (1/((2−1)!)) {(z−e^((iπ)/3) )^2 ϕ(z)}^((1)) =lim_(z→e^((iπ)/3) ) { (z^4 /((z+e^((iπ)/3) )^2 (z−e^(−((iπ)/3)) )^2 (z+e^(−((iπ)/3)) )^2 ))}^((1)) =lim_(z→e^((iπ)/3) ) ((4z^3 (z+e^((iπ)/3) )^2 (z−e^(−((iπ)/3)) )^2 (z+e^(−((iπ)/3)) )^2 −z^4 {(z+e^((iπ)/3) )^2 (z−e^(−((iπ)/3)) )^2 (z+e^(−((iπ)/3)) )^2 }^((1)) )/((z+e^((iπ)/3) )^4 (z−e^(−((iπ)/3)) )^4 (z+e^(−((iπ)/3)) )^4 )) ....be continued...

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

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