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Question Number 62805 by mathmax by abdo last updated on 25/Jun/19
calculate ∫_0 ^(+∞) ((3x^2 −2)/((x^2 +1)( x^2 −2i)^2 )) dx
$${calculate}\:\int_{\mathrm{0}} ^{+\infty} \:\:\frac{\mathrm{3}{x}^{\mathrm{2}} −\mathrm{2}}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)\left(\:{x}^{\mathrm{2}} −\mathrm{2}{i}\right)^{\mathrm{2}} }\:{dx} \\ $$
Commented by mathmax by abdo last updated on 26/Jun/19
let I =∫_0 ^∞ ((3x^2 −2)/((x^2 +1)(x^2 −2i)^2 )) dx⇒2I =∫_(−∞) ^(+∞) ((3x^2 −2)/((x^2 +1)(x^2 −2i)^2 ))dx let w(z) =((3z^2 −2)/((z^2 +1)(z^2 −2i)^2 )) ⇒w(z) =((3z^2 −2)/((z−i)(z+i)(z−(√(2i)))^2 (z+(√(2i)))^2 )) =((3z^2 −2)/((z−i)(z+i)(z−(√2)e^((iπ)/4) )^2 (z+(√2)e^((iπ)/4) )^2 )) so the poles of w are +^− i and +^− (√2)e^((iπ)/4) residus theorem give ∫_(−∞) ^(+∞) w(z)dz =2iπ { Res(w,i)+Res(w,(√2)e^((iπ)/4) )} Res(w,i) =lim_(z→i) (z−i)w(z) =((−5)/(2i(−1−2i)^2 )) =((−5)/(2i(2i+1)^2 )) =((−5)/(2i(−4+4i +1))) =((−5)/(2i(−3+4i))) Res(w,(√2)e^((iπ)/4) ) =lim_(z→(√2)e^((iπ)/4) ) (1/((2−1)!)){ (z−(√2)e^((iπ)/4) )^2 w(z)}^((1)) =lim_(z→(√2)e^((iπ)/4) ) {((3z^2 −2)/((z^2 +1)(z+(√2)e^((iπ)/4) )^2 ))}^((1)) =lim_(z→(√2)e^((iπ)/4) ) {((6z(z^2 +1)(z+(√2)e^((iπ)/4) )^2 −(3z^2 −2){2z(z+(√2)e^((iπ)/4) }^2 +2(z^2 +1)(z+(√2)e^((iπ)/4) ))/((z^2 +1)^2 (z+(√2)e^((iπ)/4) )^4 ))} =lim_(z→(√2)e^((iπ)/4) ) {((6z(z^2 +1)(z+(√2)e^((iπ)/4) )−(3z^2 −2){2z(z+(√2)e^((iπ)/4) )+2(z^2 +1)})/((z^2 +1)^2 (z+(√2)e^((iπ)/4) )^3 ))} ...be continued...
$${let}\:{I}\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{3}{x}^{\mathrm{2}} −\mathrm{2}}{\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)\left({x}^{\mathrm{2}} −\mathrm{2}{i}\right)^{\mathrm{2}} }\:{dx}\Rightarrow\mathrm{2}{I}\:=\int_{−\infty} ^{+\infty} \:\:\frac{\mathrm{3}{x}^{\mathrm{2}} −\mathrm{2}}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)\left({x}^{\mathrm{2}} −\mathrm{2}{i}\right)^{\mathrm{2}} }{dx}\:{let} \\ $$$${w}\left({z}\right)\:=\frac{\mathrm{3}{z}^{\mathrm{2}} −\mathrm{2}}{\left({z}^{\mathrm{2}} \:+\mathrm{1}\right)\left({z}^{\mathrm{2}} −\mathrm{2}{i}\right)^{\mathrm{2}} }\:\Rightarrow{w}\left({z}\right)\:=\frac{\mathrm{3}{z}^{\mathrm{2}} −\mathrm{2}}{\left({z}−{i}\right)\left({z}+{i}\right)\left({z}−\sqrt{\mathrm{2}{i}}\right)^{\mathrm{2}} \left({z}+\sqrt{\mathrm{2}{i}}\right)^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{3}{z}^{\mathrm{2}} −\mathrm{2}}{\left({z}−{i}\right)\left({z}+{i}\right)\left({z}−\sqrt{\mathrm{2}}{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)^{\mathrm{2}} \left({z}+\sqrt{\mathrm{2}}{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)^{\mathrm{2}} }\:\:{so}\:{the}\:{poles}\:{of}\:{w}\:{are}\:\overset{−} {+}{i}\:{and}\:\overset{−} {+}\sqrt{\mathrm{2}}{e}^{\frac{{i}\pi}{\mathrm{4}}} \\ $$$${residus}\:{theorem}\:{give}\:\int_{−\infty} ^{+\infty} \:{w}\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:\left\{\:{Res}\left({w},{i}\right)+{Res}\left({w},\sqrt{\mathrm{2}}{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)\right\} \\ $$$${Res}\left({w},{i}\right)\:={lim}_{{z}\rightarrow{i}} \left({z}−{i}\right){w}\left({z}\right)\:=\frac{−\mathrm{5}}{\mathrm{2}{i}\left(−\mathrm{1}−\mathrm{2}{i}\right)^{\mathrm{2}} }\:=\frac{−\mathrm{5}}{\mathrm{2}{i}\left(\mathrm{2}{i}+\mathrm{1}\right)^{\mathrm{2}} }\:=\frac{−\mathrm{5}}{\mathrm{2}{i}\left(−\mathrm{4}+\mathrm{4}{i}\:+\mathrm{1}\right)} \\ $$$$=\frac{−\mathrm{5}}{\mathrm{2}{i}\left(−\mathrm{3}+\mathrm{4}{i}\right)} \\ $$$${Res}\left({w},\sqrt{\mathrm{2}}{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)\:={lim}_{{z}\rightarrow\sqrt{\mathrm{2}}{e}^{\frac{{i}\pi}{\mathrm{4}}} } \:\:\:\frac{\mathrm{1}}{\left(\mathrm{2}−\mathrm{1}\right)!}\left\{\:\left({z}−\sqrt{\mathrm{2}}{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)^{\mathrm{2}} {w}\left({z}\right)\right\}^{\left(\mathrm{1}\right)} \\ $$$$={lim}_{{z}\rightarrow\sqrt{\mathrm{2}}{e}^{\frac{{i}\pi}{\mathrm{4}}} } \:\:\:\:\:\left\{\frac{\mathrm{3}{z}^{\mathrm{2}} −\mathrm{2}}{\left({z}^{\mathrm{2}} +\mathrm{1}\right)\left({z}+\sqrt{\mathrm{2}}{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)^{\mathrm{2}} }\right\}^{\left(\mathrm{1}\right)} \\ $$$$={lim}_{{z}\rightarrow\sqrt{\mathrm{2}}{e}^{\frac{{i}\pi}{\mathrm{4}}} } \:\:\:\:\left\{\frac{\mathrm{6}{z}\left({z}^{\mathrm{2}} +\mathrm{1}\right)\left({z}+\sqrt{\mathrm{2}}{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)^{\mathrm{2}} −\left(\mathrm{3}{z}^{\mathrm{2}} −\mathrm{2}\right)\left\{\mathrm{2}{z}\left({z}+\sqrt{\mathrm{2}}{e}^{\frac{{i}\pi}{\mathrm{4}}} \right\}^{\mathrm{2}} +\mathrm{2}\left({z}^{\mathrm{2}} +\mathrm{1}\right)\left({z}+\sqrt{\mathrm{2}}{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)\right.}{\left({z}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} \left({z}+\sqrt{\mathrm{2}}{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)^{\mathrm{4}} }\right\} \\ $$$$={lim}_{{z}\rightarrow\sqrt{\mathrm{2}}{e}^{\frac{{i}\pi}{\mathrm{4}}} } \:\:\:\left\{\frac{\mathrm{6}{z}\left({z}^{\mathrm{2}} \:+\mathrm{1}\right)\left({z}+\sqrt{\mathrm{2}}{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)−\left(\mathrm{3}{z}^{\mathrm{2}} −\mathrm{2}\right)\left\{\mathrm{2}{z}\left({z}+\sqrt{\mathrm{2}}{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)+\mathrm{2}\left({z}^{\mathrm{2}} +\mathrm{1}\right)\right\}}{\left({z}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} \left({z}+\sqrt{\mathrm{2}}{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)^{\mathrm{3}} }\right\} \\ $$$$…{be}\:{continued}… \\ $$

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