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Question Number 65130 by turbo msup by abdo last updated on 25/Jul/19

calculate ∫_0 ^∞ ((x^2 −3)/((x^4 +x^2 +2)^2 ))dx

$${calculate}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{x}^{\mathrm{2}} −\mathrm{3}}{\left({x}^{\mathrm{4}} +{x}^{\mathrm{2}} +\mathrm{2}\right)^{\mathrm{2}} }{dx} \\ $$

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

let I =∫_0 ^∞ ((x^2 −3)/((x^4 +x^2 +2)^2 )) ⇒2I =∫_(−∞) ^(+∞) ((x^2 −3)/((x^4 +x^2 +2)^2 ))dx let ϕ(z)=((z^2 −3)/((z^4 +z^2 +2)^2 )) poles of ϕ? z^4 +z^2 +2 =0 ⇒t^2 +t +2 =0 (t=z^2 ) Δ =1−8 =(i(√7))^2 ⇒t_1 =((−1+i(√7))/2) and t_2 =((−1−i(√7))/2) ∣t_1 ∣ =(√((1/4)+(7/4)))=(√2) ⇒t_1 =(√2){((−1)/(2(√2))) +((i(√7))/(2(√2)))} =(√2)e^(iarctan(−(√7))) t_2 =conj(t_1 ) =(√2)e^(iarctan((√7))) ⇒t^2 +t+2 =(t−(√2)e^(−iarctan((√7))) )(t−(√2)e^(iarctan((√7))) ) =(z^2 −(√2)e^(−iarctan((√7))) )(z^2 −(√2)e^(iarctan((√7))) ) =(z−(√(√2))e^(−(i/2)arctan((√7))) )(z+(√(√2))e^(−(i/2)arctan((√7))) )(z−(√(√2))e^((i/2)arctan((√7))) )(z+(√(√2))e^((i/2)arctan((√7))) ) ⇒ ϕ(z) =((z^2 −3)/((z−re^(−(i/2)arctan((√7))) )^2 (z+r e^(−(i/2)arctan((√7))) )^2 (z−re^((i/2)arctan((√7))) )^2 (z+r e^((i/2)arctan((√7))) )^2 )) the poles of ϕ are +^− r e^(−(i/2)arctan((√7))) and +^− r e^((i/2)arctan((√7))) with r =(√(√2)) residus theorem give ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ { Res(ϕ,r e^((i/2)arctan((√7))) ) +Res(ϕ,−r e^(−(i/2)arctan((√7))) } be continued....

$${let}\:{I}\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{x}^{\mathrm{2}} −\mathrm{3}}{\left({x}^{\mathrm{4}} \:+{x}^{\mathrm{2}} \:+\mathrm{2}\right)^{\mathrm{2}} }\:\Rightarrow\mathrm{2}{I}\:=\int_{−\infty} ^{+\infty} \:\:\frac{{x}^{\mathrm{2}} −\mathrm{3}}{\left({x}^{\mathrm{4}} \:+{x}^{\mathrm{2}} \:+\mathrm{2}\right)^{\mathrm{2}} }{dx}\:{let} \\ $$$$\varphi\left({z}\right)=\frac{{z}^{\mathrm{2}} −\mathrm{3}}{\left({z}^{\mathrm{4}} \:+{z}^{\mathrm{2}} \:+\mathrm{2}\right)^{\mathrm{2}} }\:\:\:{poles}\:{of}\:\varphi? \\ $$$${z}^{\mathrm{4}} \:+{z}^{\mathrm{2}} \:+\mathrm{2}\:=\mathrm{0}\:\Rightarrow{t}^{\mathrm{2}} \:+{t}\:+\mathrm{2}\:=\mathrm{0}\:\:\:\:\left({t}={z}^{\mathrm{2}} \right) \\ $$$$\Delta\:=\mathrm{1}−\mathrm{8}\:=\left({i}\sqrt{\mathrm{7}}\right)^{\mathrm{2}} \:\Rightarrow{t}_{\mathrm{1}} =\frac{−\mathrm{1}+{i}\sqrt{\mathrm{7}}}{\mathrm{2}}\:\:{and}\:{t}_{\mathrm{2}} =\frac{−\mathrm{1}−\boldsymbol{{i}}\sqrt{\mathrm{7}}}{\mathrm{2}} \\ $$$$\mid{t}_{\mathrm{1}} \mid\:=\sqrt{\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{7}}{\mathrm{4}}}=\sqrt{\mathrm{2}}\:\Rightarrow{t}_{\mathrm{1}} =\sqrt{\mathrm{2}}\left\{\frac{−\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\:+\frac{{i}\sqrt{\mathrm{7}}}{\mathrm{2}\sqrt{\mathrm{2}}}\right\}\:=\sqrt{\mathrm{2}}{e}^{{iarctan}\left(−\sqrt{\mathrm{7}}\right)} \\ $$$${t}_{\mathrm{2}} ={conj}\left({t}_{\mathrm{1}} \right)\:=\sqrt{\mathrm{2}}{e}^{{iarctan}\left(\sqrt{\mathrm{7}}\right)} \:\:\Rightarrow{t}^{\mathrm{2}} +{t}+\mathrm{2}\:=\left({t}−\sqrt{\mathrm{2}}{e}^{−{iarctan}\left(\sqrt{\mathrm{7}}\right)} \right)\left({t}−\sqrt{\mathrm{2}}{e}^{{iarctan}\left(\sqrt{\mathrm{7}}\right)} \right) \\ $$$$=\left({z}^{\mathrm{2}} −\sqrt{\mathrm{2}}{e}^{−{iarctan}\left(\sqrt{\mathrm{7}}\right)} \right)\left({z}^{\mathrm{2}} −\sqrt{\mathrm{2}}{e}^{{iarctan}\left(\sqrt{\mathrm{7}}\right)} \right) \\ $$$$=\left({z}−\sqrt{\sqrt{\mathrm{2}}}{e}^{−\frac{{i}}{\mathrm{2}}{arctan}\left(\sqrt{\mathrm{7}}\right)} \right)\left({z}+\sqrt{\sqrt{\mathrm{2}}}{e}^{−\frac{{i}}{\mathrm{2}}{arctan}\left(\sqrt{\mathrm{7}}\right)} \right)\left({z}−\sqrt{\sqrt{\mathrm{2}}}{e}^{\frac{{i}}{\mathrm{2}}{arctan}\left(\sqrt{\mathrm{7}}\right)} \right)\left({z}+\sqrt{\sqrt{\mathrm{2}}}{e}^{\frac{{i}}{\mathrm{2}}{arctan}\left(\sqrt{\mathrm{7}}\right)} \right)\:\Rightarrow \\ $$$$\varphi\left({z}\right)\:=\frac{{z}^{\mathrm{2}} −\mathrm{3}}{\left({z}−{re}^{−\frac{{i}}{\mathrm{2}}{arctan}\left(\sqrt{\mathrm{7}}\right)} \right)^{\mathrm{2}} \left({z}+{r}\:{e}^{−\frac{{i}}{\mathrm{2}}{arctan}\left(\sqrt{\mathrm{7}}\right)} \right)^{\mathrm{2}} \left({z}−{re}^{\frac{{i}}{\mathrm{2}}{arctan}\left(\sqrt{\mathrm{7}}\right)} \right)^{\mathrm{2}} \left({z}+{r}\:{e}^{\frac{{i}}{\mathrm{2}}{arctan}\left(\sqrt{\mathrm{7}}\right)} \right)^{\mathrm{2}} } \\ $$$${the}\:{poles}\:{of}\:\varphi\:{are}\:\overset{−} {+}{r}\:{e}^{−\frac{{i}}{\mathrm{2}}{arctan}\left(\sqrt{\mathrm{7}}\right)} \:\:{and}\:\overset{−} {+}{r}\:{e}^{\frac{{i}}{\mathrm{2}}{arctan}\left(\sqrt{\mathrm{7}}\right)} \\ $$$${with}\:{r}\:=\sqrt{\sqrt{\mathrm{2}}}\:\:\:\:{residus}\:{theorem}\:{give} \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:\left\{\:{Res}\left(\varphi,{r}\:{e}^{\frac{{i}}{\mathrm{2}}{arctan}\left(\sqrt{\mathrm{7}}\right)} \right)\:+{Res}\left(\varphi,−{r}\:{e}^{−\frac{{i}}{\mathrm{2}}{arctan}\left(\sqrt{\mathrm{7}}\right)} \right\}\right. \\ $$$${be}\:{continued}.... \\ $$

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

Res(ϕ,r e^((i/2)arctan((√7))) ) =lim_(z→r e^((i/2)arctan((√7))) ) (1/((2−1)!)) {(z−r e^((i/2)arctan((√7))) )^2 ϕ(z)}^((1)) =lim_(z→re^((i/2)arctan((√7))) ) {((z^2 −3)/((z+r e^((i/2)arctan((√7))) )^2 (z^2 −r^2 e^(−i arctan((√7))) )^2 ))}^((1)) =lim_(z→r e^((i/2)arctan((√7))) ) ((2z D−(z^2 −3)D^′ )/D^2 ) D^′ =2(z+r e^((i/2)arctan((√7))) )(z^2 −r^2 e^(−iarctan((√7))) )^2 +4z (z^2 −r^2 e^(−iarctan((√7))) )(z+r e^((i/2)arctan((√7))) )^2 =....

$${Res}\left(\varphi,{r}\:{e}^{\frac{{i}}{\mathrm{2}}{arctan}\left(\sqrt{\mathrm{7}}\right)} \right)\:={lim}_{{z}\rightarrow{r}\:{e}^{\frac{{i}}{\mathrm{2}}{arctan}\left(\sqrt{\mathrm{7}}\right)} } \:\frac{\mathrm{1}}{\left(\mathrm{2}−\mathrm{1}\right)!}\:\:\left\{\left({z}−{r}\:{e}^{\frac{{i}}{\mathrm{2}}{arctan}\left(\sqrt{\mathrm{7}}\right)} \right)^{\mathrm{2}} \varphi\left({z}\right)\right\}^{\left(\mathrm{1}\right)} \\ $$$$={lim}_{{z}\rightarrow{re}^{\frac{{i}}{\mathrm{2}}{arctan}\left(\sqrt{\mathrm{7}}\right)} } \:\:\:\:\left\{\frac{{z}^{\mathrm{2}} −\mathrm{3}}{\left({z}+{r}\:{e}^{\frac{{i}}{\mathrm{2}}{arctan}\left(\sqrt{\mathrm{7}}\right)} \right)^{\mathrm{2}} \left({z}^{\mathrm{2}} −{r}^{\mathrm{2}} \:{e}^{−{i}\:{arctan}\left(\sqrt{\mathrm{7}}\right)} \right)^{\mathrm{2}} }\right\}^{\left(\mathrm{1}\right)} \\ $$$$={lim}_{{z}\rightarrow{r}\:{e}^{\frac{{i}}{\mathrm{2}}{arctan}\left(\sqrt{\mathrm{7}}\right)} \:\:\:} \:\:\:\frac{\mathrm{2}{z}\:{D}−\left({z}^{\mathrm{2}} −\mathrm{3}\right){D}^{'} }{{D}^{\mathrm{2}} } \\ $$$${D}^{'} \:=\mathrm{2}\left({z}+{r}\:{e}^{\frac{{i}}{\mathrm{2}}{arctan}\left(\sqrt{\mathrm{7}}\right)} \right)\left({z}^{\mathrm{2}} −{r}^{\mathrm{2}} {e}^{−{iarctan}\left(\sqrt{\mathrm{7}}\right)} \right)^{\mathrm{2}} \\ $$$$+\mathrm{4}{z}\:\left({z}^{\mathrm{2}} −{r}^{\mathrm{2}} {e}^{−{iarctan}\left(\sqrt{\mathrm{7}}\right)} \right)\left({z}+{r}\:{e}^{\frac{{i}}{\mathrm{2}}{arctan}\left(\sqrt{\mathrm{7}}\right)} \right)^{\mathrm{2}} \:=.... \\ $$

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