Menu Close

find-tbe-value-of-x-3-x-2-1-x-2-x-2-2-dx-

Tinku Tara 0 Comments
Pin




Question Number 57231 by maxmathsup by imad last updated on 31/Mar/19
find tbe value of ∫_(−∞) ^(+∞) ((x−3)/((x^2 +1)(x^2 −x +2)^2 )) dx
$${find}\:{tbe}\:{value}\:{of}\:\int_{−\infty} ^{+\infty} \:\:\frac{{x}−\mathrm{3}}{\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)\left({x}^{\mathrm{2}} −{x}\:+\mathrm{2}\right)^{\mathrm{2}} }\:{dx} \\ $$
Commented by maxmathsup by imad last updated on 02/Apr/19
residus method let ϕ(z)=((z−3)/((z^2 +1)(z^2 −z +2)^2 )) poles of ϕ? z^2 −z +2 =0 →Δ =1−4(2)=−7 =(i(√7))^2 ⇒z_1 =((1+i(√7))/2) and z_2 =((1−i(√7))/2) ⇒ϕ(z) =((z−3)/((z−i)(z+i)(z−z_1 )^2 (z−z_2 )^2 )) so the poles of ϕ are +^− i(simples) and ((1+^− (√7))/2)(doubles) residus theorem give ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ { Res(ϕ,i) +Res(ϕ,z_1 )} Res(ϕ,i) =lim_(z→i) (z−i)ϕ(z) =((i−3)/((2i)(i−z_1 )^2 (i−z_1 ^− )^2 )) =((i−3)/((2i)(−1−i(z_1 +z_1 ^− ))^2 )) =((i−3)/((2i)(1+i)^2 )) =((i−3)/((2i)(1+2i−1))) =((i−3)/(−4)) =((−i+3)/4) Res(ϕ,z_1 ) =lim_(z→z_1 ) (1/((2−1)!)){(z−z_1 )^2 ϕ(z)}^((1)) =lim_(z→z_1 ) {((z−3)/((z^2 +1)(z−z_2 )^2 ))}^((1)) =lim_(z→z_1 ) (((z^2 +1)(z−z_2 )^2 −(z−3){2z(z−z_2 )^2 +2(z−z_2 )(z^2 +1)})/((z^2 +1)^2 (z−z_2 )^4 )) =lim_(z→z_1 ) (((z^2 +1)(z−z_2 )−(z−3){ 2z(z−z_2 )+2z^2 +2})/((z^2 +1)^2 (z−z_2 )^3 )) =(((z_1 ^2 +1)(z_1 −z_2 ) −(z_1 −3){2z_1 (z_1 −z_2 )+2z_1 ^2 +2})/((z_1 ^2 +1)(z_1 −z_2 )^2 )) =(((1+z_1 ^2 )(i(√7)) −(z_1 −3){2z_1 (i(√7))+2z_1 ^2 +2})/((i(√7))^2 (1+z_1 ^2 ))) 1+z_1 ^2 =1+(1/4)(1+i(√7))^2 =1+(1/4)(1+2i(√7) −7) =1+(1/4)(−6+2i(√7)) =((−2+2i(√7))/4) =((−1+i(√7))/2) ....be continued...
$${residus}\:{method}\:{let}\:\varphi\left({z}\right)=\frac{{z}−\mathrm{3}}{\left({z}^{\mathrm{2}} \:+\mathrm{1}\right)\left({z}^{\mathrm{2}} −{z}\:+\mathrm{2}\right)^{\mathrm{2}} }\:\:{poles}\:{of}\:\varphi? \\ $$$${z}^{\mathrm{2}} −{z}\:+\mathrm{2}\:=\mathrm{0}\:\rightarrow\Delta\:=\mathrm{1}−\mathrm{4}\left(\mathrm{2}\right)=−\mathrm{7}\:=\left({i}\sqrt{\mathrm{7}}\right)^{\mathrm{2}} \:\Rightarrow{z}_{\mathrm{1}} =\frac{\mathrm{1}+{i}\sqrt{\mathrm{7}}}{\mathrm{2}} \\ $$$${and}\:{z}_{\mathrm{2}} =\frac{\mathrm{1}−{i}\sqrt{\mathrm{7}}}{\mathrm{2}}\:\Rightarrow\varphi\left({z}\right)\:=\frac{{z}−\mathrm{3}}{\left({z}−{i}\right)\left({z}+{i}\right)\left({z}−{z}_{\mathrm{1}} \right)^{\mathrm{2}} \left({z}−{z}_{\mathrm{2}} \right)^{\mathrm{2}} }\:{so}\:{the}\:{poles}\:{of}\:\varphi\:{are} \\ $$$$\overset{−} {+}{i}\left({simples}\right)\:{and}\:\frac{\mathrm{1}\overset{−} {+}\sqrt{\mathrm{7}}}{\mathrm{2}}\left({doubles}\right)\:\:{residus}\:{theorem}\:{give} \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:\left\{\:{Res}\left(\varphi,{i}\right)\:+{Res}\left(\varphi,{z}_{\mathrm{1}} \right)\right\} \\ $$$${Res}\left(\varphi,{i}\right)\:={lim}_{{z}\rightarrow{i}} \left({z}−{i}\right)\varphi\left({z}\right)\:=\frac{{i}−\mathrm{3}}{\left(\mathrm{2}{i}\right)\left({i}−{z}_{\mathrm{1}} \right)^{\mathrm{2}} \left({i}−\overset{−} {{z}}_{\mathrm{1}} \right)^{\mathrm{2}} } \\ $$$$=\frac{{i}−\mathrm{3}}{\left(\mathrm{2}{i}\right)\left(−\mathrm{1}−{i}\left({z}_{\mathrm{1}} +\overset{−} {{z}}_{\mathrm{1}} \right)\right)^{\mathrm{2}} }\:=\frac{{i}−\mathrm{3}}{\left(\mathrm{2}{i}\right)\left(\mathrm{1}+{i}\right)^{\mathrm{2}} }\:=\frac{{i}−\mathrm{3}}{\left(\mathrm{2}{i}\right)\left(\mathrm{1}+\mathrm{2}{i}−\mathrm{1}\right)}\:=\frac{{i}−\mathrm{3}}{−\mathrm{4}}\:=\frac{−{i}+\mathrm{3}}{\mathrm{4}} \\ $$$${Res}\left(\varphi,{z}_{\mathrm{1}} \right)\:={lim}_{{z}\rightarrow{z}_{\mathrm{1}} } \:\:\frac{\mathrm{1}}{\left(\mathrm{2}−\mathrm{1}\right)!}\left\{\left({z}−{z}_{\mathrm{1}} \right)^{\mathrm{2}} \varphi\left({z}\right)\right\}^{\left(\mathrm{1}\right)} \\ $$$$={lim}_{{z}\rightarrow{z}_{\mathrm{1}} } \:\:\:\left\{\frac{{z}−\mathrm{3}}{\left({z}^{\mathrm{2}} \:+\mathrm{1}\right)\left({z}−{z}_{\mathrm{2}} \right)^{\mathrm{2}} }\right\}^{\left(\mathrm{1}\right)} \\ $$$$={lim}_{{z}\rightarrow{z}_{\mathrm{1}} \:} \:\:\frac{\left({z}^{\mathrm{2}} \:+\mathrm{1}\right)\left({z}−{z}_{\mathrm{2}} \right)^{\mathrm{2}} \:−\left({z}−\mathrm{3}\right)\left\{\mathrm{2}{z}\left({z}−{z}_{\mathrm{2}} \right)^{\mathrm{2}} \:+\mathrm{2}\left({z}−{z}_{\mathrm{2}} \right)\left({z}^{\mathrm{2}} \:+\mathrm{1}\right)\right\}}{\left({z}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} \left({z}−{z}_{\mathrm{2}} \right)^{\mathrm{4}} } \\ $$$$={lim}_{{z}\rightarrow{z}_{\mathrm{1}} } \frac{\left({z}^{\mathrm{2}} \:+\mathrm{1}\right)\left({z}−{z}_{\mathrm{2}} \right)−\left({z}−\mathrm{3}\right)\left\{\:\mathrm{2}{z}\left({z}−{z}_{\mathrm{2}} \right)+\mathrm{2}{z}^{\mathrm{2}} \:+\mathrm{2}\right\}}{\left({z}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} \left({z}−{z}_{\mathrm{2}} \right)^{\mathrm{3}} } \\ $$$$=\frac{\left({z}_{\mathrm{1}} ^{\mathrm{2}} \:+\mathrm{1}\right)\left({z}_{\mathrm{1}} −{z}_{\mathrm{2}} \right)\:−\left({z}_{\mathrm{1}} −\mathrm{3}\right)\left\{\mathrm{2}{z}_{\mathrm{1}} \left({z}_{\mathrm{1}} −{z}_{\mathrm{2}} \right)+\mathrm{2}{z}_{\mathrm{1}} ^{\mathrm{2}} \:+\mathrm{2}\right\}}{\left({z}_{\mathrm{1}} ^{\mathrm{2}} \:+\mathrm{1}\right)\left({z}_{\mathrm{1}} −{z}_{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$$=\frac{\left(\mathrm{1}+{z}_{\mathrm{1}} ^{\mathrm{2}} \right)\left({i}\sqrt{\mathrm{7}}\right)\:−\left({z}_{\mathrm{1}} −\mathrm{3}\right)\left\{\mathrm{2}{z}_{\mathrm{1}} \left({i}\sqrt{\mathrm{7}}\right)+\mathrm{2}{z}_{\mathrm{1}} ^{\mathrm{2}} \:+\mathrm{2}\right\}}{\left({i}\sqrt{\mathrm{7}}\right)^{\mathrm{2}} \left(\mathrm{1}+{z}_{\mathrm{1}} ^{\mathrm{2}} \right)} \\ $$$$\mathrm{1}+{z}_{\mathrm{1}} ^{\mathrm{2}} \:=\mathrm{1}+\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{1}+{i}\sqrt{\mathrm{7}}\right)^{\mathrm{2}} \:=\mathrm{1}+\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{1}+\mathrm{2}{i}\sqrt{\mathrm{7}}\:−\mathrm{7}\right)\:=\mathrm{1}+\frac{\mathrm{1}}{\mathrm{4}}\left(−\mathrm{6}+\mathrm{2}{i}\sqrt{\mathrm{7}}\right) \\ $$$$=\frac{−\mathrm{2}+\mathrm{2}{i}\sqrt{\mathrm{7}}}{\mathrm{4}}\:=\frac{−\mathrm{1}+{i}\sqrt{\mathrm{7}}}{\mathrm{2}}\:….{be}\:{continued}… \\ $$$$ \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *