Question Number 76359 by mathmax by abdo last updated on 26/Dec/19
$${calculate}\:\int_{−\infty} ^{\infty} \:\:\frac{{dx}}{\left({x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{7}\right)^{\mathrm{2}} } \\ $$
Commented by mathmax by abdo last updated on 28/Dec/19
![let I =∫_(−∞) ^(+∞) (dx/((x^2 −3x+7)^2 )) and ϕ(z)=(1/((z^2 −3z+7)^2 )) ]poles of ϕ? z^2 −3z+7 =0 →Δ=9−4.7 =9−28 =−19 z_1 =((3+i(√(19)))/2) and z_2 =((3−i(√(19)))/2) ϕ(z) =(1/((z−z_1 )^2 (z−z_2 )^2 )) residus theorem give ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ Res(ϕ,z_1 ) Res(ϕ,z_1 ) =lim_(z→z_1 ) (1/((2−1)!)){(z−z_1 )^2 ϕ(z)}^((1)) =lim_(z→z_1 ) {(z−z_2 )^(−2) }^((1)) =lim_(z→z_1 ) −2(z−z_2 )^(−3) =−2(z_1 −z_2 )^(−3 ) =−2(i(√(19)))^(−3) =((−2)/((i(√(19)))^3 )) =((−2)/(−i(19)(√(19)))) =(2/(19i(√(19)))) ⇒∫_(−∞) ^(+∞) ϕ(z)dz =2iπ×(2/(19i(√(19)))) =((4π)/(19(√(19)))) =I](https://www.tinkutara.com/question/Q76613.png)
$$\left.{let}\:{I}\:=\int_{−\infty} ^{+\infty} \:\frac{{dx}}{\left({x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{7}\right)^{\mathrm{2}} }\:\:{and}\:\varphi\left({z}\right)=\frac{\mathrm{1}}{\left({z}^{\mathrm{2}} −\mathrm{3}{z}+\mathrm{7}\right)^{\mathrm{2}} }\:\right]{poles}\:{of}\:\varphi? \\ $$$${z}^{\mathrm{2}} −\mathrm{3}{z}+\mathrm{7}\:=\mathrm{0}\:\rightarrow\Delta=\mathrm{9}−\mathrm{4}.\mathrm{7}\:=\mathrm{9}−\mathrm{28}\:=−\mathrm{19} \\ $$$${z}_{\mathrm{1}} =\frac{\mathrm{3}+{i}\sqrt{\mathrm{19}}}{\mathrm{2}}\:{and}\:{z}_{\mathrm{2}} =\frac{\mathrm{3}−{i}\sqrt{\mathrm{19}}}{\mathrm{2}} \\ $$$$\varphi\left({z}\right)\:=\frac{\mathrm{1}}{\left({z}−{z}_{\mathrm{1}} \right)^{\mathrm{2}} \left({z}−{z}_{\mathrm{2}} \right)^{\mathrm{2}} }\:{residus}\:{theorem}\:{give} \\ $$$$\int_{−\infty} ^{+\infty} \varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{Res}\left(\varphi,{z}_{\mathrm{1}} \right) \\ $$$${Res}\left(\varphi,{z}_{\mathrm{1}} \right)\:={lim}_{{z}\rightarrow{z}_{\mathrm{1}} } \frac{\mathrm{1}}{\left(\mathrm{2}−\mathrm{1}\right)!}\left\{\left({z}−{z}_{\mathrm{1}} \right)^{\mathrm{2}} \varphi\left({z}\right)\right\}^{\left(\mathrm{1}\right)} \\ $$$$={lim}_{{z}\rightarrow{z}_{\mathrm{1}} \:} \left\{\left({z}−{z}_{\mathrm{2}} \right)^{−\mathrm{2}} \right\}^{\left(\mathrm{1}\right)} \:={lim}_{{z}\rightarrow{z}_{\mathrm{1}} } −\mathrm{2}\left({z}−{z}_{\mathrm{2}} \right)^{−\mathrm{3}} \\ $$$$=−\mathrm{2}\left({z}_{\mathrm{1}} −{z}_{\mathrm{2}} \right)^{−\mathrm{3}\:} =−\mathrm{2}\left({i}\sqrt{\mathrm{19}}\right)^{−\mathrm{3}} \:=\frac{−\mathrm{2}}{\left({i}\sqrt{\mathrm{19}}\right)^{\mathrm{3}} }\:=\frac{−\mathrm{2}}{−{i}\left(\mathrm{19}\right)\sqrt{\mathrm{19}}}\:=\frac{\mathrm{2}}{\mathrm{19}{i}\sqrt{\mathrm{19}}} \\ $$$$\Rightarrow\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi×\frac{\mathrm{2}}{\mathrm{19}{i}\sqrt{\mathrm{19}}}\:=\frac{\mathrm{4}\pi}{\mathrm{19}\sqrt{\mathrm{19}}}\:={I} \\ $$$$ \\ $$