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Question Number 56931 by turbo msup by abdo last updated on 26/Mar/19
calculate ∫_(−∞) ^(+∞) ((x^2 −1)/((x^2 −x+3)^2 ))dx
$${calculate}\:\int_{−\infty} ^{+\infty} \:\:\frac{{x}^{\mathrm{2}} −\mathrm{1}}{\left({x}^{\mathrm{2}} −{x}+\mathrm{3}\right)^{\mathrm{2}} }{dx} \\ $$
Commented by maxmathsup by imad last updated on 27/Mar/19
residus method let ϕ(z) =((z^2 −1)/((z^2 −z +3)^2 )) let determine the poles of ϕ? z^2 −z+3=0 →Δ =1−12 =−11=(i(√(11)))^2 ⇒z_1 =((1+i(√(11)))/2) z_2 =((1−i(√(11)))/2) the poles of ϕ are z_1 and z_2 (doubles) and ϕ(z)=((z^2 −1)/((z−z_1 )^2 (z−z_2 )^2 )) residus theorem give ∫_(−∞) ^(+∞) ϕ(z)dz =2iπRes(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^2 −1)/((z−z_2 )^2 ))}^((1)) =lim_(z→z_1 ) {((2z(z−z_2 )^2 −2(z−z_2 )(z^2 −1))/((z−z_2 )^4 ))} =lim_(z→z_1 ) ((2z(z−z_2 )−2(z^2 −1))/((z−z_2 )^3 )) =((2z_1 (z_1 −z_2 )−2(z_1 ^2 −1))/((z_1 −z_2 )^3 )) =((2i(√(11))(((1+i(√(11)))/2))−2( (1/4)(1+2i(√(11))−11)−1))/((i(√(11)))^3 )) =((i(√(11))(1+i(√(11)))−2{((i(√(11)))/2)−(5/2)−1})/(−11i(√(11)))) =((i(√(11))−11−i(√(11))+5 +2)/(−11i(√(11)))) =((−4)/(−11i(√(11)))) =(4/(11i(√(11)))) ⇒∫_(−∞) ^(+∞) ϕ(z)dz =2iπ (4/(11i(√(11)))) =((8π)/(11(√(11)))) =((8π(√(11)))/(121)) ★∫_(−∞) ^(+∞) ((x^2 −1)/((x^2 −x+3)^2 )) dx =((8π(√(11)))/(121)) ★
$${residus}\:{method}\:\:{let}\:\varphi\left({z}\right)\:=\frac{{z}^{\mathrm{2}} −\mathrm{1}}{\left({z}^{\mathrm{2}} −{z}\:+\mathrm{3}\right)^{\mathrm{2}} }\:\:{let}\:{determine}\:{the}\:{poles}\:{of}\:\varphi? \\ $$$${z}^{\mathrm{2}} −{z}+\mathrm{3}=\mathrm{0}\:\rightarrow\Delta\:=\mathrm{1}−\mathrm{12}\:=−\mathrm{11}=\left({i}\sqrt{\mathrm{11}}\right)^{\mathrm{2}} \:\Rightarrow{z}_{\mathrm{1}} =\frac{\mathrm{1}+{i}\sqrt{\mathrm{11}}}{\mathrm{2}} \\ $$$${z}_{\mathrm{2}} =\frac{\mathrm{1}−{i}\sqrt{\mathrm{11}}}{\mathrm{2}}\:\:{the}\:{poles}\:{of}\:\varphi\:{are}\:{z}_{\mathrm{1}} {and}\:{z}_{\mathrm{2}} \left({doubles}\right)\:{and}\:\varphi\left({z}\right)=\frac{{z}^{\mathrm{2}} −\mathrm{1}}{\left({z}−{z}_{\mathrm{1}} \right)^{\mathrm{2}} \left({z}−{z}_{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$${residus}\:{theorem}\:{give}\:\:\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi{Res}\left(\varphi,{z}_{\mathrm{1}} \right) \\ $$$${Res}\left(\varphi,{z}_{\mathrm{1}} \right)\:=\:{lim}_{{z}\rightarrow{z}_{\mathrm{1}} } \:\:\:\frac{\mathrm{1}}{\left(\mathrm{2}−\mathrm{1}\right)!}\:\left\{\left({z}−{z}_{\mathrm{1}} \right)^{\mathrm{2}} \varphi\left({z}\right)\right\}^{\left(\mathrm{1}\right)} \\ $$$$={lim}_{{z}\rightarrow{z}_{\mathrm{1}} } \:\:\:\left\{\frac{{z}^{\mathrm{2}} −\mathrm{1}}{\left({z}−{z}_{\mathrm{2}} \right)^{\mathrm{2}} }\right\}^{\left(\mathrm{1}\right)} \:={lim}_{{z}\rightarrow{z}_{\mathrm{1}} } \:\:\:\left\{\frac{\mathrm{2}{z}\left({z}−{z}_{\mathrm{2}} \right)^{\mathrm{2}} −\mathrm{2}\left({z}−{z}_{\mathrm{2}} \right)\left({z}^{\mathrm{2}} −\mathrm{1}\right)}{\left({z}−{z}_{\mathrm{2}} \right)^{\mathrm{4}} }\right\} \\ $$$$={lim}_{{z}\rightarrow{z}_{\mathrm{1}} } \:\:\:\frac{\mathrm{2}{z}\left({z}−{z}_{\mathrm{2}} \right)−\mathrm{2}\left({z}^{\mathrm{2}} −\mathrm{1}\right)}{\left({z}−{z}_{\mathrm{2}} \right)^{\mathrm{3}} }\:=\frac{\mathrm{2}{z}_{\mathrm{1}} \left({z}_{\mathrm{1}} −{z}_{\mathrm{2}} \right)−\mathrm{2}\left({z}_{\mathrm{1}} ^{\mathrm{2}} −\mathrm{1}\right)}{\left({z}_{\mathrm{1}} −{z}_{\mathrm{2}} \right)^{\mathrm{3}} } \\ $$$$=\frac{\mathrm{2}{i}\sqrt{\mathrm{11}}\left(\frac{\mathrm{1}+{i}\sqrt{\mathrm{11}}}{\mathrm{2}}\right)−\mathrm{2}\left(\:\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{1}+\mathrm{2}{i}\sqrt{\mathrm{11}}−\mathrm{11}\right)−\mathrm{1}\right)}{\left({i}\sqrt{\mathrm{11}}\right)^{\mathrm{3}} } \\ $$$$=\frac{{i}\sqrt{\mathrm{11}}\left(\mathrm{1}+{i}\sqrt{\mathrm{11}}\right)−\mathrm{2}\left\{\frac{{i}\sqrt{\mathrm{11}}}{\mathrm{2}}−\frac{\mathrm{5}}{\mathrm{2}}−\mathrm{1}\right\}}{−\mathrm{11}{i}\sqrt{\mathrm{11}}}\:=\frac{{i}\sqrt{\mathrm{11}}−\mathrm{11}−{i}\sqrt{\mathrm{11}}+\mathrm{5}\:+\mathrm{2}}{−\mathrm{11}{i}\sqrt{\mathrm{11}}} \\ $$$$=\frac{−\mathrm{4}}{−\mathrm{11}{i}\sqrt{\mathrm{11}}}\:=\frac{\mathrm{4}}{\mathrm{11}{i}\sqrt{\mathrm{11}}}\:\Rightarrow\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:\:\frac{\mathrm{4}}{\mathrm{11}{i}\sqrt{\mathrm{11}}}\:=\frac{\mathrm{8}\pi}{\mathrm{11}\sqrt{\mathrm{11}}}\:=\frac{\mathrm{8}\pi\sqrt{\mathrm{11}}}{\mathrm{121}} \\ $$$$\bigstar\int_{−\infty} ^{+\infty} \:\:\frac{{x}^{\mathrm{2}} −\mathrm{1}}{\left({x}^{\mathrm{2}} −{x}+\mathrm{3}\right)^{\mathrm{2}} }\:{dx}\:=\frac{\mathrm{8}\pi\sqrt{\mathrm{11}}}{\mathrm{121}}\:\bigstar \\ $$$$ \\ $$
Answered by MJS last updated on 27/Mar/19
Ostrogradski′s Method once again ∫((p(x))/(q(x)))dx=((p_1 (x))/(q_1 (x)))+∫((p_2 (x))/(q_2 (x)))dx q_1 (x)=gcd(q(x), q′(x)) q_2 (x)=((q(x))/(q_1 (x))) p_1 (x), p_2 (x) can be found like this: ((p(x))/(q(x)))=(d/dx)[((p_1 (x))/(q_1 (x)))]+((p_2 (x))/(q_2 (x))) in our case: ∫((x^2 −1)/((x^2 −x+3)^2 ))dx=−((7x+2)/(11(x^2 −x+3)))+(4/(11))∫(dx/(x^2 −x+3))= =−((7x+2)/(11(x^2 −x+3)))+((8(√(11)))/(121))arctan (((√(11))/(11))(2x−1)) +C ∫_(−∞) ^(+∞) ((x^2 −1)/((x^2 −x+3)^2 ))dx=((8(√(11)))/(121))π
$$\mathrm{Ostrogradski}'\mathrm{s}\:\mathrm{Method}\:\mathrm{once}\:\mathrm{again} \\ $$$$\int\frac{{p}\left({x}\right)}{{q}\left({x}\right)}{dx}=\frac{{p}_{\mathrm{1}} \left({x}\right)}{{q}_{\mathrm{1}} \left({x}\right)}+\int\frac{{p}_{\mathrm{2}} \left({x}\right)}{{q}_{\mathrm{2}} \left({x}\right)}{dx} \\ $$$${q}_{\mathrm{1}} \left({x}\right)=\mathrm{gcd}\left({q}\left({x}\right),\:{q}'\left({x}\right)\right) \\ $$$${q}_{\mathrm{2}} \left({x}\right)=\frac{{q}\left({x}\right)}{{q}_{\mathrm{1}} \left({x}\right)} \\ $$$${p}_{\mathrm{1}} \left({x}\right),\:{p}_{\mathrm{2}} \left({x}\right)\:\mathrm{can}\:\mathrm{be}\:\mathrm{found}\:\mathrm{like}\:\mathrm{this}: \\ $$$$\frac{{p}\left({x}\right)}{{q}\left({x}\right)}=\frac{{d}}{{dx}}\left[\frac{{p}_{\mathrm{1}} \left({x}\right)}{{q}_{\mathrm{1}} \left({x}\right)}\right]+\frac{{p}_{\mathrm{2}} \left({x}\right)}{{q}_{\mathrm{2}} \left({x}\right)} \\ $$$$\mathrm{in}\:\mathrm{our}\:\mathrm{case}: \\ $$$$\int\frac{{x}^{\mathrm{2}} −\mathrm{1}}{\left({x}^{\mathrm{2}} −{x}+\mathrm{3}\right)^{\mathrm{2}} }{dx}=−\frac{\mathrm{7}{x}+\mathrm{2}}{\mathrm{11}\left({x}^{\mathrm{2}} −{x}+\mathrm{3}\right)}+\frac{\mathrm{4}}{\mathrm{11}}\int\frac{{dx}}{{x}^{\mathrm{2}} −{x}+\mathrm{3}}= \\ $$$$=−\frac{\mathrm{7}{x}+\mathrm{2}}{\mathrm{11}\left({x}^{\mathrm{2}} −{x}+\mathrm{3}\right)}+\frac{\mathrm{8}\sqrt{\mathrm{11}}}{\mathrm{121}}\mathrm{arctan}\:\left(\frac{\sqrt{\mathrm{11}}}{\mathrm{11}}\left(\mathrm{2}{x}−\mathrm{1}\right)\right)\:+{C} \\ $$$$\underset{−\infty} {\overset{+\infty} {\int}}\frac{{x}^{\mathrm{2}} −\mathrm{1}}{\left({x}^{\mathrm{2}} −{x}+\mathrm{3}\right)^{\mathrm{2}} }{dx}=\frac{\mathrm{8}\sqrt{\mathrm{11}}}{\mathrm{121}}\pi \\ $$
Commented by turbo msup by abdo last updated on 27/Mar/19
thanks sir mjs.
$${thanks}\:{sir}\:{mjs}. \\ $$

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