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Question Number 56827 by maxmathsup by imad last updated on 24/Mar/19

find the value of ∫_(−∞) ^(+∞) (dx/(x^4 −x^2 +3))

$${find}\:{the}\:{value}\:{of}\:\int_{−\infty} ^{+\infty} \:\:\:\frac{{dx}}{{x}^{\mathrm{4}} −{x}^{\mathrm{2}} +\mathrm{3}} \\ $$

Commented by maxmathsup by imad last updated on 25/Mar/19

residus method let ϕ(z)=(1/(z^4 −z^2 +3)) poles of ϕ z^4 −z^2 +3 =0 ⇒t^2 −t+3=0 with t=z^2 Δ =1−12 =−11 ⇒t_1 =((1+i(√(11)))/2) and t_2 =((1−i(√(11)))/2) ⇒ϕ(z)=(1/((z^2 −t_1 )(z^2 −t_2 ))) = (1/((z−(√t_1 ))(z+(√t_1 ))(z−(√t_2 ))(z+(√t_2 )))) we have ∣t_1 ∣ =(√((1/4)+((11)/4)))=(√3) ⇒t_1 =(√3)((1/(2(√3))) +((i(√(11)))/(2(√3))))=r e^(iθ) ⇒cosθ =(1/(2(√3))) and sinθ=((√(11))/(2(√3))) ⇒tanθ =(√(11)) ⇒θ=arctan((√(11))) ⇒t_1 =(√3)e^(iarctan((√(11)))) ⇒ϕ(z) =(1/((z−(√(√3)) e^((i/2)arctan((√(11)))) )(z+(√(√3))e^((i/2)arctan((√(11)))) )(z−(√(√3))e^(−(i/2)arctan((√(11)))) )(z+(√(√3))e^(−(i/2)arctan((√(11)))) ))) residus theorem give ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ{ Res(ϕ,(√(√3))e^((i/2)arctan((√(11)))) +Res(ϕ,−(√(√3))e^(−(i/2)arctan((√(11)))) )} Res(ϕ,(√(√3))e^((i/2)arctan((√(11)))) )= (1/(2(√(√3))e^((i/2)arctan((√(11)))) 2i(√3) sin(((arctan((√(11))))/2))2cos(((arctan((√(11))))/2)))) = (e^(−(i/2)arctan((√(11)))) /(4i(√3)(√(√3))sin(arctan((√(11)))))) Res(ϕ,−(√(√3))e^(−(i/2)arctan((√(11)))) ) =(1/(−2(√(√3))e^(−(i/2)arctan((√(11)))) (−(√(√3)))2cos(((arctan((√(11))))/2))2i(√(√3))sin(((arctan((√(11))))/2)))) = (e^((i/2)arctan((√(11)))) /(4i(√3)(√(√3))sin(arctan((√(11)))))) ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz =((2iπ)/(4i(√3)(√(√3))sin(arctan((√(11)))))) { e^((i/2)arctan((√(11)))) +e^(−(i/2)arctan((√(11)))) } =(π/(2(√3)(√(√3))sin(arctan((√(11))))) .2cos(((arctan((√(11))))/2)) ⇒ I =(π/((√3)(√(√3)))) ((cos(((arctan((√(11))))/2)))/(2sin(((arctan((√(11))))/2))cos(((arctan((√(11))))/2)))) I =(π/(2(√3)(√(√3))sin(((arctan((√(11))))/2)))) .

$${residus}\:{method}\:{let}\:\varphi\left({z}\right)=\frac{\mathrm{1}}{{z}^{\mathrm{4}} −{z}^{\mathrm{2}} \:+\mathrm{3}}\:\:{poles}\:{of}\:\varphi \\ $$$${z}^{\mathrm{4}} −{z}^{\mathrm{2}} \:+\mathrm{3}\:=\mathrm{0}\:\:\Rightarrow{t}^{\mathrm{2}} −{t}+\mathrm{3}=\mathrm{0}\:{with}\:{t}={z}^{\mathrm{2}} \\ $$$$\Delta\:=\mathrm{1}−\mathrm{12}\:=−\mathrm{11}\:\Rightarrow{t}_{\mathrm{1}} =\frac{\mathrm{1}+{i}\sqrt{\mathrm{11}}}{\mathrm{2}}\:\:{and}\:{t}_{\mathrm{2}} =\frac{\mathrm{1}−{i}\sqrt{\mathrm{11}}}{\mathrm{2}}\:\Rightarrow\varphi\left({z}\right)=\frac{\mathrm{1}}{\left({z}^{\mathrm{2}} −{t}_{\mathrm{1}} \right)\left({z}^{\mathrm{2}} −{t}_{\mathrm{2}} \right)} \\ $$$$=\:\frac{\mathrm{1}}{\left({z}−\sqrt{{t}_{\mathrm{1}} }\right)\left({z}+\sqrt{{t}_{\mathrm{1}} }\right)\left({z}−\sqrt{{t}_{\mathrm{2}} }\right)\left({z}+\sqrt{{t}_{\mathrm{2}} }\right)}\:\:{we}\:{have} \\ $$$$\mid{t}_{\mathrm{1}} \mid\:=\sqrt{\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{11}}{\mathrm{4}}}=\sqrt{\mathrm{3}}\:\Rightarrow{t}_{\mathrm{1}} =\sqrt{\mathrm{3}}\left(\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}}\:+\frac{{i}\sqrt{\mathrm{11}}}{\mathrm{2}\sqrt{\mathrm{3}}}\right)={r}\:{e}^{{i}\theta} \:\Rightarrow{cos}\theta\:=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}}\:{and}\:{sin}\theta=\frac{\sqrt{\mathrm{11}}}{\mathrm{2}\sqrt{\mathrm{3}}} \\ $$$$\Rightarrow{tan}\theta\:=\sqrt{\mathrm{11}}\:\Rightarrow\theta={arctan}\left(\sqrt{\mathrm{11}}\right)\:\Rightarrow{t}_{\mathrm{1}} =\sqrt{\mathrm{3}}{e}^{{iarctan}\left(\sqrt{\mathrm{11}}\right)} \\ $$$$\Rightarrow\varphi\left({z}\right)\:=\frac{\mathrm{1}}{\left({z}−\sqrt{\sqrt{\mathrm{3}}}\:{e}^{\frac{{i}}{\mathrm{2}}{arctan}\left(\sqrt{\mathrm{11}}\right)} \right)\left({z}+\sqrt{\sqrt{\mathrm{3}}}{e}^{\frac{{i}}{\mathrm{2}}{arctan}\left(\sqrt{\mathrm{11}}\right)} \right)\left({z}−\sqrt{\sqrt{\mathrm{3}}}{e}^{−\frac{{i}}{\mathrm{2}}{arctan}\left(\sqrt{\mathrm{11}}\right)} \right)\left({z}+\sqrt{\sqrt{\mathrm{3}}}{e}^{−\frac{{i}}{\mathrm{2}}{arctan}\left(\sqrt{\mathrm{11}}\right)} \right)} \\ $$$${residus}\:{theorem}\:{give}\:\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\left\{\:{Res}\left(\varphi,\sqrt{\sqrt{\mathrm{3}}}{e}^{\frac{{i}}{\mathrm{2}}{arctan}\left(\sqrt{\mathrm{11}}\right)} +{Res}\left(\varphi,−\sqrt{\sqrt{\mathrm{3}}}{e}^{−\frac{{i}}{\mathrm{2}}{arctan}\left(\sqrt{\mathrm{11}}\right)} \right)\right\}\right. \\ $$$${Res}\left(\varphi,\sqrt{\sqrt{\mathrm{3}}}{e}^{\frac{{i}}{\mathrm{2}}{arctan}\left(\sqrt{\mathrm{11}}\right)} \right)=\:\frac{\mathrm{1}}{\mathrm{2}\sqrt{\sqrt{\mathrm{3}}}{e}^{\frac{{i}}{\mathrm{2}}{arctan}\left(\sqrt{\mathrm{11}}\right)} \mathrm{2}{i}\sqrt{\mathrm{3}}\:{sin}\left(\frac{{arctan}\left(\sqrt{\mathrm{11}}\right)}{\mathrm{2}}\right)\mathrm{2}{cos}\left(\frac{{arctan}\left(\sqrt{\mathrm{11}}\right)}{\mathrm{2}}\right)} \\ $$$$=\:\frac{{e}^{−\frac{{i}}{\mathrm{2}}{arctan}\left(\sqrt{\mathrm{11}}\right)} }{\mathrm{4}{i}\sqrt{\mathrm{3}}\sqrt{\sqrt{\mathrm{3}}}{sin}\left({arctan}\left(\sqrt{\mathrm{11}}\right)\right)} \\ $$$${Res}\left(\varphi,−\sqrt{\sqrt{\mathrm{3}}}{e}^{−\frac{{i}}{\mathrm{2}}{arctan}\left(\sqrt{\mathrm{11}}\right)} \right)\:=\frac{\mathrm{1}}{−\mathrm{2}\sqrt{\sqrt{\mathrm{3}}}{e}^{−\frac{{i}}{\mathrm{2}}{arctan}\left(\sqrt{\mathrm{11}}\right)} \left(−\sqrt{\sqrt{\mathrm{3}}}\right)\mathrm{2}{cos}\left(\frac{{arctan}\left(\sqrt{\mathrm{11}}\right)}{\mathrm{2}}\right)\mathrm{2}{i}\sqrt{\sqrt{\mathrm{3}}}{sin}\left(\frac{{arctan}\left(\sqrt{\mathrm{11}}\right)}{\mathrm{2}}\right)} \\ $$$$=\:\frac{{e}^{\frac{{i}}{\mathrm{2}}{arctan}\left(\sqrt{\mathrm{11}}\right)} }{\mathrm{4}{i}\sqrt{\mathrm{3}}\sqrt{\sqrt{\mathrm{3}}}{sin}\left({arctan}\left(\sqrt{\mathrm{11}}\right)\right)}\:\Rightarrow \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\frac{\mathrm{2}{i}\pi}{\mathrm{4}{i}\sqrt{\mathrm{3}}\sqrt{\sqrt{\mathrm{3}}}{sin}\left({arctan}\left(\sqrt{\mathrm{11}}\right)\right)}\:\left\{\:{e}^{\frac{{i}}{\mathrm{2}}{arctan}\left(\sqrt{\mathrm{11}}\right)} \:+{e}^{−\frac{{i}}{\mathrm{2}}{arctan}\left(\sqrt{\left.\mathrm{11}\right)}\right.} \right\} \\ $$$$=\frac{\pi}{\mathrm{2}\sqrt{\mathrm{3}}\sqrt{\sqrt{\mathrm{3}}}{sin}\left({arctan}\left(\sqrt{\mathrm{11}}\right)\right.}\:.\mathrm{2}{cos}\left(\frac{{arctan}\left(\sqrt{\mathrm{11}}\right)}{\mathrm{2}}\right)\:\Rightarrow \\ $$$${I}\:=\frac{\pi}{\sqrt{\mathrm{3}}\sqrt{\sqrt{\mathrm{3}}}}\:\:\frac{{cos}\left(\frac{{arctan}\left(\sqrt{\mathrm{11}}\right)}{\mathrm{2}}\right)}{\mathrm{2}{sin}\left(\frac{{arctan}\left(\sqrt{\mathrm{11}}\right)}{\mathrm{2}}\right){cos}\left(\frac{{arctan}\left(\sqrt{\mathrm{11}}\right)}{\mathrm{2}}\right)} \\ $$$${I}\:=\frac{\pi}{\mathrm{2}\sqrt{\mathrm{3}}\sqrt{\sqrt{\mathrm{3}}}{sin}\left(\frac{{arctan}\left(\sqrt{\mathrm{11}}\right)}{\mathrm{2}}\right)}\:. \\ $$$$ \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 25/Mar/19

(1/(x^4 −x^2 +3)) =((1/x^2 )/(x^2 −1+(3/x^2 ))) =(1/(2(√3)))((((2(√3))/x^2 )/(x^2 −1+(3/x^2 )))) =(1/(2(√3)))(((1+((√3)/x^2 )−(1−((√3)/x^2 )))/(x^2 +(3/x^2 )−1))) =(1/(2(√3)))(((d(x−((√3)/x)))/((x−((√3)/x))^2 +2(√3) −1)))−(1/(2(√3)))(((d(x+((√3)/x)))/((x+((√3)/x))^2 −2(√3) −1))) (1/(2(√3)))∫_(−∞) ^(+∞) ((d(x−((√3)/x)))/((x−((√3)/x))^2 +((√(2(√3) −1)) )^2 ))←I_1 (1/(2(√3)))×(1/(√(2(√3) −1)))×∣tan^(−1) (((x−((√3)/x))/(√(2(√3) −1))))∣_(−∞) ^(+∞) =(1/(2(√3)))×(1/(√(2(√3) −1)))×{((π/2))−(((−π)/2))} =(π/(2(√3)))×(1/(√(2(√3) −1)))←value of I_1 (1/(2(√3)))∫_(−∞) ^(+∞) ((d(x+((√3)/x)))/((x+((√3)/x))^2 −(2(√3) +1))) ←I_2 (1/(2(√3)))×(1/(2((√(2(√3) +1)) )))∣ln(((x+((√3)/x)−(√(2(√3) +1)))/(x+((√3)/x)+(√(2(√3) +1)))))∣_(−∞) ^(+∞) now ∣ln(((x+(a/x)−b)/(x+(a/x)+b)))∣_(−∞) ^(+∞) =∣ln(((x^2 +a−bx)/(x^2 +a+bx)))∣_(−∞) ^(+∞) =∣ln(((1+(a/x^2 )−(b/x))/(1+(a/x^2 )+(b/x))))∣_(−∞) ^(+∞) =0 so I_2 =0 hence answer is (π/(2(√3)))×(1/(√(2(√3) −1))) this is the answer

$$\frac{\mathrm{1}}{{x}^{\mathrm{4}} −{x}^{\mathrm{2}} +\mathrm{3}} \\ $$$$=\frac{\frac{\mathrm{1}}{{x}^{\mathrm{2}} }}{{x}^{\mathrm{2}} −\mathrm{1}+\frac{\mathrm{3}}{{x}^{\mathrm{2}} }} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}}\left(\frac{\frac{\mathrm{2}\sqrt{\mathrm{3}}}{{x}^{\mathrm{2}} }}{{x}^{\mathrm{2}} −\mathrm{1}+\frac{\mathrm{3}}{{x}^{\mathrm{2}} }}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}}\left(\frac{\mathrm{1}+\frac{\sqrt{\mathrm{3}}}{{x}^{\mathrm{2}} }−\left(\mathrm{1}−\frac{\sqrt{\mathrm{3}}}{{x}^{\mathrm{2}} }\right)}{{x}^{\mathrm{2}} +\frac{\mathrm{3}}{{x}^{\mathrm{2}} }−\mathrm{1}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}}\left(\frac{{d}\left({x}−\frac{\sqrt{\mathrm{3}}}{{x}}\right)}{\left({x}−\frac{\sqrt{\mathrm{3}}}{{x}}\right)^{\mathrm{2}} +\mathrm{2}\sqrt{\mathrm{3}}\:−\mathrm{1}}\right)−\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}}\left(\frac{{d}\left({x}+\frac{\sqrt{\mathrm{3}}}{{x}}\right)}{\left({x}+\frac{\sqrt{\mathrm{3}}}{{x}}\right)^{\mathrm{2}} −\mathrm{2}\sqrt{\mathrm{3}}\:−\mathrm{1}}\right) \\ $$$$\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}}\int_{−\infty} ^{+\infty} \frac{{d}\left({x}−\frac{\sqrt{\mathrm{3}}}{{x}}\right)}{\left({x}−\frac{\sqrt{\mathrm{3}}}{{x}}\right)^{\mathrm{2}} +\left(\sqrt{\mathrm{2}\sqrt{\mathrm{3}}\:−\mathrm{1}}\:\right)^{\mathrm{2}} }\leftarrow\boldsymbol{{I}}_{\mathrm{1}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}}×\frac{\mathrm{1}}{\sqrt{\mathrm{2}\sqrt{\mathrm{3}}\:−\mathrm{1}}}×\mid{tan}^{−\mathrm{1}} \left(\frac{{x}−\frac{\sqrt{\mathrm{3}}}{{x}}}{\sqrt{\mathrm{2}\sqrt{\mathrm{3}}\:−\mathrm{1}}}\right)\mid_{−\infty} ^{+\infty} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}}×\frac{\mathrm{1}}{\sqrt{\mathrm{2}\sqrt{\mathrm{3}}\:−\mathrm{1}}}×\left\{\left(\frac{\pi}{\mathrm{2}}\right)−\left(\frac{−\pi}{\mathrm{2}}\right)\right\} \\ $$$$=\frac{\pi}{\mathrm{2}\sqrt{\mathrm{3}}}×\frac{\mathrm{1}}{\sqrt{\mathrm{2}\sqrt{\mathrm{3}}\:−\mathrm{1}}}\leftarrow{value}\:{of}\:{I}_{\mathrm{1}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}}\int_{−\infty} ^{+\infty} \frac{{d}\left({x}+\frac{\sqrt{\mathrm{3}}}{{x}}\right)}{\left({x}+\frac{\sqrt{\mathrm{3}}}{{x}}\right)^{\mathrm{2}} −\left(\mathrm{2}\sqrt{\mathrm{3}}\:+\mathrm{1}\right)}\:\leftarrow\boldsymbol{{I}}_{\mathrm{2}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}}×\frac{\mathrm{1}}{\mathrm{2}\left(\sqrt{\mathrm{2}\sqrt{\mathrm{3}}\:+\mathrm{1}}\:\right)}\mid{ln}\left(\frac{{x}+\frac{\sqrt{\mathrm{3}}}{{x}}−\sqrt{\mathrm{2}\sqrt{\mathrm{3}}\:+\mathrm{1}}}{{x}+\frac{\sqrt{\mathrm{3}}}{{x}}+\sqrt{\mathrm{2}\sqrt{\mathrm{3}}\:+\mathrm{1}}}\right)\mid_{−\infty} ^{+\infty} \\ $$$${now}\:\mid{ln}\left(\frac{{x}+\frac{{a}}{{x}}−{b}}{{x}+\frac{{a}}{{x}}+{b}}\right)\mid_{−\infty} ^{+\infty} \\ $$$$=\mid{ln}\left(\frac{{x}^{\mathrm{2}} +{a}−{bx}}{{x}^{\mathrm{2}} +{a}+{bx}}\right)\mid_{−\infty} ^{+\infty} \\ $$$$=\mid{ln}\left(\frac{\mathrm{1}+\frac{{a}}{{x}^{\mathrm{2}} }−\frac{{b}}{{x}}}{\mathrm{1}+\frac{{a}}{{x}^{\mathrm{2}} }+\frac{{b}}{{x}}}\right)\mid_{−\infty} ^{+\infty} =\mathrm{0} \\ $$$$\boldsymbol{{so}}\:\boldsymbol{{I}}_{\mathrm{2}} =\mathrm{0} \\ $$$$\boldsymbol{{hence}}\:\boldsymbol{{answer}}\:\boldsymbol{{is}} \\ $$$$\frac{\pi}{\mathrm{2}\sqrt{\mathrm{3}}}×\frac{\mathrm{1}}{\sqrt{\mathrm{2}\sqrt{\mathrm{3}}\:−\mathrm{1}}}\:\boldsymbol{{this}}\:\boldsymbol{{is}}\:\boldsymbol{{the}}\:\boldsymbol{{answer}} \\ $$$$ \\ $$

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