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Question Number 146197 by mathmax by abdo last updated on 11/Jul/21
calculate ∫_(−∞) ^(+∞) (dx/((x^2 −x+1)^3 ))
$$\mathrm{calculate}\:\int_{−\infty} ^{+\infty} \:\frac{\mathrm{dx}}{\left(\mathrm{x}^{\mathrm{2}} −\mathrm{x}+\mathrm{1}\right)^{\mathrm{3}} } \\ $$
Answered by Olaf_Thorendsen last updated on 12/Jul/21
Ω = ∫_(−∞) ^(+∞) (dx/((x^2 −x+1)^3 )) f(a) = ∫_(−∞) ^(+∞) (dx/(x^2 −x+a)), a>(1/4) (1) f(a) = ∫_(−∞) ^(+∞) (dx/((x−(1/2))^2 +a−(1/4))) f(a) = (1/( (√(a−(1/4)))))[arctan(((x−(1/2))/( (√(a−(1/4))))))]_(−∞) ^(+∞) f(a) = (π/( (√(a−(1/4))))) = π(a−(1/4))^(−1/2) (2) (1) : f′(a) = −∫_(−∞) ^(+∞) (dx/((x^2 −x+a)^2 )) f′′(a) = 2∫_(−∞) ^(+∞) (dx/((x^2 −x+a)^3 )) ⇒ Ω = (1/2)f′′(1) (3) (2) : f′(a) = −(π/2)(a−(1/4))^(−3/2) f′′(a) = ((3π)/4)(a−(1/4))^(−5/2) (3) : Ω = (1/2)×((3π)/4)(1−(1/4))^(−5/2) Ω = ((4π)/(3(√3)))
$$\Omega\:=\:\int_{−\infty} ^{+\infty} \frac{{dx}}{\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)^{\mathrm{3}} } \\ $$$${f}\left({a}\right)\:=\:\int_{−\infty} ^{+\infty} \frac{{dx}}{{x}^{\mathrm{2}} −{x}+{a}},\:{a}>\frac{\mathrm{1}}{\mathrm{4}}\:\:\:\:\:\left(\mathrm{1}\right) \\ $$$${f}\left({a}\right)\:=\:\int_{−\infty} ^{+\infty} \frac{{dx}}{\left({x}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +{a}−\frac{\mathrm{1}}{\mathrm{4}}} \\ $$$${f}\left({a}\right)\:=\:\frac{\mathrm{1}}{\:\sqrt{{a}−\frac{\mathrm{1}}{\mathrm{4}}}}\left[\mathrm{arctan}\left(\frac{{x}−\frac{\mathrm{1}}{\mathrm{2}}}{\:\sqrt{{a}−\frac{\mathrm{1}}{\mathrm{4}}}}\right)\right]_{−\infty} ^{+\infty} \\ $$$${f}\left({a}\right)\:=\:\frac{\pi}{\:\sqrt{{a}−\frac{\mathrm{1}}{\mathrm{4}}}}\:=\:\pi\left({a}−\frac{\mathrm{1}}{\mathrm{4}}\right)^{−\mathrm{1}/\mathrm{2}} \:\:\:\:\:\left(\mathrm{2}\right) \\ $$$$\left(\mathrm{1}\right)\::\:{f}'\left({a}\right)\:=\:−\int_{−\infty} ^{+\infty} \frac{{dx}}{\left({x}^{\mathrm{2}} −{x}+{a}\right)^{\mathrm{2}} } \\ $$$${f}''\left({a}\right)\:=\:\mathrm{2}\int_{−\infty} ^{+\infty} \frac{{dx}}{\left({x}^{\mathrm{2}} −{x}+{a}\right)^{\mathrm{3}} } \\ $$$$\Rightarrow\:\Omega\:=\:\frac{\mathrm{1}}{\mathrm{2}}{f}''\left(\mathrm{1}\right)\:\:\:\:\:\left(\mathrm{3}\right) \\ $$$$\left(\mathrm{2}\right)\::\:{f}'\left({a}\right)\:=\:−\frac{\pi}{\mathrm{2}}\left({a}−\frac{\mathrm{1}}{\mathrm{4}}\right)^{−\mathrm{3}/\mathrm{2}} \\ $$$${f}''\left({a}\right)\:=\:\frac{\mathrm{3}\pi}{\mathrm{4}}\left({a}−\frac{\mathrm{1}}{\mathrm{4}}\right)^{−\mathrm{5}/\mathrm{2}} \\ $$$$\left(\mathrm{3}\right)\::\:\Omega\:=\:\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{3}\pi}{\mathrm{4}}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}}\right)^{−\mathrm{5}/\mathrm{2}} \\ $$$$\Omega\:=\:\frac{\mathrm{4}\pi}{\mathrm{3}\sqrt{\mathrm{3}}} \\ $$
Answered by mathmax by abdo last updated on 12/Jul/21
Υ=∫_(−∞) ^(+∞) (dx/((x^2 −x+1)^3 )) we considere ϕ(z)=(1/((z^2 −z+1)^3 )) poles of ϕ z^2 −z+1=0→Δ=−3 ⇒z_1 =((1+i(√3))/2)=e^((iπ)/3) and z_2 =e^(−((iπ)/3)) ⇒ϕ(z)=(1/((z−e^((iπ)/3) )^3 (z−e^(−((iπ)/3)) )^3 )) residus theorem ∫_(−∞) ^(+∞) ϕ(z)dz=2iπRes(ϕ,e^((iπ)/3) ) Res(ϕ,e^((iπ)/3) )=lim_(z→e^((iπ)/3) ) (1/((3−1)!)){(z−e^((iπ)/3) )^3 ϕ(z)}^((2)) =(1/2)lim_(z→e^((iπ)/3) ) {(z−e^(−((iπ)/3)) )^(−3) }^((2)) =(1/2)lim_(z→e^((iπ)/3) ) {−3(z−e^(−((iπ)/3)) )^(−4) }^((1)) =(1/2)lim_(z→e^((iπ)/3) ) 12(z−e^(−((iπ)/3)) )^(−5) =6(e^((iπ)/3) −e^(−((iπ)/3)) )^(−5) =6(2i×((√3)/2))^(−5) =(6/(i^5 ((√3))^5 ))=(6/(i9(√3))) (2/(3i(√3))) ∫_(−∞) ^(+∞) ϕ(z)dz=2iπ×(2/(3i(√3))) =((4π)/(3(√3))) ⇒Υ=((4π)/(3(√3)))
$$\Upsilon=\int_{−\infty} ^{+\infty} \:\frac{\mathrm{dx}}{\left(\mathrm{x}^{\mathrm{2}} −\mathrm{x}+\mathrm{1}\right)^{\mathrm{3}} }\:\mathrm{we}\:\mathrm{considere}\:\varphi\left(\mathrm{z}\right)=\frac{\mathrm{1}}{\left(\mathrm{z}^{\mathrm{2}} −\mathrm{z}+\mathrm{1}\right)^{\mathrm{3}} }\:\mathrm{poles}\:\mathrm{of}\:\varphi \\ $$$$\mathrm{z}^{\mathrm{2}} −\mathrm{z}+\mathrm{1}=\mathrm{0}\rightarrow\Delta=−\mathrm{3}\:\Rightarrow\mathrm{z}_{\mathrm{1}} =\frac{\mathrm{1}+\mathrm{i}\sqrt{\mathrm{3}}}{\mathrm{2}}=\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{3}}} \:\mathrm{and}\:\mathrm{z}_{\mathrm{2}} =\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{3}}} \\ $$$$\Rightarrow\varphi\left(\mathrm{z}\right)=\frac{\mathrm{1}}{\left(\mathrm{z}−\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{3}}} \right)^{\mathrm{3}} \left(\mathrm{z}−\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{3}}} \right)^{\mathrm{3}} }\:\:\mathrm{residus}\:\mathrm{theorem} \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left(\mathrm{z}\right)\mathrm{dz}=\mathrm{2i}\pi\mathrm{Res}\left(\varphi,\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{3}}} \right) \\ $$$$\mathrm{Res}\left(\varphi,\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{3}}} \right)=\mathrm{lim}_{\mathrm{z}\rightarrow\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{3}}} } \:\:\frac{\mathrm{1}}{\left(\mathrm{3}−\mathrm{1}\right)!}\left\{\left(\mathrm{z}−\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{3}}} \right)^{\mathrm{3}} \:\varphi\left(\mathrm{z}\right)\right\}^{\left(\mathrm{2}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{lim}_{\mathrm{z}\rightarrow\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{3}}} } \:\:\left\{\left(\mathrm{z}−\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{3}}} \right)^{−\mathrm{3}} \right\}^{\left(\mathrm{2}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{lim}_{\mathrm{z}\rightarrow\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{3}}} } \:\:\:\:\:\left\{−\mathrm{3}\left(\mathrm{z}−\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{3}}} \right)^{−\mathrm{4}} \right\}^{\left(\mathrm{1}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{lim}_{\mathrm{z}\rightarrow\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{3}}} } \:\:\:\:\mathrm{12}\left(\mathrm{z}−\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{3}}} \right)^{−\mathrm{5}} \\ $$$$=\mathrm{6}\left(\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{3}}} −\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{3}}} \right)^{−\mathrm{5}} \:=\mathrm{6}\left(\mathrm{2i}×\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{−\mathrm{5}} \:=\frac{\mathrm{6}}{\mathrm{i}^{\mathrm{5}} \left(\sqrt{\mathrm{3}}\right)^{\mathrm{5}} }=\frac{\mathrm{6}}{\mathrm{i9}\sqrt{\mathrm{3}}}\:\frac{\mathrm{2}}{\mathrm{3i}\sqrt{\mathrm{3}}} \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left(\mathrm{z}\right)\mathrm{dz}=\mathrm{2i}\pi×\frac{\mathrm{2}}{\mathrm{3i}\sqrt{\mathrm{3}}}\:=\frac{\mathrm{4}\pi}{\mathrm{3}\sqrt{\mathrm{3}}}\:\Rightarrow\Upsilon=\frac{\mathrm{4}\pi}{\mathrm{3}\sqrt{\mathrm{3}}} \\ $$

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