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Question Number 62806 by mathmax by abdo last updated on 25/Jun/19

$$findthevalueof{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{x+1}{{(x^4+x^2+1)}^3}dx$$

Commented by mathmax by abdo last updated on 27/Jun/19

$$letA={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{x+1}{{(x^4+x^2+1)}^3}dxletW\left(z\right)=\frac{z+1}{{(z^4+z^2+1)}^3}?polesofW?$$$$z^4+z^2+1=0\Rightarrow t^2+t+1=0(with{z}^2=t)$$$$\mathrm{\Delta }=1-4={\left(i\sqrt{3}\right)}^2\Rightarrow t_1=\frac{-1+i\sqrt{3}}{2}=e^{\frac{i2\pi }{3}}(=j)and{t}_2=\frac{-1-i\sqrt{3}}{2}=e^{-\frac{i2\pi }{3}}\left(\stackrel{-}{j}\right)\Rightarrow$$$$t^2+t+1=(t-e^{\frac{i2\pi }{3}})(t-e^{-i\frac{2\pi }{3}})=(z^2-e^{\frac{i2\pi }{3}})(z^2-e^{-\frac{i2\pi }{3}})$$$$\Rightarrow W\left(z\right)=\frac{z+1}{{\left\{(z-e^{\frac{i\pi }{3}})(z+e^{\frac{i\pi }{3}})(z-e^{-\frac{i\pi }{3}})(z+e^{-\frac{i\pi }{3}})\right\}}^3}$$$$=\frac{z+1}{{(z-e^{\frac{i\pi }{3}})}^3{(z+e^{\frac{i\pi }{3}})}^3{(z-e^{-\frac{i\pi }{3}})}^3{(z+e^{-\frac{i\pi }{3}})}^3}sothepolesof\phi are\stackrel{-}{+}{e}^{\frac{i\pi }{3}}and\stackrel{-}{+}{e}^{-\frac{i\pi }{3}}\left(triples\right)$$$$residustheoremgive$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}W\left(z\right)dz=2i\pi \{Res(W,e^{\frac{i\pi }{3}})+Res(W,-e^{-\frac{i\pi }{3}})\}$$$$Res(W,e^{\frac{i\pi }{3}})={lim}_{z\to e^{\frac{i\pi }{3}}}\frac{1}{(3-1)!}{\left\{{(z-e^{\frac{i\pi }{3}})}^{3}W\left(z\right)\right\}}^{\left(2\right)}$$$$={lim}_{z\to e^{\frac{i\pi }{3}}}\frac{1}{2}{\left\{\frac{z+1}{{(z+e^{\frac{i\pi }{3}})}^3(z^2-e^{-i\frac{2\pi }{3}})}\right\}}^{\left(2\right)}...becontinued....$$

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