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Question Number 67673 by Abdo msup. last updated on 30/Aug/19

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

Commented by Abdo msup. last updated on 30/Aug/19

$$letA={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{dx}{(x^2-x+1)(x^2+x+1)}$$$$and\phi \left(z\right)=\frac{1}{(z^2-z+1)(z^2+z+1)}polesof\phi ?$$$$z^2-z+1=0\to \mathrm{\Delta }=-3={\left(i\sqrt{3}\right)}^2\Rightarrow$$$$z_1=\frac{1+i\sqrt{3}}{2}and{z}_2=\frac{1-i\sqrt{3}}{2}$$$$z^2+z+1=0\to \mathrm{\Delta }=-3={\left(i\sqrt{3}\right)}^2\Rightarrow$$$${\alpha }_1=\frac{-1+i\sqrt{3}}{2}and{\alpha }_2=\frac{-1-i\sqrt{3}}{2}$$$$\phi \left(z\right)=\frac{1}{(x-z_1)(x-z_2)(x-{\alpha }_1)(x-{\alpha }_2)}$$$$residustheoremgive$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(z\right)dz=2i\pi \{Res(\phi ,z_1)+Res(\phi ,{\alpha }_1)\}$$$$Res(\phi ,z_1)=\frac{1}{(z_1-z_2)(z_1^2+z_1+1)}$$$$=\frac{1}{i\sqrt{3}(z_1^2+z_1+1)}$$$$Res(\phi ,{\alpha }_1)=\frac{1}{({\alpha }_1-{\alpha }_2)({\alpha }_1^2-{\alpha }_1+1)}=\frac{1}{i\sqrt{3}({\alpha }_1^2-{\alpha }_1+1)}$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(z\right)dz=2i\pi \{\frac{1}{i\sqrt{3}(z_1^2+z_1+1)}+\frac{1}{i\sqrt{3}({\alpha }_1^2-{\alpha }_1+1)}\}$$$$=\frac{2\pi }{\sqrt{3}}\{\frac{1}{z_1^2+z_1+1}+\frac{1}{{\alpha }_1^2-{\alpha }_1+1}\}$$$$$$

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