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Question Number 31107 by abdo imad last updated on 02/Mar/18

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

Commented byabdo imad last updated on 06/Mar/18

$$letput{A}_n={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{dx}{{(x^2+x+1)}^n}withn⩾2wehave$$ $$x^2+x+1=x^2+2x\frac{1}{2}+\frac{1}{4}+\frac{3}{4}={(x+\frac{1}{2})}^2+\frac{3}{4}thech.$$ $$x+\frac{1}{2}=\frac{\sqrt{3}}{2}t\Rightarrow I_n={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{1}{{\left(\frac{3}{4}(1+t^2)\right)}^n}\frac{\sqrt{3}}{2}dt$$ $$={\left(\frac{4}{3}\right)}^n\frac{\sqrt{3}}{2}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{dt}{{(1+t^2)}^n}=\sqrt{3}{\left(\frac{4}{3}\right)}^n{\int }_0^{\mathrm{\infty }}\frac{dt}{{(1+t^2)}^n}but$$ $$wehaveprovedthat{\int }_0^{\mathrm{\infty }}\frac{dt}{{(1+t^2)}^n}=\pi \frac{(2n-2)!}{2^{2n-1}{((n-1)!)}^2}\Rightarrow$$

Commented byabdo imad last updated on 06/Mar/18

$$\Rightarrow A_n=\pi \sqrt{3}{\left(\frac{4}{3}\right)}^n\frac{(2n-2)!}{2^{2n-1}{((n-1)!)}^2}.$$

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