calculate-dx-x-2-x-1-n-with-n-gt-1-

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

c a l c u l a t e \int_{- \infty}^{+ \infty} \frac{d x}{{(x^{2} + x + 1)}^{n}} w i t h n > 1 .

Commented by abdo imad last updated on 06/Mar/18

l e t p u t A_{n} = \int_{- \infty}^{+ \infty} \frac{d x}{{(x^{2} + x + 1)}^{n}} w i t h n ⩾ 2 w e h a v e

x^{2} + x + 1 = x^{2} + 2 x \frac{1}{2} + \frac{1}{4} + \frac{3}{4} = {(x + \frac{1}{2})}^{2} + \frac{3}{4} t h e c h .

x + \frac{1}{2} = \frac{\sqrt{3}}{2} t \Rightarrow I_{n} = \int_{- \infty}^{+ \infty} \frac{1}{{(\frac{3}{4} (1 + t^{2}))}^{n}} \frac{\sqrt{3}}{2} d t

= {(\frac{4}{3})}^{n} \frac{\sqrt{3}}{2} \int_{- \infty}^{+ \infty} \frac{d t}{{(1 + t^{2})}^{n}} = \sqrt{3} {(\frac{4}{3})}^{n} \int_{0}^{\infty} \frac{d t}{{(1 + t^{2})}^{n}} b u t

w e h a v e p r o v e d t h a t \int_{0}^{\infty} \frac{d t}{{(1 + t^{2})}^{n}} = π \frac{(2 n - 2)!}{2^{2 n - 1} {((n - 1)!)}^{2}} \Rightarrow

Commented by abdo imad last updated on 06/Mar/18

\Rightarrow A_{n} = π \sqrt{3} {(\frac{4}{3})}^{n} \frac{(2 n - 2)!}{2^{2 n - 1} {((n - 1)!)}^{2}} .