calculate-1-dx-x-2-x-1-9-

Question Number 133537 by mathmax by abdo last updated on 22/Feb/21

calculate \int_{1}^{\infty} \frac{dx}{{(x^{2} + x + 1)}^{9}}

Answered by Ñï= last updated on 23/Feb/21

\int_{1}^{\infty} \frac{d x}{{(1 + x + x^{2})}^{9}}

= \int_{1}^{\infty} \frac{d x}{{[{(x + \frac{1}{2})}^{2} + \frac{3}{4}]}^{9}}

= \sqrt{\frac{3}{4}} \int_{\frac{π}{3}}^{\frac{π}{2}} \frac{{s e c}^{2} θ d θ}{{(\frac{3}{4} {t a n}^{2} θ + \frac{3}{4})}^{9}}

= \frac{\sqrt{\frac{3}{4}}}{{(\frac{3}{4})}^{9}} \int_{π / 3}^{π / 2} {c o s}^{16} θ d θ

= {(\frac{3}{4})}^{- 17 / 2} R e (\int_{π / 3}^{π / 2} e^{16 i θ} d θ)

= {(\frac{3}{4})}^{- 17 / 2} R e (\frac{1}{16 i} (e^{8 i π} - e^{16 i π / 3})

= \frac{2^{12}}{3^{8}}

Answered by mathmax by abdo last updated on 24/Feb/21

Φ = \int_{1}^{\infty} \frac{dx}{{(x^{2} + x + 1)}^{9}} \Rightarrow Φ = \int_{1}^{\infty} \frac{dx}{{({(x + \frac{1}{2})}^{2} + \frac{3}{4})}^{9}}

=_{x + \frac{1}{2} = \frac{\sqrt{3}}{2} t} \int_{\sqrt{3}}^{\infty} {(\frac{4}{3})}^{9} \frac{1}{{(t^{2} + 1)}^{9}} \frac{\sqrt{3}}{2} dt = \frac{\sqrt{3}}{2} {(\frac{4}{3})}^{9} \int_{\sqrt{3}}^{\infty} \frac{dt}{{(t^{2} + 1)}^{9}}

wehave \int_{\sqrt{3}}^{\infty} \frac{dt}{{(t^{2} + 1)}^{9}} = \int_{\sqrt{3}}^{\infty} \frac{dt}{{(t - i)}^{9} {(t + i)}^{9}} = \int_{\sqrt{3}}^{\infty} \frac{dt}{{(\frac{t - i}{t + i})}^{9} {(t + i)}^{18}}

changement \frac{t - i}{t + i} = z give t - i = zt + iz \Rightarrow (1 - z) t = i (1 + z) \Rightarrow

t = i \frac{1 + iz}{1 - iz} \Rightarrow \frac{dt}{dz} = i \frac{i (1 - iz) - (1 + iz) (- i)}{{(1 - iz)}^{2}} = i \frac{i + z + i - z}{{(1 - iz)}^{2}} = \frac{- 2}{{(1 - iz)}^{2}}

and t + i = \frac{i - z}{1 - iz} + i = \frac{i - z + i + z}{(1 - iz)} = \frac{2 i}{(1 - iz)} \Rightarrow

Φ = \int_{\frac{\sqrt{3} - i}{\sqrt{3} + i}}^{1} \frac{- 2}{{(1 - iz)}^{2} z^{9} {(\frac{2 i}{1 - iz})}^{18}} dz

= \frac{2}{{(2 i)}^{18}} \int_{1}^{\frac{\sqrt{3} - i}{\sqrt{3} + i}} \frac{{(1 - iz)}^{16}}{z^{9}} dz = \frac{2}{{(2 i)}^{18}} \int_{1}^{\frac{\sqrt{3} - i}{\sqrt{3} + i}} \frac{\sum_{k = 0}^{16} C_{16}^{k} {(iz)}^{k} {(- 1)}^{16 - k}}{z^{9}} dz

= - \frac{2}{2^{18}} \int_{1}^{\frac{\sqrt{3} - i}{\sqrt{3} + i}} \sum_{k = 0}^{16} {(- 1)}^{k} C_{16}^{k} i^{k} z^{k - 9} dz

= - \frac{1}{2^{17}} \sum_{k = 0, k \neq 8}^{16} {(- i)}^{k} C_{16}^{k} {[\frac{1}{k - 8} z^{k - 8}]}_{1}^{\frac{\sqrt{3} - i}{\sqrt{3} + i}}

- \frac{1}{2^{17}} C_{16}^{8} {[lnz]}_{1}^{\frac{\sqrt{3} - i}{\sqrt{3} + i}}

= - \frac{1}{2^{17}} \sum_{k = 0 and k \neq 8}^{16} \frac{{(- 1)}^{k} C_{16}^{k}}{k - 8} {{(\frac{\sqrt{3} - i}{\sqrt{3} + i})}^{k - 8} - 1}

- \frac{1}{2^{17}} C_{16}^{8} \ln (\frac{\sqrt{3} - i}{\sqrt{3} + i})