let-I-n-0-dx-1-x-3-n-find-a-relation-etween-I-n-and-I-n-1-2-calculate-I-1-and-I-2-

Question Number 40159 by maxmathsup by imad last updated on 16/Jul/18

l e t I_{n} = \int_{0}^{\infty} \frac{d x}{{(1 + x^{3})}^{n}}

f i n d a r e l a t i o n e t w e e n I_{n} a n d I_{n + 1}

2) c a l c u l a t e I_{1} a n d I_{2}

Commented by prof Abdo imad last updated on 17/Jul/18

1) w e h a v e I_{n} = \int_{0}^{\infty} \frac{1 + x^{3}}{{(1 + x^{3})}^{n + 1}} d x

= I_{n + 1} + \int_{0}^{\infty} \frac{x^{3}}{{(1 + x^{3})}^{n + 1}} d x b y p a r t s

\int_{0}^{\infty} \frac{x^{3}}{{(1 + x^{3})}^{3}} d x = \frac{1}{3} \int_{0}^{\infty} x (\frac{3 x^{2}}{{(1 + x^{3})}^{n + 1}}) d x

= \frac{1}{3} \int_{0}^{\infty} x (3 x^{2} {(1 + x^{3})}^{- n - 1}) d x

= \frac{1}{3} {[- \frac{x}{n} {(1 + x^{3})}^{- n}]}_{0}^{+ \infty} - \frac{1}{3} \int_{0}^{\infty} 1 (- \frac{1}{n}) \frac{1}{{(1 + x^{3})}^{n}} d x

= \frac{1}{3 n} I_{n} \Rightarrow I_{n} = I_{n + 1} + \frac{1}{3 n} I_{n} \Rightarrow

(1 - \frac{1}{3 n}) I_{n} = I_{n + 1} \Rightarrow I_{n + 1} = \frac{3 n - 1}{3 n} I_{n}

2) w e h a v e I_{1} = \int_{0}^{\infty} \frac{d x}{1 + x^{3}}

=_{x^{} = t^{\frac{1}{3}}} \int_{0}^{\infty} \frac{1}{1 + t} \frac{1}{3} t^{\frac{1}{3} - 1} d t = \frac{1}{3} \int_{0}^{\infty} \frac{t^{\frac{1}{3} - 1}}{1 + t} d t

= \frac{1}{3} \frac{π}{s i n (\frac{π}{3})} = \frac{π}{3 . \frac{\sqrt{3}}{2}} = \frac{2 π}{3 \sqrt{3}}

I_{2} = \frac{2}{3} I_{1} = \frac{2}{3} \frac{2 π}{3 \sqrt{3}} = \frac{4 π}{9 \sqrt{3}} .