0-x-n-1-x-6-dx-

Question Number 130594 by Lordose last updated on 27/Jan/21

\int_{0}^{\infty} \frac{x^{n}}{1 + x^{6}} dx

Answered by mindispower last updated on 27/Jan/21

e x i s t e i f n \in] - 1, 5 [= \int_{0}^{\infty} \frac{t^{\frac{n - 5}{6}}}{6 (1 + t)} d t = \frac{1}{6} \int_{0}^{\infty} \frac{t^{\frac{n + 1}{6} - 1}}{{(1 + t)}^{\frac{n + 1}{6} + \frac{- n + 5}{6}}}

= \frac{1}{6} β (\frac{n + 1}{6}, \frac{5 - n}{6}) = \frac{1}{6} . \frac{π}{s i n (\frac{(n + 1) π}{6})}

Commented by Lordose last updated on 27/Jan/21

Gratitude

Answered by mathmax by abdo last updated on 27/Jan/21

A_{n} = \int_{0}^{\infty} \frac{x^{n}}{1 + x^{6}} dx we do the cha 7 gement x^{6} = t \Rightarrow

A_{n} = \int_{0}^{\infty} \frac{{(t^{\frac{1}{6}})}^{n}}{1 + t} . \frac{1}{6} t^{\frac{1}{6} - 1} dt = \frac{1}{6} \int_{0}^{\infty} \frac{t^{\frac{n}{6} + \frac{1}{6} - 1}}{1 + t} dt

= \frac{1}{6} \int_{0}^{\infty} \frac{t^{\frac{n + 1}{6} - 1}}{1 + t} dt = \frac{1}{6} \times \frac{π}{\sin (\frac{(n + 1) π}{6})}

here i have used \int_{0}^{\infty} \frac{t^{a - 1}}{1 + t} dt = \frac{π}{\sin (π a)}

convergence of A_{n} at + \infty \frac{x^{n}}{x^{6} + 1} \sim \frac{1}{x^{6 - n}} and \int_{1}^{\infty} \frac{dx}{x^{6 - n}} cv \Leftrightarrow 6 - n > 1

5 - n > 1 \Rightarrow n < 5 \Rightarrow n ⩽ 4