give-the-integralA-n-1-dt-1-x-n-with-n-integr-and-n-2-at-form-of-serie-

Question Number 65352 by mathmax by abdo last updated on 28/Jul/19

g i v e t h e {i n t e g r a l A}_{n} = \int_{1}^{+ \infty} \frac{d t}{1 + x^{n}} w i t h n i n t e g r a n d n ⩾ 2

a t f o r m o f s e r i e .

Commented by mathmax by abdo last updated on 29/Jul/19

A_{n} = \int_{1}^{+ \infty} \frac{d x}{1 + x^{n}} c h a n g e m e n t x = \frac{1}{t} g i v e

A_{n} = - \int_{0}^{1} \frac{1}{1 + \frac{1}{t^{n}}} (- \frac{d t}{t^{2}}) = \int_{0}^{1} \frac{t^{n}}{(1 + t^{n}) t^{2}} d t = \int_{0}^{1} \frac{t^{n - 2}}{1 + t^{n}} d t

= \int_{0}^{1} t^{n - 2} (\sum_{k = 0}^{\infty} {(- 1)}^{k} t^{k n}) d t = \sum_{k = 0}^{\infty} {(- 1)}^{k} \int_{0}^{1} t^{n - 2 + k n} d t

= \sum_{k = 0}^{\infty} {(- 1)}^{k} {[\frac{1}{n - 2 + k n + 1} t^{n - 2 + k n + 1}]}_{0}^{1}

= \sum_{k = 0}^{\infty} \frac{{(- 1)}^{k}}{n (k + 1) - 1} = \sum_{k = 1}^{\infty} \frac{{(- 1)}^{k - 1}}{n k - 1} \Rightarrow

A_{n} = \frac{1}{n - 1} - \frac{1}{2 n - 1} + \frac{1}{3 n - 1} - \dots . .