Integrating $\frac{1}{x^r-1}$ with r>1

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Discussion Overview

The discussion revolves around the integral of the function $\frac{1}{x^r-1}$, where $r$ is a real number greater than 1. Participants explore the potential methods for integrating this expression, including series expansions and the conditions under which these methods apply.

Discussion Character

  • Exploratory, Technical explanation, Debate/contested

Main Points Raised

  • Some participants suggest that the integral may not be expressible in terms of elementary functions.
  • One participant proposes that for $|x|<1$, the function can be expressed as a series: $\frac{1}{x^{r}-1} = - \sum_{n=0}^{\infty} x^{n\ r}$, allowing for term-by-term integration.
  • Another participant extends this idea, stating that for $|x|>1$, a substitution can be made to transform the integral into a similar series form, which can also be integrated term by term.
  • There is a request for clarification regarding the source of the series formula used in the discussion.
  • One participant mentions that the series is derived from the sum of an infinite geometric series, which they consider common knowledge.

Areas of Agreement / Disagreement

Participants express differing views on the expressibility of the integral in elementary terms, and while some agree on the validity of the series expansion for specific ranges of $x$, the overall discussion remains unresolved regarding the general case of the integral.

Contextual Notes

The discussion includes assumptions about the convergence of series and the conditions under which the integral can be evaluated. There are unresolved aspects related to the applicability of the proposed methods across different ranges of $x$.

bincy
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Hii Everyone,[math] \int\frac{1}{x^{r}-1}dx
[/math] where r is a real no. greater than 1regards,
Bincy
 
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bincybn said:
Hii Everyone,[math] \int\frac{1}{x^{r}-1}dx
[/math] where r is a real no. greater than 1regards,
Bincy

Hi Bincy, :)

This integral may not be expressible in terms of elementary functions. See this.

Kind Regards,
Sudharaka.
 
If $|x|<1$ then is...

$\displaystyle \frac{1}{x^{r}-1} = - \sum_{n=0}^{\infty} x^{n\ r}$ (1)

... and You can integrate the expression (1) 'term by term'...

Kind regards

$\chi$ $\sigma$
 
chisigma said:
If $|x|<1$ then is...

$\displaystyle \frac{1}{x^{r}-1} = - \sum_{n=0}^{\infty} x^{n\ r}$ (1)

... and You can integrate the expression (1) 'term by term'...

... and if $|x|>1$ then You can set $\displaystyle t=\frac{1}{x}$ and the function to be integrated becomes...

$\displaystyle \frac{1}{t^{2}\ \{1-(\frac{1}{t})^{r}\}}\ = - \sum_{n=0}^{\infty} t^{(n+1)\ r-2}$ (1)

... and also in this case You can integrate 'term by term'...

Kind regards

$\chi$ $\sigma$
 
Can you please explain me the source of these formula?
 
bincybn said:
Can you please explain me the source of these formula?

chisigma is just using the sum of an infinite geometric series:

\[\sum_{n=0}^{\infty} x^n = \frac{1}{1-x}, \ \ \text{for } |x|<1\]

which should be common knowledge.

CB
 

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