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

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The integral of $\frac{1}{x^r-1}$ for real numbers r greater than 1 may not be expressible in elementary functions. For values of |x| less than 1, it can be represented as a negative infinite series, allowing for term-by-term integration. Conversely, for |x| greater than 1, a substitution can transform the integral into a similar series form. The discussion emphasizes the use of the infinite geometric series formula as a foundational concept for these transformations. Understanding these series expansions is crucial for integrating the given function effectively.
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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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