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- Thread starter bincybn
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- Thread starter
- #1

- Feb 5, 2012

- 1,621

Hi Bincy,Hii Everyone,

\(\displaystyle \int\frac{1}{x^{r}-1}dx

\) where r is areal no.greater than 1

regards,

Bincy

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

Kind Regards,

Sudharaka.

- Feb 13, 2012

- 1,704

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

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

Kind regards

$\chi$ $\sigma$

- Feb 13, 2012

- 1,704

... and if $|x|>1$ then You can set $\displaystyle t=\frac{1}{x}$ and the function to be integrated becomes...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'...

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

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

Kind regards

$\chi$ $\sigma$

- Thread starter
- #5

- Jan 26, 2012

- 890

chisigma is just using the sum of an infinite geometric series:Can you plz explain me the source of these formula?

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

which should be common knowledge.

CB