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Integration

bincybn

Member
Apr 29, 2012
36
Hii Everyone,


\(\displaystyle \int\frac{1}{x^{r}-1}dx
\) where r is a real no. greater than 1


regards,
Bincy
 

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
Hii Everyone,


\(\displaystyle \int\frac{1}{x^{r}-1}dx
\) where r is a real no. greater than 1


regards,
Bincy
Hi Bincy, :)

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

Kind Regards,
Sudharaka.
 

chisigma

Well-known member
Feb 13, 2012
1,704
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

Well-known member
Feb 13, 2012
1,704
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

$\chi$ $\sigma$
 

bincybn

Member
Apr 29, 2012
36
Can you plz explain me the source of these formula?
 

CaptainBlack

Well-known member
Jan 26, 2012
890
Can you plz 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