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
 

Thread 'Proving that convexity implies second order derivative being positive'

There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...
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