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I did a quick search here but couldn't find a similar problem (it's probably out there somewhere...)
I'm stuck with this rather simple improper integral:

$\int_{1}^{\infty} \frac{1}{x^{\alpha}-1} \mathrm dx$

Now, I see that it could be divided into two seperate integrals, so as to tackle one "problem" at a time:

$\int_{1}^{\infty} \frac{1}{x^{\alpha}-1}dx=\int_{1}^{2}\frac{1}{x^{\alpha}-1}dx+\int_{2}^{\infty}\frac{1}{x^{\alpha}-1}dx$

The rightmost integral, $\int_{2}^{\infty}\frac{1}{x^{\alpha}-1}dx$, is convergent if and only if $\alpha>1$, that's easy to see.

My problem seems to be with the other one, from 1 to 2. I mean, I realize that if $\alpha\ge1$ it is divergent (I hope...). I can't seem to formally show that. It's possible to use the comparison test (well, for absolute convergence), but with what function?

This is probably very simple and I'm just having a "blank" moment. I'd appreciate any kind of help.

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2 Answers 2

First it is easy to see that for $a \leq 1$, the integral diverges by comparing it with $\dfrac1{x-1}$.

For $a>1$, we can proceed as follows. Replace $x$ by $1/x$, we then get that $$\int_1^{\infty} \dfrac{dx}{x^a - 1} = \int_1^0 \dfrac1{1/x^a-1} \dfrac{-dx}{x^2} = \int_0^1 \dfrac{x^{a-2} dx}{1-x^a}$$ Let $f_n(x) = x^{a-2} \left(\displaystyle \sum_{k=0}^n x^{ak}\right) = \left(\displaystyle \sum_{k=0}^n x^{ak+a-2}\right)$. We have $$f(x) = \dfrac{x^{a-2}}{1-x^a} > f_n(x) \,\,\,\,\, \forall x \in (0,1)$$ Now $$\int_0^1 f_n(x)dx = \sum_{k=0}^n \dfrac1{ak+a-1} > \dfrac{H_{n+1}}{a}$$ Hence, we have that $$\int_0^1 f(x) dx > \dfrac{H_{n+1}}{a}$$ for all $n$. Hence, $$\int_1^{\infty} \dfrac{dx}{x^a - 1}$$ diverges for all $a\geq 0$

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Hint: We only care about $\alpha\gt 0$, else there is bad trouble at $\infty$. Let $x=1+t$. We are interested in $(1+t)^\alpha-1$ for positive $t$ near $0$. Usin Taylor's Theorem, or otherwise, we have $(1+t)^\alpha-1=\alpha t+O(t^2)$. So if $t$ is positive and close enough to $0$, $(1+t)^\alpha -1\lt 2\alpha t$. Then use the Comparison Test to conclude that the integral diverges.

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I get it now. Thank you! –  Guy Feb 7 '13 at 10:29

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