Does $\int_{1}^{\infty} \frac{\mathrm dx}{x^{\alpha}-1} $ converge? 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.
 A: 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. 
A: 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 $a\geq 0$
A: One might use the direct comparison test.  If $a=0$, then the integrand is undefined everywhere.
Else, one might note that for $a>1$,
$$1<x<2\implies\frac1{x^a-1}>\underbrace{\frac1{a(x-1)}}_{\text{diverges}}+\frac1{2n}-\frac12\\\int_1^2\frac1{|a|(x-1)}~\mathrm dx=\infty$$
For $0<a<1$,
$$1<x<2\implies\frac1{x^a-1}>\frac1{x^{a+1}-1}\\(\text{see above for why it diverges})$$
For $a<0$, note that
$$x>2\implies\frac1{x^a-1}>\frac1{2^a-1}\\\int_2^\infty\frac1{2^a-1}~\mathrm dx=\infty$$
