convergence of an integral with logarithm for which values of $\alpha$ and $\beta$ the integral
$\int_2^\infty \frac{dx}{x^\alpha ln^\beta x}$
 converge?
I'm try to do the following integral
$\int_2^n \frac{dx}{x^\alpha ln^\beta x}$
for then calculate the limite when n tends to infinity, but I'm stuck.
 A: Hints: The only problem for the integral is at $\infty.$ Take $\beta = 0$ at first. You should get convergence iff $\alpha > 1.$ Now toss in $\beta$ and see if you can show that for $\alpha > 1,$ no power of $\ln x$ can hurt this convergence, and for  $\alpha < 1$ no power of $\ln x$ can change divergence. For $\alpha =1,$ interesting things happen, because $(\ln x)' = 1/x.$
A: By comparison we start with
$$\frac{1}{x^{\alpha}\ln^{\beta}(x)} \leq \frac{1}{x^{\alpha}}$$
Now you know that 
$$\sum_{k = 0}^{+\infty} \frac{1}{x^k}$$
Does converge iff $k \geq 2$ hence by comparison your first condition is
$$\alpha \geq 2$$
For the remaining part, we can also compare your integral with the sum
$$\sum_{n = 2}^{+\infty} \frac{1}{n^{\alpha}\ln^{\beta}(n)}$$
Which is convergent for $\beta > 1$.
Indeed for $\beta > 1$ we have
$$\int_2^{+\infty} \frac{\text{d}x}{x\ln^{\beta}(x)} = \frac{1}{1 - \beta}\ln^{1-\beta}(x)\bigg|_2^{+\infty} < \infty$$
Whilst for $\beta \leq 1$ you can check it does diverge.
The same things holds for the general case $x^{\alpha}$ for $\alpha \geq 2$.
A: Case 1. For $\alpha > 1$ and all $\beta \in \mathbb R$, you have
$$\lim\limits_{x \to \infty} \frac{1}{x^{\frac{\alpha-1}{2}} \ln^\beta x} = 0$$ hence for $x > 2$ large enough
$$0 \le \frac{1}{x^{\frac{\alpha-1}{2}} \ln^\beta x} \le 1$$ hence
$$0 \le \frac{1}{x^\alpha \ln^\beta x} \le \frac{1}{x^{1+\frac{\alpha-1}{2}}}$$
As $1+\frac{\alpha-1}{2} > 1$, and $\displaystyle \int_2^\infty \frac{dx}{x^{1+\frac{\alpha-1}{2}}}$ converges, $\displaystyle \int_2^\infty \frac{dx}{x^\alpha \ln^\beta x}$ also converges for $\alpha > 1$.
Case 2. For $\alpha <1$, you have
$$\frac{1}{x^\alpha \ln^\beta x} \ge \frac{1}{x} \ge 0$$ for $x$ large enough. As $\displaystyle \int_2^\infty \frac{dx}{x}$ diverges, therefore $\displaystyle \int_2^\infty \frac{dx}{x^\alpha \ln^\beta x}$ also diverges in that case.
Case 3. $\alpha = 1$ has been already covered by Alan Turing, his doppleganger more precisely. The integral converges for $\beta >1$ and diverges otherwise.
