Prove that $\int_{2}^{\infty} \frac{dx}{x^{1-\epsilon}\cdot \ln(x)^{\beta}}$ diverges for every $\beta$. let $\beta,\epsilon\in \mathbb R$, such that $\epsilon>0$. prove that for every $
\beta$:
$$\int_{2}^{\infty}  \frac{dx}{x^{1-\epsilon}\cdot \ln(x)^{\beta}}$$ 
Diverges.
SOLUTION ATTEMPT: 


*

*if $\beta<0$:


Using comparison test we get that:
$\lim_{x\to\infty} \frac{\frac{1}{x^{1-\epsilon}\cdot \ln(x)^{\beta}}}{\frac{1}{x}}=\lim_{x\to\infty} \frac{x^{\epsilon}}{ln(x)^{\beta}}=\infty.$ 
which means that when $x$ reaches infinity: $0<\frac{1}{x}<\frac{1}{x^{1-\epsilon}\cdot \ln(x)^{\beta}} $. we know that $\int_{2}^{\infty} \frac{1}{x}$ diverges then $\int_{2}^{\infty} \frac{1}{x^{1-\epsilon}\cdot \ln(x)^{\beta}}$ diverges.


*

*if $\beta=0$ its easy to prove that by comparing with $\int_{2}^{\infty} \frac{1}{x}$. 


$\frac{1}{x} < \frac{1}{x^{1-\epsilon}}$, $\int_{2}^{\infty} \frac{1}{x}$ diverges then 
$ \int_{2}^{\infty} \frac{1}{x^{1-\epsilon}}$ diverges.
I some how got stuck by the $\beta >0$ case. I always get a contradiction. I would like to see a detailed example for how to do this case, I got really frustrated after dealing with this question for 2 hours now. any kind of help would be appreciated.
 A: 
The case for $\beta\le 0$ should be evident.  Therefore, the ensuing discussion is relegated to the case for which $\beta>0$.


In THIS ANSWER, I showed using standard, non-calculus-based tools only that the logarithm function satisfies the inequalities
$$\frac{z}{z+1}\le \log (1+z)\le z \tag 1$$
for $1-<z$.  Letting $z=x-1$ in $(1)$ yields
$$\log (x)\le x-1<x \tag 2$$
for $x>0$.

Next, we recall for any number $\alpha$, 
$$\log(x^{\alpha})=\alpha \log(x) \tag 3$$ 
for $x>0$.  So, from $(2)$ and $(3)$ we see that for any $\alpha >0$ and $x>0$
$$\log(x)\le \frac{x^{\alpha}}{\alpha}$$
Therefore for $\beta>0$, $\alpha >0$, and $x>0$ we have
$$\log^\beta(x)\le \frac{x^{\alpha \beta}}{\alpha^{\beta}} \tag 4$$

Now, given any $\epsilon>0$ and $\beta>0$ we choose any $\alpha<\epsilon/\beta$.  Let us arbitrarily choose $\alpha = \epsilon/2\beta$.  Then, using $(4)$ with this choice of $\alpha$ reveals 
$$\frac{1}{x^{1-\epsilon}\log^\beta(x)}\ge \frac{\alpha^{\beta}}{x^{1-\epsilon+\alpha \beta}}=\frac{\left(\frac{\epsilon}{2\beta}\right)^{\beta}}{x^{1-\epsilon/2}}$$

Finally, we have
$$\int_2^L \frac{1}{x^{1-\epsilon}\log^\beta(x)}\,dx\ge\int_2^L \frac{\left(\frac{\epsilon}{2\beta}\right)^{\beta}}{x^{1-\epsilon/2}}\,dx\to \infty \,\,\text{as}\,\,L\to \infty$$
By the comparison test, the integral diverges for all $\beta>0$
