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.


  • 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.

  • 1
    $\begingroup$ Put $\ln x = y$. Should make your analysis easier. $\endgroup$ Jan 13, 2016 at 23:57
  • 1
    $\begingroup$ Here is a brief outline: The log term is small when compared to the power function term; make this precise with an estimate that looks like $\ln x < x^{\epsilon / (2\beta)}$ for very large $x$. Then your integrand can be estimated by $$\frac{1}{x^{1 - \epsilon} (\ln x)^{\beta}} > \frac{1}{x^{1 - \epsilon/2}}$$ after a line of algebra. This is now a pure power function of $x$, which is easily seen to have divergent integral. $\endgroup$
    – user296602
    Jan 14, 2016 at 0:04
  • $\begingroup$ That's good enough to be the answer. $\endgroup$ Jan 14, 2016 at 0:14
  • $\begingroup$ @user254665 I wrote an answer that was very similar to that (+ an explanation of how to guarantee that this works for large x), and it was quickly sent to -2. $\endgroup$
    – user296602
    Jan 14, 2016 at 0:33

1 Answer 1


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 $\beta>0$

  • $\begingroup$ Couldn't think of a better way to understand this! :) $\endgroup$
    – F1sargyan
    Jan 14, 2016 at 11:59
  • 1
    $\begingroup$ @f1sargyan Thank you! Much appreciative and pleased I was able to help. - Mark $\endgroup$
    – Mark Viola
    Jan 14, 2016 at 14:56

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