I'll have to explain to my students in an exercise class for what values of $\alpha,\beta\in \mathbb{R}$ the so-called "Bertrand integral" converges
$$ \int_e^\infty \frac{dt}{t^\alpha (\ln(t))^\beta} $$
I was wondering if this integral has any applications or if it appears naturally somewhere?
Bertrand's integral points at the gap which exists between slow growing increasing functions on neighborhoods of $+\infty$ which tend do $+\infty$ and those who tend to a finite limit. I will try to explain this here.
For $n \in \mathbb{N}^{>0}$, write $\log_n$ for the $n$-fold iterate of the natural logarithm $\log$, defined on $I_n:= (\exp_{n-1}(0),+\infty)$ (where $\exp_n$ is the $n$-fold iterate of $\exp$). And set $i_0=(0,+\infty)$ and $\log_0=\operatorname{id}_{I_0}$. Write $g_n$ for the function $\frac{-1}{\log_n}$ on $I_n$.
For each $n\in \mathbb{N}$, the function $\log_n$ is a very slow growing function which tends to $+\infty$ whereas $g_n$ is a very slow growing function which tends to $0$. In fact most functions $f$ which naturally arise in analysis will satisfy $f<_{\infty}g_n$ for a certain $n$ if they tend to $0$, or $\log_n <_{\infty} f$ for a certain $n$ if they tend to $+\infty$. Here $g<_{\infty}h$ means that $g(x)<h(x)$ for sufficiently large $x$. So there is a gap $\Lambda$ suggestively illustarted as follows:
$g_0<_{\infty}g_1<_{\infty}...<_{\infty} g_n <_{\infty} \ \color{red}{...\Lambda ...}\ <_{\infty} \log_n < ... <_{\infty} \log <_{\infty} \log_0$
In fact, a better representation of a gap is obtained by taking derivatives. The idea there is that a strictly increasing (and differentiable) function $f$ which tends to $+\infty$ can only have a derivative which tends to zero so fast, until it fails to overcome any finite limit. If the derivative eventually falls below some $g_n'=\frac{1}{\log_0 \log_1 \cdot \cdot \cdot \log_{n-1} {\log_n}^2}$, then the function tends to a finite limit. If the derivative stays above some $\log_n'=\frac{1}{\log_0 \log_1 \cdot \cdot \cdot \log_n}$, then the function tends to $+\infty$. So you have another gap $\Gamma$ shown here
$g_0'<_{\infty}g_1'<_{\infty}...<_{\infty} g_n' <_{\infty} \ \color{red}{...\Gamma ...}\ <_{\infty} \log_n' < ... <_{\infty} \log' <_{\infty} \log_0'$
(I suggest you try to find something that fits in this gap, you'll see that those things don't really occur in undergraduate analysis.)
Now consider the lexicographic order $\prec$ on $\mathbb{R}^{n+1}$ given for $\overline{a}=(a_0,...,a_n),\overline{b}=(b_0,...,b_n) \in \mathbb{R}^{n+1}$ by $\overline{a}\prec \overline{b}$ iff $a_i<b_i$ where $i \in \{0,...,n\}$ is minimal with $a_i\neq b_i$. Write also $1_{n+1}=(1,1,...,1) \in \mathbb{R}^{n+1}$. For $\overline{a} \in \mathbb{R}^{n+1}$, write $B_{\overline{a}}$ for the function $x\mapsto \frac{1}{{\log_0(x)}^{a_0} \cdot \cdot \cdot {\log_n(x)}^{a_n}}: I_n \longrightarrow \mathbb{R}$.
We have the generalized "Bertrand's rule":
This can be proved by noticing that one has $B_{\overline{a}}<_{\infty} g_{n+1}'$ if $1\prec_{n+1}\overline{a}$, and $\log_{n+1}'<_{\infty}B_{\overline{a}}$ if $\overline{a}\preceq 1_{n+1}$.
Disclaimer: I write as if every function were differentiable, strictly monotonous, and as if for each two functions $f \neq g$, one had $f<_{\infty} g$ or $g<_{\infty} f$. Of course this is not the case. But there is a class of problems (or classes of functions) for which this is the case, so this type of asymptotic arguments can be illustrative in general, and demonstrative in certain cases.