Proposition 3 ("logarithmic criterion of order 1")

Let f be a regulated function $\geq 0$ on an interval $[a, +\infty[$; if $f(x) \preccurlyeq 1/x(\log x)^\mu$ for some $\mu > 1$ then the integral $\int_a^{+\infty} f(t) \, dt$ converges; if $f(x) \succcurlyeq 1/x(\log x)^\mu$ for some $\mu \leq 1$ then the integral $\int_a^{+\infty} f(t) \, dt$ is infinite.

Generally, for every integer $n \geq 0$, let us denote by $l_n(x)$ the function defined inductively (for x large enough) by the relations $l_0(x) = x$, $l_n(x) = \log(l_{n-1}(x))$ for $n \geq 1$; one says that $l_n(x)$ is the nth iterated logarithm of x (cf. Appendix). One verifies immediately that $\frac{1}{1-\mu} (l_n(x))^{1-\mu}$ is a primitive of

for $\mu \neq 1$, and $l_{n+1}(x)$ is a primitive of $\frac{1}{x l_1(x) l_2(x) \ldots l_{n-1}(x) l_n(x)}$. Hence: