Proposition 2 ("logarithmic criterion of order 0")

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

This criterion is not decisive when $1/x^{1+\alpha} \ll f(x) \ll 1/x$ for every exponent $\alpha > 0$, as, for example, when $f(x) = 1/x(\log x)^\mu$ ($\mu > 0$). But in this last case f has primitive $\frac{1}{1-\mu} (\log x)^{1-\mu}$ if $\mu \neq 1$ and $\log \log x$ when $\mu = 1$. Thus: