Indeed, by hypothesis there exist a $b \geqslant a$ and a number $c' > 0$ such that

whence since, moreover, one may suppose $b$ so large that $\int_a^b g(t)\,dt > 0$, there exists a $c'' > 0$ such that $\left\| \int_a^b \mathbf{f}(t)\,dt \right\| \leqslant c'' \int_a^b g(t)\,dt$; on putting $c = \max(c', c'')$ one thus has for every $x \geqslant b$, whence the proposition.

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:

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:

Each logarithmic criterion is thus applicable to functions for which the criteria of lower order are not decisive (cf. V, p. 264, exerc. 5 b) and V, p. 265, exerc. 8).

Because of its usefulness we translate the criterion of order 0 for integrals $\int_{\alpha}^{a} f(t) \, dt$ where $f$ is regulated and $\geq 0$ on a non-compact interval $]\alpha, a]$:

We leave it to the reader to translate the logarithmic criterion of order $n$ similarly.

The application of the logarithmic criteria is immediate if one knows how to obtain the principal part of $f$ with respect to a comparison scale containing the functions that feature in these criteria: if $f_1$ is the principal part, the integral $\int_{\alpha}^{+\infty} f(t) \, dt$ converges or is infinite together with $\int_{\alpha}^{+\infty} f_1(t) \, dt$, and the logarithmic criteria apply immediately to this last integral.

1. *If the integral $\int_{a}^{+\infty} g(t) \, dt$ converges then the relation $\mathbf{f} \ll g$ (resp. $\mathbf{f} \sim c g$, where $c$ is constant) implies that $\int_{x}^{+\infty} \mathbf{f}(t) \, dt \ll \int_{x}^{+\infty} g(t) \, dt$ (resp. $\int_{x}^{+\infty} \mathbf{f}(t) \, dt \sim c \int_{x}^{+\infty} g(t) \, dt$).

2. If the integral $\int_{a}^{+\infty} g(t) \, dt$ is infinite then the relation $\mathbf{f} \ll g$ (resp. $\mathbf{f} \sim c g$) implies that

for any $\alpha$ and $\beta$ in $[a, +\infty[$.

It is enough to prove the proposition for the relation $\mathbf{f} \ll g$ since, if $c \neq 0$, the relation $\mathbf{f} \sim c g$ is equivalent to $\mathbf{f} - c g \ll g$.

The first part is an immediate consequence of the mean value theorem, since if $\| \mathbf{f}(x) \| \leq \varepsilon g(x)$ for $x \geq x_0$ one deduces that

In the second place, suppose that $\int_{a}^{+\infty} g(t) \, dt = +\infty$. If $\| \mathbf{f}(x) \| \leq \varepsilon g(x)$ for $x \geq x_0 \geq \max(\alpha, \beta)$, one has

Now there exists an $x_1 \geq x_0$ such that for every $x \geq x_1$

whence, for $x \geq x_1$

which completes the proof, since $\varepsilon > 0$ is arbitrary.

In other words, one can integrate the two terms of a strong relation $\mathbf{f} \ll g$, $\mathbf{f} \sim \mathbf{a}g$, when $g$ is positive on an interval $[a, +\infty[$, and the relation persists between the primitives of the two terms, provided one takes care to integrate from $x$ to $+\infty$ if $\int_{a}^{+\infty} g(t) \, dt$ converges, and from $\alpha$ to $x$ (for any $\alpha$ in $[a, +\infty[$) in the opposite case.

Note that props. 1 (V, p. 228) and 6 (V, p. 230) remain valid when $\mathfrak{F}$ is a base for the trace filter of the intervals $[t, +\infty[$ (where $t > a$) on the complement of a countable set (cf. I, p. 15, th. 2).

Now, since $f'$ and $g'$ are of constant sign on an interval $[x_0, +\infty[$, both $f$ and $g$ are monotone on this interval, so tend to a finite or infinite limit as $x$ tends to $+\infty$. It is clear that $f$ and $g$ are comparable as $x$ tends to $+\infty$ if one of these limits is finite and $\neq 0$, or if one is zero and the other infinite. If both $f$ and $g$ tend to 0 one can write $f(x) = -\int_x^{+\infty} f'(t) \, dt,\ g(x) = -\int_x^{+\infty} g'(t) \, dt$; since $f'$ and $g'$ are comparable the same is true for $f$ and $g$ and the comparison relation between $f$ and $g$ is the same as that between $f'$ and $g'$, by prop. 6 (V, p. 20). Similarly, if $f$ and $g$ both have an infinite limit one has $f(x) = f(x_0) + \int_{x_0}^x f'(t) \, dt,\ g(x) = g(x_0) + \int_{x_0}^x g'(t) \, dt$; again prop. 6 (V, p. 230) shows that $f$ and $g$ are comparable and that the comparison relation between $f$ and $g$ is the same as that between $f'$ and $g'$. To complete the proof it remains to consider the case where $g$ tends to $\pm \infty$ and $f$ to a constant: one cannot then have $f' \succcurlyeq g'$, since then one could deduce from prop. 1 (V, p. 218) that the integral $\int_{x_0}^{+\infty} g'(t) \, dt$ was convergent; since $f'$ and $g'$ have been assumed comparable, one must have $f' \ll g'$.

Indeed, since $f^{(k)}$ and $g^{(k)}$ have constant sign on an interval $[x_0, +\infty[$, it follows that $f^{(k-1)}$ and $g^{(k-1)}$ are monotone on this interval, so have constant sign on a neighbourhood of $+\infty$; further, prop. 7 of V, p. 232, shows that $f^{(k-1)}$ and $g^{(k-1)}$ are comparable, so the corollary follows from applying prop. 7 recursively.

1 If $f$ is of finite order $\mu \neq -1$ with respect to $x$ one has

(1)

2 If $f$ is of infinite order relative to $x$ and if $f/f'$ and $x$ are comparable of order 1, then one has

(2)

One should note that the hypothesis implies that $f(x)$ has a determinate order relative to $x$ (V, p. 219).

1 If $f$ is of order $\mu \neq 0$ relative to $x$ one has $\log|f| \sim \mu \log x$, so, since $\log|f|$ and $\log x$ are comparable of order 1, by prop. 7 of V, p. 232 one has $f'/f \sim$ μ/x, whence $x f' \sim \mu f$. If $\mu > -1$ one has $f(x) \gg x^{\mu-\varepsilon}$ for every $\varepsilon > 0$, and hence (V, p. 229, prop. 2) the integral $\int_a^{+\infty} f(t)\,dt$ is infinite. One can write $F(x) = \int_a^x f(t)\,dt = x f(x) - af(a) - \int_a^x t f'(t)\,dt$, whence again

since $\mu \neq -1$ we have $f(x) + x f'(x) \sim (\mu + 1) f(x)$, so (V, p. 230, prop. 6) which proves (1) in this case. If $\mu = 0$ one has likewise that $x f'(x) \ll f(x)$, which again gives $f(x) + x f'(x) \sim f(x)$. One argues similarly when $\mu < -1$, in the case where $\int_a^{+\infty} f(t)\,dt$ converges.

2) If $f$ is of order $+\infty$ relative to $x$ one has $\log |f| \gg \log x$, so (V, p. 232, prop. 7) $f'/f \gg 1/x$, or again, putting $g(x) = f(x)/f'(x)$, $g(x) \ll x$; further, since $f(x) \gg x^\alpha$ for every $\alpha > 0$ the integral $\int_a^{+\infty} f(t)\,dt$ is infinite. One can write

since $g$ and $x$ are comparable of order 1, from the relation $g(x) \ll x$ one can deduce (V, p. 232, prop. 7) that $g'(x) \ll 1$, hence that $f g' \ll f$, and consequently (V, p. 230, prop.6) which establishes the relation (2). The proof is similar when $f$ is of order $-\infty$ relative to $x$, in the case where $\int_a^{+\infty} f(t)\,dt$ converges.

Let $\mathscr{E}$ be a comparison scale (for real $x$ tending to $+\infty$) formed of nonzero real functions that are of constant sign on a neighbourhood of $+\infty$, such that $x \in \mathscr{E}$ and such that the product and quotient of two functions in $\mathscr{E}$ again belongs to $\mathscr{E}$ (V, p. 221 and p. 224). If a regulated function $f$ with constant sign on a neighbourhood of $+\infty$ has a principal part $c g$ relative to $\mathscr{E}$, then $\int_x^{+\infty} f(t)\,dt$ (resp. $\int_a^x f(t)\,dt$ according to the case) will be equivalent to $c \int_x^{+\infty} g(t)\,dt$ (resp. $c \int_a^x g(t)\,dt$); if the function $g$ satisfies the hypotheses of prop. 8 of V, p. 233, and if (when the formula (2) of V, p. 233, applies) one knows a principal part of $g'$ relative to $\mathscr{E}$, one will thus have a principal part of $\int_x^{+\infty} f(t)\,dt$ (resp. $\int_a^x f(t)\,dt$) relative to $\mathscr{E}$.