Put F(x)=∫x+∞f(t)dt if ∫a+∞f(t)dt converges, and F(x)=∫axf(t)dt if ∫a+∞f(t)dt is infinite. Suppose that log∣f∣ and logx are comparable of order 1.
1 If f is of finite order μ=−1 with respect to x one has
F(x)∼∣μ+1∣1xf(x). (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
F(x)∼∣f′(x)∣(f(x))2. (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 μ=0 relative to x one has log∣f∣∼μlogx, so, since log∣f∣ and logx are comparable of order 1, by prop. 7 of V, p. 232 one has f′/f∼ μ/x, whence xf′∼μf. If μ>−1 one has f(x)≫xμ−ε for every ε>0, and hence (V, p. 229, prop. 2) the integral ∫a+∞f(t)dt is infinite. One can write F(x)=∫axf(t)dt=xf(x)−af(a)−∫axtf′(t)dt, whence again
∫ax(f(t)+tf′(t))dt=xf(x)−af(a); since
μ=−1 we have
f(x)+xf′(x)∼(μ+1)f(x), so (V, p. 230, prop. 6)
∫ax(f(t)+tf′(t))dt∼(μ+1)F(x), which proves (1) in this case. If
μ=0 one has likewise that
xf′(x)≪f(x), which again gives
f(x)+xf′(x)∼f(x). One argues similarly when
μ<−1, in the case where
∫a+∞f(t)dt converges.
2) If f is of order +∞ relative to x one has log∣f∣≫logx, so (V, p. 232, prop. 7) f′/f≫1/x, or again, putting g(x)=f(x)/f′(x), g(x)≪x; further, since f(x)≫xα for every α>0 the integral ∫a+∞f(t)dt is infinite. One can write
F(x)=∫axf(t)dt=∫axg(t)f′(t)dt=g(x)f(x)−g(a)f(a)−∫axf(t)g′(t)dt; since
g and
x are comparable of order 1, from the relation
g(x)≪x one can deduce (V, p. 232, prop. 7) that
g′(x)≪1, hence that
fg′≪f, and consequently (V, p. 230, prop.6)
∫axf(t)g′(t)dt≪F(x), which establishes the relation (2). The proof is similar when
f is of order
−∞ relative to
x, in the case where
∫a+∞f(t)dt converges.
Let E be a comparison scale (for real x tending to +∞) formed of nonzero real functions that are of constant sign on a neighbourhood of +∞, such that x∈E and such that the product and quotient of two functions in E again belongs to E (V, p. 221 and p. 224). If a regulated function f with constant sign on a neighbourhood of +∞ has a principal part cg relative to E, then ∫x+∞f(t)dt (resp. ∫axf(t)dt according to the case) will be equivalent to c∫x+∞g(t)dt (resp. c∫axg(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 E, one will thus have a principal part of ∫x+∞f(t)dt (resp. ∫axf(t)dt) relative to E.
Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.