Remark

Tag 08FG, permanent. Cite it as [Bourbaki, Tag 08FG].

Functions of a Real Variable, chapter V, § 3. ASYMPTOTIC EXPANSIONS OF FUNCTIONS OF A REAL VARIABLE

When $g$ is not assumed to remain $\geq 0$ on an interval $[a, +\infty[$ (or to remain $\leq 0$ on such an interval), and $\int_{a}^{+\infty} g(t) \, dt$ is not convergent, the relation $f \sim g$ does not necessarily imply that $\int_{a}^{x} f(t) \, dt \sim \int_{a}^{x} g(t) \, dt$ , as is shown by the example where $g(x) = \sin x$ and $f(x) = \left( 1 + \frac{\sin x}{x} \right) \sin x$ ; here

\int_{n\pi}^{(n+1)\pi} \frac{\sin^2 t}{t} \, dt \geq \frac{1}{(n+1)\pi} \int_{0}^{\pi} \sin^2 t \, dt \geq \frac{1}{2} \int_{n+1}^{n+2} \frac{dt}{t},

whence

\int_{\pi}^{n\pi} \frac{\sin^2 t}{t} \, dt \geq \frac{1}{2} \int_2^{n+1} \frac{dt}{t}

and the integral

\int_1^{+\infty} dt/t

is infinite, though

\int_{\frac{\pi}{2}}^x g(t) \, dt = -\cos x

remains bounded (see V, p. 260, exerc. 4).

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

Cited by

Nothing.

Cites

V, p. 260