# Integrating an asymptotic

Let $f \sim g$ mean that $f/g \rightarrow 1$ as $x \rightarrow \infty$. Does it follow that $\int_{1}^{x} f(t)\, dt \sim \int_{1}^{x}g(t)\, dt$?

No. Take $f(x) = \frac{1}{x^2}$ and $g(x) =\frac{1}{x^2} +\frac{1}{x^3}$.

• Well, don't both integrals converge at the same time? Apr 7, 2021 at 17:14

This question is related to the comparison and limit comparison tests for establishing the convergence or divergence of a series or integral. Throughout my answer, I will assume that both $f$ and $g$ are (eventually) positive and continuous everywhere.

As BebopButUnsteady's answer shows, the conclusion in question is not always true. However, one could still say a little more:

Divergent case: If either one of the integrals $\int_1^x f(t) ~\mathrm dt$ and $\int_1^x ~\mathrm g(t) dt$ diverges, then the other diverges as well, thanks to the limit comparison test. In this case, it is true that $$\int_1^x f(t) ~\mathrm dt \sim \int_1^x g(t) ~\mathrm dt.$$ [Both sides approach $\infty$ as $x \to \infty$.] See this post for a proof.

Convergent case: If either one of the integrals $\int_1^x f(t) ~\mathrm dt$ and $\int_1^x g(t) ~\mathrm dt$ converges, then the other converges as well, again thanks to the limit comparison test. In this case, just from the definition of the improper integrals $\int_1^{\infty} f(t) ~\mathrm dt$ and $\int_1^{\infty} g(t) ~\mathrm dt$, we have $$\int_1^x f(t) ~\mathrm dt \to \int_1^{\infty} f(t) ~\mathrm dt$$ and $$\int_1^x g(t) ~\mathrm dt \to \int_1^{\infty} g(t) ~\mathrm dt.$$ However, of course, the two integrals might converge to different limits; i.e., we need not necessarily have the equality $$\int_1^{\infty} f(t) ~\mathrm dt \stackrel{\color{Red}{??}}{=} \int_1^{\infty} g(t) ~\mathrm dt.$$ Therefore, it is not necessarily true that $$\int_1^{x} f(t) ~\mathrm dt \stackrel{\color{Red}{??}}{\sim} \int_1^{x} g(t) ~\mathrm dt.$$

• Your second bullet point is not correct unless you impose some continuity condition since one of the function may have a non-integrable singularity at finite $x$ which does not change its asymptotic properties at all. Dec 6, 2011 at 1:03
• @BebopButUnsteady You're right. I will assume the functions are continuous everywhere. Thanks. Dec 6, 2011 at 2:12