Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question
up vote 5 down vote accepted

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 $$ \int_1^{x} f(t) ~\mathrm dt \stackrel{\color{Red}{??}}{\sim} \int_1^{x} g(t) ~\mathrm dt. $$

share|cite|improve this answer
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. – BebopButUnsteady Dec 6 '11 at 1:03
@BebopButUnsteady You're right. I will assume the functions are continuous everywhere. Thanks. – Srivatsan Dec 6 '11 at 2:12

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

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.