Let $f \sim g$ mean that $f/g \rightarrow 1$ as $x \rightarrow \infty$. Does it follow that $\int_{1}^{x} f(t)\, dt \sim \int_{1}^{x}g(t)\, dt$?
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2 Answers
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No. Take $f(x) = \frac{1}{x^2}$ and $g(x) =\frac{1}{x^2} +\frac{1}{x^3}$.
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$\begingroup$ Well, don't both integrals converge at the same time? $\endgroup$ Apr 7, 2021 at 17:14
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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 $$ \int_1^{x} f(t) ~\mathrm dt \stackrel{\color{Red}{??}}{\sim} \int_1^{x} g(t) ~\mathrm dt. $$
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$\begingroup$ 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. $\endgroup$ Dec 6, 2011 at 1:03
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$\begingroup$ @BebopButUnsteady You're right. I will assume the functions are continuous everywhere. Thanks. $\endgroup$ Dec 6, 2011 at 2:12