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Given a positive function $f(x)$ that diverges as $x \rightarrow \infty$, and an asymptotic expression $\tilde{f}(x)$ for $f$ around $x=+\infty$, an asymptotic expansion for $I(x) = \int_0^x f(u) du$ is given by $\tilde{I}(x) = \int_0^x \tilde{f}(u) du$. But up to what order?

Suppose for instance that $I(x) \sim x^2$. The integral $\tilde{I}(x) $ may yields terms that are $\mathcal{O}(x^2)$ but also $\mathcal{O}(x)$ and $\mathcal{O}(x^0)$. How do we know which terms are part of a correct asymptotic series?

I would expect that terms of order $\mathcal{O}(x^0)$ are not reliable -- because moving the lower bound of integration would change those terms but not the diverging part of the integral. Is that correct? If it is, how would you proceed to compute corrections of order $\mathcal{O}(x^0)$?

EDIT: I can see how to prove that $I(x) \sim \tilde{I}(x)$ (by finding $\forall \epsilon >0$ an $M(δ)$ such that for all $x > M(\delta)$ etc). Where I'm really lost is what to make of the extra terms that are part of $\tilde{I}$.

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    Define asymptote, then use that. – 2017-01-09
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    Google what asymptote (mathematics) is. :D – 2017-01-09
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    By "diverges as $x\to \infty$", do you mean $f(x) \to +\infty$ (or $\to -\infty$) as $x\to +\infty$, or just that $f(x)$ doesn't have a limit as $x\to +\infty$? – 2017-01-09
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    Yes, it tends to $\infty$. Say it tends to $+ \infty$. – 2017-01-09
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    You probably mean $M(\epsilon)$ instead of $M(\delta)$. Yes, that or something equivalent. Now you use that $\frac{f(x)}{\tilde{f}(x)} \to 1$, so given $\eta > 0$, for all $x \geqslant X(\eta)$ you have $(1-\eta)\tilde{f}(x) < f(x) < (1+\eta)\tilde{f}(x)$. You can pretty much ignore the integrals up to $X(\eta)$, those only give some constant whose contribution vanishes in the limit. Estimate $$\frac{\int_{X(\eta)}^{x} f(u)\,du}{\int_{X(\eta)}^{x} \tilde{f}(u)\,du}.$$ – 2017-01-09
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    @Daniel Fischer I can see how $I(x) \sim \tilde{I}(x)$. But using that fact yields some extra terms, and that's really what I'm wondering about. I've made my question a bit more specific. – 2017-01-09
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    Well, it depends on the difference $e(x) = f(x) - \tilde{f}(x)$. If you have a bound $\lvert e(x)\rvert \leqslant h(x)$, then you know that $$\Biggl\lvert \int_0^x f(u) - \tilde{f}(u)\,du\Biggr\rvert \leqslant \int_0^x h(u)\,du.$$ Typically, only terms of $\int_0^x \tilde{f}(u)\,du$ that tend to $\infty$ as $x\to \infty$ survive. Everything that doesn't grow faster than $\int_0^x h(u)\,du$ is uncertain. – 2017-01-09

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