It's a common exercise in measure theory courses to solve problems like the following:

Compute $$\lim_{n \to \infty} \int_0^\infty (1+ (x/n))^{-n} \sin(x/n)\,\mathrm{d}x.$$

(cf. Folland Ch 2.4, exercise 28a).

The dominated convergence theorem makes short work of things like this, since (for large $n$) we get roughly exponential decay. A little bit of work turns this intuition into a pointwise bound, and then we swap $\lim \int = \int \lim$ to solve the problem.

Often in computer science, combinatorics, etc. this is not enough. We're interested in the rate of convergence. That is, it's not enough to know

$$ \lim_{n \to \infty} \int_0^\infty (1+ (x/n))^{-n} \sin(x/n)\ dx = \int_0^\infty 0\ dx = 0 $$

We want to say that

$$\int_0^\infty (1+ (x/n))^{-n} \sin(x/n)\ dx = O(1/n),$$ or something similar.

I'm not well versed with computing integrals while using estimates of this kind, but I'm sure that people have thought about this. Is there a way to get these kinds of error estimates from the dominated (resp. monotone) convergence theorem? I would be happy to use this problem (or any others that you may feel are more emblematic of the technique) as a case study.

I suspect one could argue by integrating some error bounds between $f_n$ and $f$, and it may be helpful to look at, say $f_n \cdot \chi_{[0,n]}$ in order to make the error bounds more useful. This feels like it is kind of sidestepping the convergence theorems entirely, though, and I'm curious if there's a more elegant approach.

Thanks in advance! ^_^

  • $\begingroup$ In this case, using $|\sin\frac{x}{n}|\leq \frac{x}{n}$ does the job, doesn't it? You can apply the DCT on the upper bound, and you get the $1/n$ factor "for free". $\endgroup$
    – Clement C.
    Mar 21, 2021 at 22:38
  • $\begingroup$ @ClementC. it doesn't seem like a very close upper bound considering the domain $\endgroup$
    – Henry Lee
    Mar 21, 2021 at 22:44
  • $\begingroup$ @HenryLee it will give the convergence rate the OP is looking for, which is optimal up to constant factors. $\endgroup$
    – Clement C.
    Mar 21, 2021 at 22:48
  • $\begingroup$ @ClementC. -- For what it's worth, I'm not interested in estimating this precise integral. I'm more interested in general techniques for computing asymptotic estimates from places where one might normally use the D/MCT. That said, I should have probably picked a better example, because you're right -- that bound looks like it solves this problem. $\endgroup$ Mar 21, 2021 at 23:00
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    $\begingroup$ @HallaSurvivor Yes, I figured this was not your ultimate goal -- just wanted to mention this ad hoc bound, in this specific case. $\endgroup$
    – Clement C.
    Mar 21, 2021 at 23:01

2 Answers 2


Asymptotic analysis- especially of integrals with infinite bounds- is somewhat of an art. There exists a proliferation of techniques whose goal is to estimate the leading behavior of an integral with respect to some interesting limit. A good place to start diving into the field is the book "Introduction to asymptotics" by D.S. Jones (and of course there are other readily googlable choices).

The convergence theorems you mention unfortunately cannot provide any information on how to perform asymptotic estimates on integrals. Sometimes, procedures and lemmata used to prove that DCT or MCT are applicable may hint at the right asymptotic behavior (this was hinted in the comment section).

In this particular example, it is very simple to obtain arbitrarily high orders of an asymptotic formula involving only powers of $n$. Simply rewrite $(1+x/n)^{-n}=e^{-n\ln(1+x/n)}$ and Taylor expand the integrand in powers of $1/n$:

$$\sin(x/n)e^{-n\ln(1+x/n)}=\frac{xe^{-x}}{n} + \frac{x^3e^{-x} }{2 n^2} + \frac{x^3 e^{-x} (-4 - 8 x + 3 x^2)}{ 24 n^3}+\mathcal{O}(n^{-4})$$

Due to the presence of $e^{-x}$ on every single one of the coefficients, the integrals can be computed in this case and we end up with the asymptotic estimate


You can verify that you can obtain asymptotic estimates of any order, and that order by order, the limit $n^{m+1}(I(n)-\sum_{m=1}^N c_m/n^m)$ exists and is finite. However, you may complain that this series does not converge, and you would be right, but that's the nature of most asymptotic series. As this list comprised of exactly one technique for solving asymptotics is far from being conclusive, I really do hope this helps!

  • $\begingroup$ Great, thank you! For the book by Jones, do you mean the one which uses nonstandard techniques? $\endgroup$ Apr 8, 2021 at 0:23
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    $\begingroup$ Sure, it's not really that different than any ordinary treatise. "Asymptotic methods" by de Bruijn is also good, and "Advanced Mathematical Methods" by Carl Bender, Chapter 6-8 $\endgroup$ Apr 8, 2021 at 1:52
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    $\begingroup$ I wasn't doubting it! I just wanted to make sure, because I'm always a bit surprised when I encounter nonstandard analysis in the wild :P Thanks again ^_^ $\endgroup$ Apr 8, 2021 at 1:53
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    $\begingroup$ @HallaSurvivor you're welcome! Great blog by the way, I learned something from your talk on decidability so I pressed the subscribe button hard on your youtube channel :3 Keep up the enthusiasm! $\endgroup$ Apr 8, 2021 at 3:05
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    $\begingroup$ That's very kind of you <3 I'm glad you enjoyed ^_^ $\endgroup$ Apr 8, 2021 at 5:29

I think the answer in total generality has to be no. For any sequence $a_n$ with a limit of $0$, Let $f_n$ be the function which is equal to $a_n$ on the closed unit interval $[0, 1]$ and 0 elsewhere.

Then you could apply the dominated convergence theorem to get $\lim_{n \to \infty} \int f_n = 0$, but of course you can also just compute the integral, which is $a_n$. But $a_n$ can decay as slowly as you like, e.g. take $a_n = 1/\log(\log(n))$ (for $n$ large).


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