Finding Limit of an Integral: $\lim_{n\to\infty}\int_a^b f(x)\sin^3{(nx)} \:dx$

Parisina 07/31/2017. 3 answers, 492 views
integration analysis limits continuity uniform-convergence

Suppose $f:[a,b]\to\mathbb{R}$ is continuous. Determine if the following limit exists

$$\lim_{n\to\infty}\int_a^b f(x)\sin^3{(nx)} \:dx.$$

As $f(x)$ and $\sin^3{(nx)}$ are continuous, so their product is Riemann integrable. However $\lim_{n\to\infty} f(x)\sin^3{(nx)} $ does not exist, so it's not uniformly convergence and we cannot pass the limit inside the integral. It also doesn't satisfy in the conditions of Dini Theorem. I don't know how to make a valid argument for this problem, but I think by what I said the limit doesn't exist. I appreciate any help.

3 Answers


Robert Israel 07/31/2017.

Riemann-Lebesgue lemma. Note that $\sin^3(nx) = \frac{3}{4} \sin(nx) - \frac{1}{4} \sin(3nx)$.

2 comments
Parisina 07/31/2017
Thanks, I think, I can complete it now
Teepeemm 07/31/2017
That seems to be more advanced than the problem is calling for.

Sangchul Lee 07/31/2017.

A slightly different way of solving this is to use the following observation.

Proposition. If $f : [a, b] \to \mathbb{R}$ is continuous, $g : \mathbb{R} \to \mathbb{R}$ is continuous and $L$-periodic, then

$$ \lim_{n\to\infty} \int_{a}^{b} f(x)g(nx) \, dx = \left( \int_{a}^{b} f(x) \, dx \right)\left( \frac{1}{L}\int_{0}^{L} g(x) \, dx \right). $$

  1. Assuming this statement, the answer follows immediately since $x \mapsto \sin^3 x$ is $2\pi$-periodic and

    $$ \int_{0}^{2\pi} \sin^3 x \, dx = 0. $$

  2. The intuition is very clear: If $n$ is very large, then on subinterval $[c,c+\frac{L}{n}] \subset [a, b]$ we have

    $$ \int_{c}^{c+\frac{L}{n}} f(x)g(nx) \, dx \approx f(c) \int_{c}^{c+\frac{L}{n}} g(nx) \, dx = f(c) \cdot \frac{1}{n} \int_{0}^{L} g(x) \, dx. $$

    So ignoring details, we would have

    $$ \int_{a}^{b} f(x)g(nx) \, dx \approx \left( \sum_{k=1}^{\lfloor n(b-a)/L \rfloor} f\left(a+\frac{kL}{n}\right) \frac{1}{n} \right)\left( \int_{0}^{L} g(x) \, dx \right) $$

    and taking limit as $n\to\infty$, the right-hand side converges to the desired value. Filling in the details is quite routine.

  3. The assumption on continuity is just a technical setting for simple proof, and you can relax them to certain degrees by paying more effort.


Michael Hartley 07/31/2017.

You can't conclude $$\lim_{n\rightarrow\infty} \int_a^b g(x,n)dx$$ doesn't exist just because $$\lim_{n\rightarrow\infty} g(x,n)$$ doesn't. For example, $$\lim_{n\rightarrow\infty} \sin(nx)$$ doesn't exist, but $$\lim_{n\rightarrow\infty} \int_0^\pi \sin(nx) dx = 0,$$ since the integral is zero for all $n$.

I'm afraid my usefulness runs out at this point, though I think the limit exists: you should, if nothing else, be able to find some epsilon-delta argument expressing the integral as the sum of a bunch of integrals on intervals of length $\frac{2\pi}{n}$. This may be a very bad way to tackle the problem.

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