Finding Limit of an Integral: $\lim_{n\to\infty}\int_a^b f(x)\sin^3{(nx)} \:dx$ 
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.
 A: 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.
A: Riemann-Lebesgue lemma.  Note that $\sin^3(nx) = \frac{3}{4} \sin(nx) - \frac{1}{4} \sin(3nx)$.
A: 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
$$ \lim_{n\to\infty} \int_{a}^{b} f(x)g(nx) \, dx = \left( \int_{a}^{b} f(x) \, dx \right)\left( \frac{1}{L}\int_{0}^{L} g(x) \, dx \right). $$



*

*Assuming this statement, the answer follows immediately since $x \mapsto \sin^3 x$ is $2\pi$-periodic and
$$ \int_{0}^{2\pi} \sin^3 x \, dx = 0. $$

*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.

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