What is $\lim_{t\rightarrow \infty }\int_{a}^{b}f(x,\sin(tx))dx$？

Question

Let f $\in C([a,b]\times [-1,1]) $,What is the value of this integral: $$\lim_{t\to\infty}\int_{a}^{b}f(x,\sin(tx))dx.$$ This is a topic I encountered when I was studying mathematical analysis recently, I want to know how to solve this problem, is there some theoretical background?
First of all, f is a bounded continuous binary function, so we can get that this integral must also be bounded.Then I tried the commutation method so that tx=u, but I couldn't push the final result.
This is the result of my appeal operation： $$\lim_{t\to\infty}\frac{1}{t}\int_{at}^{bt}f(\frac{u}{t},\sin(u))du$$ I would appreciate it if you could solve it.

Answer 1

I think that $$ \lim_{t \to \infty}\int_a^b f(x,\sin(tx)) dx = \frac{1}{2\pi}\int_a^b \int_0^{2\pi} f(x,\sin(s)) ds dx. $$

First, let $g \in C^1([a,b])$ and note that for $n \in \mathbb{Z}\setminus \{0\}$ $$ \int_a^b g(x)\exp(itnx) dx \to 0 \quad (t \to \infty), $$ which can be proved by partial integration. Let $n \in \mathbb{N}_0$ and consider $$ \int_a^b g(x) (\sin(tx))^n dx = \int_a^b g(x) \frac{1}{(2i)^n}(\exp(itx) - \exp(-itx))^n dx $$ $$ =\int_a^bg(x) \frac{1}{(2i)^n} \sum_{k=0}^n {n\choose k} (-1)^k\exp(it(n-2k)x) dx $$ $$ \to\left\{ \begin{array}{cc} 0, & 2 \nmid n \\ \frac{(-1)^{n/2}}{(2i)^n}{n\choose n/2}\int_a^bg(x) dx,& 2 \mid n \end{array} \right. \quad (t \to \infty). $$ With a similar reasoning we get $$ \frac{1}{2\pi}\int_0^{2\pi} (\sin(x))^n dx = \frac{1}{2\pi}\int_0^{2\pi} \frac{1}{(2i)^n} \sum_{k=0}^n {n\choose k}(-1)^k \exp(i(n-2k)x) dx $$ $$ =\left\{ \begin{array}{cc} 0, & 2 \nmid n \\ \frac{(-1)^{n/2}}{(2i)^n}{n\choose n/2}, & 2 \mid n \end{array} \right. $$ Thus $$ (\ast) \quad \int_a^b g(x) (\sin(tx))^n dx \to \int_a^b \frac{1}{2\pi}\int_0^{2\pi} g(x)(\sin(s))^n ds dx \quad (t \to \infty). $$ Now let $f \in C([a,b] \times [-1,1])$ and let $\varepsilon > 0$. By the Weierstrass Approximation Theorem there is some polynomial $p(x,y)$ such that $$ |f(x,y)-p(x,y)| \le \varepsilon \quad ((x,y) \in [a,b] \times [-1,1]). $$ Application of $(\ast)$ to the monomials $x^m(\sin(tx))^n$ leads to $$ \int_a^b p(x,\sin(tx)) dx \to \int_a^b \frac{1}{2\pi}\int_0^{2\pi} p(x,\sin(s)) ds dx \quad (t \to \infty). $$ Thus, there is some $t_0 > 0$ such that for $t \ge t_0$: $$ |\int_a^b p(x,\sin(tx)) dx -\int_a^b \frac{1}{2\pi}\int_0^{2\pi} p(x,\sin(s)) ds dx| < \varepsilon, $$ hence $$ |\int_a^b f(x,\sin(tx)) dx - \int_a^b \frac{1}{2\pi}\int_0^{2\pi} f(x,\sin(s)) ds dx | $$ $$ \le |\int_a^b f(x,\sin(tx)) dx - \int_a^b p(x,\sin(tx)) dx| $$ $$ + |\int_a^b p(x,\sin(tx)) dx -\int_a^b \frac{1}{2\pi}\int_0^{2\pi} p(x,\sin(s)) ds dx | $$ $$ +|\int_a^b \frac{1}{2\pi}\int_0^{2\pi} p(x,\sin(s)) ds dx - \int_a^b \frac{1}{2\pi}\int_0^{2\pi} f(x,\sin(s)) ds dx | $$ $$ \le (b-a) \varepsilon + \varepsilon + (b-a) \varepsilon = (2(b-a)+1)\varepsilon. $$