TITLE: How to prove $\lim\limits_{x\to\infty}\int_a^bf(t)\sin(xt)\,dt=0$
QUESTION [5 upvotes]: I need help to prove:

Suppose $f\in C$. Prove that
  $$
\lim\limits_{x\to\infty}\int_a^bf(t)\sin(xt)\,dt=0
$$

My idea is to use substitute of $xt=u$ and prove
$$
\lim\limits_{x\to\infty}\dfrac1{x}\int_{ax}^{bx}f\left(\dfrac{y}{x}\right)\sin(y)\,dy=0
$$
But I am not sure how to do it. Thanks.

REPLY [4 votes]: This is essentially a simplification of the idea presented in hermes answer.

Note that
$$
\int_{a+(k-1)\frac{2\pi}x}^{a+k\frac{2\pi}x}\sin(xt)\,\mathrm{d}t=0\tag{1}
$$
With $n=\left\lfloor(b-a)\frac{x}{2\pi}\right\rfloor$, we have $\left|b-a-n\frac{2\pi}x\right|\le\frac{2\pi}x$ and
$$
\begin{align}
\left|\int_a^bf(t)\sin(xt)\,\mathrm{d}t\right|
&\le\left|\sum_{k=1}^n\int_{a+(k-1)\frac{2\pi}x}^{a+k\frac{2\pi}x}f(t)\sin(xt)\,\mathrm{d}t\right|\\
&+\left|\int_{a+n\frac{2\pi}x}^bf(t)\sin(xt)\,\mathrm{d}t\right|\\
&=\left|\sum_{k=1}^n\int_{a+(k-1)\frac{2\pi}x}^{a+k\frac{2\pi}x}\left(f(t)-f\left(a+k\tfrac{2\pi}x\right)\right)\sin(xt)\,\mathrm{d}t\right|\\
&+\left|\int_{a+n\frac{2\pi}x}^bf(t)\sin(xt)\,\mathrm{d}t\right|\\[12pt]
&\le|b-a|\max_{|s-t|\le\frac{2\pi}x}|f(s)-f(t)|\\
&+\frac{2\pi}x\max_{x\in[a,b]}|f(x)|\tag{2}
\end{align}
$$
Since $f$ is continuous on $[a,b]$ it is uniformly continuous, and therefore,
$$
\lim_{x\to\infty}\max_{|s-t|\le\frac{2\pi}x}|f(s)-f(t)|=0\tag{3}
$$
Thus, $(2)$ and $(3)$ imply that
$$
\lim_{x\to\infty}\int_a^bf(t)\sin(xt)\,\mathrm{d}t\tag{4}
$$