TITLE: How to prove that $\lim\limits_{n\to\infty}\int\limits _{a}^{b}\sin\left(nt\right)f\left(t\right)dt=0\text { ? }$
QUESTION [5 upvotes]: Let $f:\left[a,b\right]\to\mathbb{R}$ be a function that is derivative so that $f'$ is continuous then
$$
\lim_{n\to\infty}\int\limits _{a}^{b}\sin\left(nt\right)f\left(t\right)dt=0
$$ 
My attempt: I want to show that $\forall\epsilon>0 \ \exists \ n_0 \in  \mathbb{N}$ so that $\forall n>n_0 \ \ \ \left|\int\limits _{a}^{b}\sin\left(nt\right)f\left(t\right)dt\right|<\epsilon$.
Let there be $\epsilon>0$  because $f$ and $\sin$ are derivative and $f'$ is continuous we can use integration by parts so that
$$
\left|\int\limits _{a}^{b}f'\left(t\right)\sin\left(nt\right)dt\right|=\left|\left[f\left(t\right)\sin\left(nt\right)\right]_{a}^{b}-n\int\limits _{a}^{b}f\left(t\right)\cos\left(nt\right)\right|\leq\left|\left[f\left(t\right)\sin\left(nt\right)\right]_{a}^{b}\right|+\left|n\int\limits _{a}^{b}f\left(t\right)\cos\left(nt\right)\right|
$$
That is my main idea. any suggestions?

REPLY [4 votes]: For future references, this is a special case of the Riemann Lebesgue Lemma (well, this is the Riemann Lebesgue Lemma, really). It states:

T Let $f:[a,b]\to\Bbb R$ be Riemann integrable over $[a,b]$. Then $$\lim_{\lambda\to\infty}\int_a^b f(t)\sin\lambda t dt=0$$

P First, observe this is true for constant functions, since $$\int_a^b K\sin\lambda tdt=K\frac{-\cos(\lambda b)+\cos(\lambda a)}{\lambda}\to 0$$
It follows that it is also true for a step function, that is, a piecewise constant function, say $$s(x)=\sum_{i=1}^n s_i{\bf 1}_{X_i}$$ where the $X_i=[x_{i-1},x_i]$ are subintervals of $[a,b]$, since $$\int_a^b s(t) \sin\lambda tdt=\sum_{i=1}^n s_i\int_{x_{i-1}}^{x_i} \sin\lambda tdt$$
Let $\epsilon >0$ be given. Since $f$ is Riemann integrable there exists a step function $s\leq f$ defined on $[a,b]$ such that $$\int_a^b f-\int_a^b s<\frac{\epsilon}2$$ 
Since the lemma is true for step functions, there exists $M$ such that $\lambda>M$ implies $$\left| {\int_a^b {s\left( t \right)\sin \lambda tdt} } \right|<\frac{\epsilon}2$$
But $$\begin{align}
  \left| {\int_a^b f (t)\sin \lambda tdt} \right| &= \left| {\int_a^b {\left( {f(t) - s\left( t \right) + s\left( t \right)} \right)\sin \lambda tdt} } \right| \\ 
   &\leq  \left| {\int_a^b {\left( {f(t) - s\left( t \right)} \right)\sin \lambda tdt} } \right| + \left| {\int_a^b {s\left( t \right)\sin \lambda tdt} } \right| \\ 
   &\leq  \int_a^b {\left| {f(t) - s\left( t \right)} \right|dt}  + \left| {\int_a^b {s\left( t \right)\sin \lambda tdt} } \right| \\
   &= \int_a^b {\left( {f(t) - s\left( t \right)} \right)dt}  + \left| {\int_a^b {s\left( t \right)\sin \lambda tdt} } \right| \\ 
   &< \frac{\epsilon }{2} + \frac{\epsilon }{2} = \epsilon  \end{align} $$
so the theorem is proven. $\blacktriangle$.