$\int_{a}^{b}f(x)\cos(kx)dx\rightarrow 0 (k\rightarrow \infty)$ only need $f\in L([a,b])$? This is strange result $$\int_{a}^{b}f(x)\cos(kx)dx\rightarrow 0$$
when $k\rightarrow \infty$.
Similarly  under the same condition,$\int_{a}^{b}f(x)\sin(kx)dx\rightarrow 0 (k\rightarrow \infty)$ .Why will have this?
Appreciate your help!
 A: The result is not that strange. What the lemma states is that when the oscillations become faster and faster, the overall area will cancel almost perfectly, and when $\lambda \to\infty$, the cancellation will be ideal. Consider an integrable function over $[a,b]$. This means that for each $\epsilon >0$  there exist step functions $s_1,s_2$ such that $s_1\leq f\leq s_2$ and $$\int_a^b s_2-\int_a^b f<\epsilon$$
$$\int_a^b f-\int_a^b s_1<\epsilon$$
$(1)$ Note that in the particular case $f\equiv \text{constant}$, the lemma is easy to prove.
$$\lim_{\lambda\to\infty}\int_a^b \kappa \cos\lambda x dx=\kappa  \lim_{\lambda\to\infty} \frac{\sin\lambda b }{\lambda}- \frac{\sin\lambda a }{\lambda}=0$$
$(2)$ Similarily, for any step function $s$ with an associated partition $P=\{t_0,\dots,t_n\}$ and constants $\{\sigma_1,\dots,\sigma_n\}$ we have $$\begin{align}
  \mathop {\lim }\limits_{\lambda  \to \infty } \int\limits_a^b {s\left( x \right)\cos \lambda xdx}  & = \mathop {\lim }\limits_{\lambda  \to \infty } \int\limits_{{t_{k - 1}}}^{{t_k}} {\sum\limits_{k = 1}^n {{\sigma _k}\cos \lambda xdx} }   \cr 
  \\ &   = \mathop {\lim }\limits_{\lambda  \to \infty } \sum\limits_{k = 1}^n {\int\limits_{{t_{k - 1}}}^{{t_k}} {{\sigma _k}\cos \lambda xdx} }   \cr 
  \\ &   = \mathop {\lim }\limits_{\lambda  \to \infty } \sum\limits_{k = 1}^n {\frac{{\sin \lambda {t_k} - \sin \lambda {t_{k - 1}}}}{\lambda }}   \cr 
  \\ &   = \sum\limits_{k = 1}^n {\mathop {\lim }\limits_{\lambda  \to \infty } \frac{{\sin \lambda {t_k} - \sin \lambda {t_{k - 1}}}}{\lambda }}   \cr 
  \\ &   = 0 \end{align} $$
$(3)$ Finally, the general case is deduced from $f$ being integrable. For $\epsilon>0$ given choose a suitable $s_1\geq f$. Then
$$\begin{align}
   \left| {\int\limits_a^b {f\cos \lambda xdx} } \right|   &= \left| {\int\limits_a^b {\left( {f + {s_1} - {s_1}} \right)\cos \lambda xdx} } \right|  
   \\&  = \left| {\int\limits_a^b {\left( {f - {s_1}} \right)\cos \lambda xdx}  + \int\limits_a^b {{s_1}\cos \lambda xdx} } \right|  \cr 
    \\&  \leqslant \left| {\int\limits_a^b {\left( {f - {s_1}} \right)\cos \lambda xdx} } \right| + \left| {\int\limits_a^b {{s_1}\cos \lambda xdx} } \right|  \cr 
   \\ &  \leqslant \int\limits_a^b {\left( {f - {s_1}} \right)\left| {\cos \lambda x} \right|dx}  + \left| {\int\limits_a^b {{s_1}\cos \lambda xdx} } \right|  \cr 
   \\ &  \leqslant \int\limits_a^b {\left( {f - {s_1}} \right)dx}  + \left| {\int\limits_a^b {{s_1}\cos \lambda xdx} } \right|  \cr 
   \\ &  <\epsilon  + \epsilon =2\epsilon \end{align} $$
The first $\epsilon$ comes from integrability, and the second from $(2)$.
A: The Riemann-Lebesgue Lemma says that if $f\in L^1(\mathbb{R})$, and
$$
\hat{f}(k)=\int_{\mathbb{R}}f(x)e^{-2\pi ikx}\,\mathrm{d}x
$$
then
$$
\lim_{k\to\infty}\hat{f}(k)=0
$$
Your statements follow from looking at the real and imaginary parts.
A: Actually, it is connected to the coefficients of Fourier series(just the items of series) and that is the reason why your above result is right. 
