# $\int_{a}^{b}f(x)\cos(kx)dx\rightarrow 0 (k\rightarrow \infty)$ 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!

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Maybe... Riemann Lebesgue Lemma? In that case some integrability conditions are needed.. $f\in L^1([a,b])$ or similar.. en.wikipedia.org/wiki/Riemann%E2%80%93Lebesgue_lemma –  uforoboa Nov 15 '12 at 15:17
The proof isn't very pretty - more technical. So don't waste time looking for a pretty one... –  Peter Sheldrick Nov 15 '12 at 17:27

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 $(2)$.

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It looks as if your proof is for Riemann integrable functions. The case for Lebesgue integrable functions is more general; it allows the domain to be $\mathbb{R}$. –  robjohn Nov 15 '12 at 20:12
@robjohn Yes, my proof is for Riemann integrable functions, which is what the OP, I guess, was asking. =) –  Pedro Tamaroff Nov 15 '12 at 21:59

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.

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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.

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