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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.. – 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
up vote 5 down vote accepted

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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