Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!

share|improve this question
2  
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
add comment

3 Answers

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 $(2)$.

share|improve this answer
    
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
add comment

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.

share|improve this answer
add comment

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.