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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?

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Integrate the original by parts. –  Ted Shifrin May 9 '13 at 22:44
    
For something that is nearly identical, check out the Riemann-Lebesgue lemma. Essentially what this is saying is that you can cover $[a,b]$ with intervals on which $f(t)\sin(nt)$ goes through an entire period, and hence integrates to zero. As $n$ gets larger these periods get smaller, and so the "left over" part that you're not covering is a smaller and smaller set. Its not hard to see what happens in the limit. –  Dhruv Ranganathan May 9 '13 at 22:46
    
Deja-vu I asked it yesterday :P math.stackexchange.com/questions/386273/… –  Gastón Burrull May 10 '13 at 0:18

3 Answers 3

up vote 5 down vote accepted

Good idea. You just did the wrong integration by parts. Instead $$ \int_a^b\sin(nt)f(t)dt=\frac{-\cos(nt)f(t)}{n}\Big|_a^b+\frac{1}{n}\int_a^b\cos(nt)f'(t)dt. $$ Now it is easy to see why the rhs tends to $0$.

Note: this follows from Riemann-Lebesgue lemma as soon as $f$ is Lebesgue integrable over the interval. But actually, a convenient proof of the latter goes by proving it for $C^1$ functions and then using their density in $L^1$. Riemann-Lebesgue holds over unbounded intervals also. Then one can consider compactly supported $C^1$ functions.

Fourier coefficients: if you integrate over an interval of length $2\pi$, you get a Fourier coefficient of $f$. So in general, when $f$ is integrable, the Fourier coefficients tend to $0$. But what you see here is that when $f$ is $C^1$, the coefficients are $o\big(\frac{1}{n}\big)$. And going on like this, the more regular the function, the faster the coefficients converge to $0$. If $f$ is $C^k$, the coefficients are $o\big(\frac{1}{n^k} \big)$.

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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. $\blacktriangle$.

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Note that $\lambda\to\infty$ is taken as a "limit in $\Bbb R$", not as a sequential limit, i.e. "$n\to\infty$". –  Pedro Tamaroff May 9 '13 at 23:20
1  
Since $f$ is differentiable, this can be proven more easily than proving the Riemann-Lebesgue Lemma. Nice proof however. –  robjohn May 9 '13 at 23:27
    
@robjohn Thanks. Yes, the point was that this was true in more generality, and maybe the OP will bump into this again. –  Pedro Tamaroff May 9 '13 at 23:28
    
Note however that a classic proof goes by doing the $C^1$ case first, and then obtain the result by $L^1$ density of the latter. What you did is another classic argument, by density of step functions. –  1015 May 9 '13 at 23:30

You're integrating the wrong thing by parts. Use the technique on the actual expression you want the limit of, differentiating $f$ and integrating $\sin$ to get a $1/n$ which is small.

Then have a go at bounding the thing it multiplies by something finite. A very crude estimate will do!

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