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Let $f$ be a real-valued, $L^1$ integrable function on the interval $[a,b]$. Then the Riemann-Lebesgue Lemma tells us that: $\int_a^bf(x)\sin(2\pi nx)dx\rightarrow0 \text{ as } n\rightarrow\infty.$

Does this have any asymptotic estimate attached to it? i.e. for sufficiently nice $f$ (say continuously differentiable), do we have the estimate that, say: $\int_a^bf(x)\sin(2\pi nx)dx= O(1/n)$ or something similar?

Any kind of reference would also be appreciated!

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    Please see also my [answer here](http://math.stacke$x$change.com/questions/43467/proving-that-the-fourier-coefficients-for-a-pretty-smooth-function-are-pretty-sma/43509#43509). Eric's excellent answer is of course sufficient but the answer to which I linked is for a similar question.2011-07-19

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For convenience, lets work on the interval $[-\pi,\pi]$. Riemann Lesbegue just says that the Fourier coefficients of $f$ go to zero for an integrable function. If $f$ is in $C^1(\mathbb{T})$ then we have that $\hat{f'}(n)=in\hat{f}(n)$. (Here $\mathbb{T}$ refers to $[-\pi,\pi]$ with the endpoints identified.) Since $f'$ is continuous it will be integrable, so Riemann Lesbegue implies the coefficients are $o(1)$. Consequently the coefficients of $f$ are $o\left(\frac{1}{n}\right)$, and hence $\int_{-\pi}^\pi f(x)\sin(nx)dx=o\left(\frac{1}{n}\right).$

For a function $f\in C^k(\mathbb{T})$ we get

$\int_{-\pi}^\pi f(x)\sin(nx)dx=o\left(\frac{1}{n^k}\right).$

What if $f\in C^1[-\pi,\pi]$, but $f(-\pi)\neq f(\pi)$?

Since the coefficients of the Sawtooth Function have order $\frac{1}{n}$, we will not have a result as strong as before. (The sawtooth function is $C^{\infty}[-\pi,\pi]$).

We can prove that if $f\in C^1[-\pi,\pi]$, but $f(-\pi)\neq f(\pi)$, then the fourier coefficients will be of order $\frac{1}{n}$.

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    A remark, Theorem 4.1 in Katznelson's book tells us that R-L can not be improved http://tailieuhoctap.files.wordpress.com/2007/01/katznelson-y-an-introduction-to-harmonic-analysis-cup-2004299s_mcf_.pdf2011-12-15