Asymptotic estimate for Riemann-Lebesgue Lemma 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!
 A: 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 $\frac{1}{n}$.
