Highly Oscillating Integrals I'd like to know the behavior of integrals of the form:
$$ 
\int_0^1 f(x) \cos(k x) dx
$$
as $ k \rightarrow \infty $ where f is a smooth function. It is easy to see, by expanding f in power series, that the integral is bounded by $ \approx \frac{1}{k} $, but I'd like to know the coefficient. Is there some way to know this easily?
 A: An important principle of Fourier analysis is that the decay of Fourier coefficients is tied to the smoothness of the function you seek the Fourier coefficients of. The same is true of the Fourier transform.
If $f$ is $k$ times continuously differentiable and periodic (or, equivalently, extends smoothly to a periodic function), then the Fourier coefficients decay like $o(1/|n|^k)$, i.e. $|n|^ka_n \to 0$ as $|n|\to\infty$. This comes out of the Riemann-Lebesgue lemma and an easy induction.
If $f$ is merely Lipschitz continuous, i.e. there exists $C>0$ such that $|f(x)-f(y)|\leq C|x-y|$ for all $x,y$, then its Fourier coefficients decay like $o(1/|n|)$. If $f$ is Holder continuous of order $\alpha$, $0<\alpha\leq 1$, i.e.  there exists $C>0$ such that $|f(x)-f(y)|\leq C|x-y|^\alpha$ for all $x,y$, then its Fourier coefficients decay like $o(1/|n|^\alpha)$. (Lipschitz is the special case of Holder for $\alpha=1$.)
If $f$ is Riemann integrable, then by Parseval's theorem all we get is $\sum a_n^2<\infty$ and therefore $a_n$ are bounded, i.e. $a_n=o(1)$.
Note that mere continuity is a pretty weak condition to put on a function when it comes to Fourier coefficients. There is even a continuous function whose Fourier series diverges at a point. It comes as a modification of the sawtooth function. Of course, it can't be differentiable since Fourier series of a function converge to the function where it is differentiable.
