Evil Fourier Coefficients Let $f:[0,1]\to[0,1]$ be the classical devil's staircase.
Has anybody ever computed (or studied) the fourier coefficient of  $f(x)$?
Related question: is the fourier series of $f(x)-x$ normally convergent (with respect to uniform norm)?
 A: The Fourier transform of the derivative $\mu$ of the Devil staircase is explicitely stated on the wikipedia page of the Cantor distribution, in the table at the right,
under the heading "cf" (characteristic function). Its value is
$$ \int_0^1 e^{itx} d\mu(x) = e^{it/2}\ \ \prod_{k=1}^\infty \cos(t/3^k)$$ 
Just multiply by $-1/it$, add $1/it$, and you get the Fourier transform of the Devil staircase.
A word on the proof. The Cantor distribution is the weak limit of the functions obtained by summing the indicator functions of the 2^n intervals generating the Cantor set at the nth step
(after renormalization). The Fourier transform of these sums can be computed explicitely. Then let n goes to infinity.
A: I might start by thinking about the Riemann--Stieltjes integral $\varphi(t) = \int_0^1 e^{itx} \; df(x)$.  Since $f$ is cumulative probability distribution, the $n$th moment of that distribution would be $E(X^n) = \varphi^{(n)}(t)$ where $X$ is a random variable so distributed.  The $n$th moment depends in a well-understood way on the first $n$ cumulants.  Then I'd try to use self-similarity together with the law of total cumulance to figure out what the cumulants are.
Having written that, I see at this article that I knew the cumulants several years ago; I think I added them to that Wikipedia article.  (The odd-order cumulants are zero because of symmetry.)
