Fourier coefficients of sin(1/k), i.e. essential discontinuity Consider on $[-\pi,\pi]$ the function
$$f(k) = sin \big(\frac{\pi^2} k\big).$$
This function has an essential discontinuity at $k=0$ and is smooth otherwise. My question is what are the Fourier coefficients of f, i.e.
$$a_n = \frac 1 {2\pi} \int_{-\pi}^\pi d k\; e^{i n k} f(k) =\ ???.$$
Can $a_n$ be computed explicitly? What is the bahaviour of $a_n $ as $n\to\infty$.
Of course one can ask the general question: What is the behaviour of the Fourier coefficents/Fourier transform if the function has an essential discontinuity? Any reference?
 A: One can show in this case that $|a_n|=C_n n^{-3/4}+O(1/n),|C_n| \le C$ by applying some theorems from the exponential sums theory.
The idea is that it is enough to compute $a_n = \frac i {\pi} \int_{0}^\pi \sin nx f(x)dx=\frac i {2\pi} \int_{0}^\pi (\cos (nx -\pi^2/x)-\cos (nx +\pi^2/x))dx, n \ge 1$ and split the integral at $1/n$ estimating trivially by $1/n$ the integral from $0$ to $1/n$
But now $nx +\pi^2/x$ has a critical point at $y^2 = \pi^2/n, y \ge 1/n$, while $nx -\pi^2/x$ has no critical points so by the second mean theorem
$\int_{1/n}^\pi \cos (nx -\pi^2/x)dx=$
$=\int_{1/n}^\pi (\cos (nx -\pi^2/x))'/(n+\pi^2/x^2)dx=1/(n+1)\int_{c}^\pi (\cos (nx -\pi^2/x))'dx=O(1/n)$
while for the other integral we split into three parts from $1/n$ to $y/2$, from  $y/2$ to $2y$ and from $2y$ to $\pi$; for $\int_{2y}^\pi \cos (nx +\pi^2/x)dx$ we can again apply the second mean theorem since the derivative $n-\pi^2/x^2$ is monotonic and non zero now and get again an $O(1/n)$ as above and we can do same for $\int_{1/n}^{y/2} \cos (nx +\pi^2/x)dx$ and get same $O(1/n)$
For the integral around the critical point we apply the standard exponential integral estimate:
$\int_{y/2}^{2y} e^{i(nx +\pi^2/x)}dx=e^{(\pi/4+ny+\pi^2/y)i}/(4\pi^2 (1/y^3))^{1/2}+O(R)$ and $R <<y(1/y^5) y^6 + y (1/y^8) y^9 +y^3(1/y) << 1/n$ and take the real part to get the estimate $ 1/(2\pi) \cos (\pi/4+2 \pi \sqrt n)(n/\pi)^{-3/4}+O(1/n) $ as claimed
Putting all together we get the required estimate $|a_n| =C_n n^{-3/4}+O(1/n)$
Note that this result depends decisively on the form of $f(x)$ and would change if $f(x)=\sin (1/x^2)$ for example, but the above method would still work and get some similar estimates from the critical point integral estimate (for this latter example $y^3$ is about $n$, the second derivative is about $y^{-4}$ so the estimate gives a main term $C_ny^2$ which is $C_n n^{-2/3}$ plus a remainder still $O(1/n)$)
