Fourier transform of a triangle Consider a 2-dim regular n-gon whose vertices lie on the unit circle.
Let $\chi_n$ denote the characteristic function of this polygon and
$\widehat{\chi}_n$ its Fourier transform.
The special case n = 4 lends itself particularly well to calculation.
Namely, without much loss of generality, rotate the square so that its
sides are parallel to the coordinate axes.  The result is a product
interval which leads to an immediate determination of
$\widehat{\chi}_4$ as a product of sinc functions.
How does this compare to other values of n ?
Questions: (1) What is the simplest way to evaluate $\widehat{\chi}_3$,
the transform of an equilateral triangle?
(2)  Have the$\;$ $\widehat{\chi}_n$$\,$ been explicitly worked out for small n ?
(3)  Denote by$\,$ $\chi_\infty$ $\,$ the characteristic function of the
unit disk and let$\,$  $\widehat{\chi}_\infty$  be its Fourier transform
(essentially a Bessel function).$\,$  Are there sharp bounds - using any convenient
norm - for the difference $\,$ $\Vert$ $\widehat{\chi}_\infty$ -
$\widehat{\chi}_n$$\Vert$ $\,$ ?   $\;$ Thanks
 A: One way to do this is to set up an integral over a triangle having vertices at the points $(0,1)$, $(-\sqrt{3}/2,-1/2)$, $(\sqrt{3}/2,-1/2)$.  The FT may then be written as
$$\int_{-\sqrt{3}/2}^0 dx \, e^{i k_x x} \, \int_{-1/2}^{-\sqrt{3} x+1} dy \, e^{i k_y y} + \int_0^{\sqrt{3}/2} dx \, e^{i k_x x} \, \int_{-1/2}^{\sqrt{3} x+1} dy \, e^{i k_y y}$$
which is messy but straightforward.  
A: For a $n$-gon $P$ whose boundary in positive orientation is $p_1 \to p_2 \to\ldots \to p_n \to p_1, \ \  p_j = (x_j,y_j)$  and indicator function $\displaystyle\chi(x,y) = 1_{(x,y) \in P}$, then the distributional derivative $\partial_x\chi$  is the distribution $$\partial_x \chi = -\sum_{j=1}^n a_j \delta_{[p_j \to p_{j+1}]}$$ indicating the boundary of $P$, where $a_j = \frac{y_{j+1}-y_j}{\|p_{j+1}-p_j\|}$ is the $\frac{dy}{d\|.\|}$ slope of the edge $[p_j \to p_{j+1}]$ and $\delta_{[p_j \to p_{j+1}]}$ is the distribution defined by $\langle \delta_{[p_j \to p_{j+1}]},\phi \rangle=  \int_{p_j}^{p_{j+1}} \phi(x) d\|x\|$.
Thus $$i \omega_x \widehat{\chi}(\omega) = -\sum_{j=1}^n a_j\int_{p_j}^{p_{j+1}} e^{-i (\omega,u)} d \|u\|= -\sum_{i=1}^n a_i  \frac{e^{-i (\omega,p_{j+1})}-e^{-i (\omega,p_{j})}}{-i(\omega ,p_{j+1}-p_j)}$$
$$ \widehat{\chi}(\omega) = \iint_P e^{-i (\omega,u)} d^2u= \frac{-1}{ \omega_x}\sum_{i=1}^n \frac{x_{j+1}-x_j}{\|p_{j+1}-p_j\|} \frac{e^{-i (\omega,p_{j+1})}-e^{-i (\omega,p_{j})}}{(\omega ,p_{j+1}-p_j)}$$
(ie. reproving Green's theorem but also making the contribution of the edges clear. And I didn't check correctness on an example)
