TITLE: Is there a $C_c^{\infty}( \mathbb{R}^d)$ function whose Fourier transform we can explicitly write down?
QUESTION [15 upvotes]: I noticed that although $C_c^{\infty}$-functions are dense in some quite large spaces and  well understood (especially their Fourier transform) I have never encountered an explicit example of a Fourier transform $F(f)$ for $f \in C_c^{\infty}$ in my life (there are good reasons for this, as most explicit expression for $C_C^{\infty}$ look rather difficult to integrate). Does anybody know an example, where the Fourier transform can be explicitly calculated or are there just no examples?

REPLY [7 votes]: The simplest infinitely differentiable function with compact support is
$f(x)=\exp\left(-1/x(1-x)\right), \; 0<x<1$. This is the first example given in textbooks.
Its Fourier transform is 
$$F(z)=\int_0^1e^{-izx}f(x)dx.$$
Is this an "explicit" formula? In not, why is the infinite product in Terence Tao's answer "explicit"?
(I think the word "explicit" (and "analytic expression", which some people think is synonimous) are very much abused on this site. These words always require a further explanation of their meaning).
Back to $C^\infty$ functions with compact support. The inverse Fourier transform
of $\mathrm{sinc}$ is the characteristic function of an interval.
Thus the example in Tao's answer is an infinite convolution of these
characteristic functions. (Is this explicit or not?) H/"ormander, Analysis of partial differential operators, vol. I, Chap I, sect 1.3 discusses at great length how to construct $C^\infty$ functions with compact support by taking infinite convolutions of characteristic functions, and surprisingly, obtains a proof of the Denjoy-Carleman quasianalyticity theorem by elementary methods:-)