How do you take the Fourier transform of a function when the integral that would define its transform doesn’t converge? The answer is similar to how you can differentiate a non-differentiable function: you take a theorem from ordinary functions and make it a definition for generalized functions. I’ll outline how this works below.
Generalized functions are linear functionals on smooth (ordinary) functions. Given an ordinary function f, you can create a generalized function that maps a smooth test function φ to the integral of fφ.
There are other kinds of generalized functions. For example, the Dirac delta “function” is really the generalized function δ that maps φ to φ(0). This is the formalism behind the hand-wavy nonsense about a function infinitely concentrated at 0 and integrating to 1. “Integrating” the product δφ is really applying the linear functional δ to φ.
Now for absolutely integrable functions f and g, we have
In words, the integral of the Fourier transform of f times g equals the integral of f times the Fourier transform of g. This is the theorem we use as motivation for our definition.
Now suppose f is a function that doesn’t have a classical Fourier transform. We make f into a generalized function and define its Fourier transform as the linear function that maps a test function φ to the integral of f times the Fourier transform of φ.
More generally, the Fourier transform of a generalized function f is the linear function that maps a test function φ to the action of f on the Fourier transform of φ.
This allows us to say, for example, that the Fourier transform of the constant function f(x) = 1 is 2πδ, an exercise left for the reader.
The Heisenberg uncertainty principle for ordinary functions says that the flatter a function is, the more concentrated its Fourier transform and vice versa. Generalized Fourier transforms take this to an extreme. The Fourier transform of the flattest functions, i.e. constant functions, are multiples of the most concentrated function, the delta (generalized) function.