One of the requirements for the existence of Fourier transform of $f(x)$ is that:
$\int_{-\infty}^{\infty} |f(x)| dx $ exists.
However, the table says that the Fourier transform of constant functions (\emph{i.e.}, $f(x)=1$) do exist and it is $\delta(k)$ although $\int_{-\infty}^{\infty} 1 dx = \infty$ .
Could anyone can help me to understand this? Thanks in advance.
Best Answer
The Fourier transform defined by an ordinary (Riemann or) Lebesgue integral only exists when $f \in L^1$.
It is however possible to extend the definition to tempered distributions (for example, every locally integrable function that "doesn't grow too fast" can be identified with a tempered distribution). The Fourier transform of such a thing is not in general a function though, as witnessed by your example.