Fourier Transform of $sgn(t)$ I'm trying to find the Fourier transform of $sgn(t)$ where
$$
sgn(t) = \begin{cases}
1, & t > 0 \\
0, & t = 0 \\
-1, & t < 0.
\end{cases}
$$
By definition,
$$
X(\omega) = \int_{-\infty}^{\infty} sgn(t) e^{-i\omega t} \, dt = \int_0^{\infty} e^{-i\omega t} \, dt + \int_{-\infty}^0 -e^{-i\omega t} \, dt.
$$
However, that second integral diverges. Where is the flaw in my methodology?
 A: You have correctly deduced that taking the necessary integral in any usual fashion does not give an answer that is a function.  
However, there is a way to get the answer as long as you've defined the fourier transform of the functions $f(x) = 1$ and $f(x) = \delta(x)$.  In particular, we note that
$$
\operatorname{sgn}(t) = -1 + 2\int_{-\infty}^t \delta(\tau)\,d\tau
$$
A: Echoing @Omnonmnomnm's point: the notion of "integral" in Fourier transforms very often needs to be an "extension by continuity" (in a topology that may vary, depending on context), so is in any case absolutely not any kind of Riemann, Lebesgue, or other "integral". Nevertheless, if/when we know what kind of limit the integrand is, and in what sense we want the outcome to have a sense... it is often possible, in real-life situations, to attach useful meaning to "divergent" integrals or sums which in a strict, late 19th-century interpretation, would be meaningless.
That is, "integral" is not literal integral, but is a linear map from one topological vector space to another, sometimes just to scalars. It should be continuous in relevant topologies... though the relevant topologies are not always obvious, and not always classical, not always related to "pointwise convergence", etc.
So it's not that one "defines" the Fourier transform of $1$ and $\delta$, but that one determines what these are, in the extended sense of Fourier transform, and then @Omn.'s computation is what you want.
