What can be a Fourier transform's domain? As I understand it, Fourier inversion theorem states that, for a Schwartz function $f: \mathbb{R} \to \mathbb{C}$, Fourier transform 
$$\mathcal{F}f(\omega) = \int_{\mathbb{R}} f(t) e^{i \omega t} dt $$
and its inverse
$$\mathcal{F}^{-1}f(t) = \frac{1}{2\pi} \int_{\mathbb{R}} f(\omega) e^{-i \omega t} d\omega$$
the following holds:
$$ \mathcal{F}^{-1}\mathcal{F}f(t) = f(t). $$
Yet Fourier transforms of generalized functions like Dirac delta are considered. Is that justified? Why?
Also, what about considering a function $f: \mathbb{C} \to \mathbb{C}$ and taking pointwise limit as $\Im[t] \to 0$? Would a Fourier inversion theorem be true in this limit?
 A: Addressing your first question:
Denote the space of Schwartz functions, $S(\mathbb{R})$, and consider the dual space, $S(\mathbb{R})^*$. If $T$ is in $S(\mathbb{R})^*$, then $T: S(\mathbb{R}) \mapsto \mathbb{R}$ is a linear functional. If $\phi \in S(\mathbb{R})$, you often you will see $T$ acting on $\phi$ written as $
\langle T, \phi \rangle$.
Define $\delta$ to be such that:
$$
\langle \delta, \phi \rangle = \phi(0)
$$
Clearly, $\delta \in S(\mathbb{R})^*$. Also, note that if $f$ is a locally integrable function, you can define $T_f$ such that:
$$
\langle T_f, \phi \rangle = \int_{\mathbb{R}} f(x) \phi(x)dx
$$
and $T_f \in S(\mathbb{R})^*$.

Now then, the Fourier transform on $S(\mathbb{R})^*$ is defined as 
$$
\langle FT, \phi \rangle = \langle T, F\phi \rangle
$$
Why? Consider when $f: \mathbb{R} \mapsto \mathbb{R}$ and $Ff$ "makes sense". Then:
\begin{align*}
\langle FT_f, \phi \rangle &= \int_{\mathbb{R}} \left[ \int_{\mathbb{R}} f(y) \exp(-ixy)dy\right]\phi(x)dx \\
&= \int_{\mathbb{R}} \left[ \int_{\mathbb{R}} \phi(x) \exp(-ixy)dx \right] f(y)dy \\
&= \langle T_f, F\phi \rangle
\end{align*}

Consider again the Dirac delta $\delta$ defined above.
\begin{align*}
\langle F\delta, \phi \rangle &= \langle \delta , F\phi \rangle \\
&= (F\phi)(0) \\
&= \int_{\mathbb{R}} \phi(x) dx \\
&= \int_{\mathbb{R}} 1 \cdot \phi(x) dx \\
&= \langle T_1, \phi \rangle
\end{align*}
So, the Fourier transform of $\delta$ is the linear functional acting on the space $S(\mathbb{R})$ induced by the constant, $1$. Now, here's the cool part that we can't do by treating things in the "normal" way. We can find the Fourier transform of 1:
\begin{align*}
\langle FT_1, \phi \rangle &= \langle T_1, F\phi \rangle \\
&= \int_{\mathbb{R}} (F\phi)(x) dx \\
&= 2\pi \frac{1}{2\pi} \int_{\mathbb{R}} (F\phi)(x) \exp(ix\cdot 0) dx \\
&= 2\pi \phi(0) \\
&= \langle 2\pi\delta, \phi \rangle \\
\end{align*}
So $FT_1 = 2\pi \delta$.
A: Concerning the second question, here's the (small)${}^{[1]}$ result mentioned in comments. Assume that the function $f\colon \mathbb{R}\to \mathbb{C}$ is integrable on the real line and satisfies the estimate 
\begin{equation}
\left\lvert f(x)\right\rvert\le C e^{-a \lvert x \rvert}
\end{equation}
where $a>0$. (The value of $C>0$ is not important). Then the Fourier transform 
$$\tag{1} \mathcal{F}(f)(z)=\int_{-\infty}^\infty f(x) e^{-ix z}\, dx$$ 
is well-defined and analytic for $z\in S_a$, the latter denoting the strip
$$
S_a=\left\{ z\in\mathbb{C}\ :\ \left\lvert \Im z\right\rvert < a\right\}.
$$
To prove this one writes $z=\xi+i\eta$ and observes that 
$$
\left\lvert f(x) e^{-izx}\right\rvert\le Ce^{-\lvert x \rvert (a-\lvert \eta\rvert)}, $$ 
so the integral in $(1)$ is convergent for $\lvert \eta\rvert < a$. The convergence is also uniform on compact subsets of $S_a$, so it defines an analytic function because $f(x)e^{-ixz}$ is an analytic function of $z$. Indeed, one can decompose $f(x)e^{-ixz}$ into a Taylor series and the locally uniform convergence allows for termwise integration. $\square$
The following example${}^{[2]}$ shows that in general one cannot take a strip bigger than $S_a$: 
$$
\mathcal{F}\left( \frac{e^{-a\lvert x\rvert}}{2} \right) (z)= \frac1a\frac{1}{1+(z/a)^2}.$$
The right hand side has poles at $\pm ai$. Therefore the biggest strip of analyticity that contains the real axis is $S_a$, which is the one predicted by the result above. 

Notes.
${}^{[1]}$ It is indeed the "baby" version of a family of famous results in analysis known as Paley-Wiener's theorems.
${}^{[2]}$ This computation arises when solving the Laplace equation on the half-plane. Look for the keyword "Poisson kernel" for more information. It is also easy enough to compute directly.
