Fourier Transform, for entire function On THIS site, Alexandre used Fourier transform to solve the problem.
If we use Fourier transform, how to define it to ensure any entire function has a FT?
Classical FT is defined by
$$ \mathcal{F}[f] = F(\xi) = \frac{1}{\sqrt{2 \pi}}\int_{-\infty}^{+\infty}f(z)\mathrm{e}^{-\mathrm{i} \xi z} \mathrm{d} z. $$
This only work for $f \in L^1(\mathbb{R})$. (If improved, it can work for $f \in L^2(\mathbb{R})$.)
I know $\mathcal{F}[\mathrm{e}^{sz}] = \sqrt{2 \pi} \delta(\xi - \mathrm{i}s)$, but I'm not sure about a general definition.
 A: There are two questions. 


*

*For the specific functional equation considered in 
this question On equation f(z+1)-f(z)=f'(z), the formula I gave covers all entire solutions. I added the references there.

*On the general question about "Fourier transform" of entire functions or functions on the real
line which are not in $L^p$. One usually replaces Fourier transform with various versions of Laplace
transform. There are many versions, for various problems. I recommend Hormander, Analysis of differential operators..., Chap. 9, or the paper 
MR0199747 Ljubič, Ju. I.; Tkačenko, V. A. Theory and certain applications of the local Laplace transform. (Russian) Mat. Sb. (N.S.) 70 (112) 1966 416–437. There is an English translation
in Math USSR Sbornik. There is also a nice little book by Carleman of Fourier transform (in French).
Edit. See also On linear independence of exponentials for an example how Laplace transform of entire functions is used.
A: The Fourier Transform $\mathcal{F}$ is at first defined on the Schwartz space $\mathcal{S}$ and is a linear isomorphism there. As always, there is a dual operator $\mathcal{F}^\prime$ that is an isomorphism on the dual space $\mathcal{S}^\prime$ of tempered distributions.
This operator $\mathcal{F}^\prime$ is the extension you are looking for, as $\mathcal{S}$ and also more general functions (for example polynomials) can be regarded as subset of $\mathcal{S}^\prime$.
However, as far as I know, not ANY entire function has a Fourier transform, only those that also lie in $\mathcal{S}^\prime$.
A: There is a definition of Fourier transforms for distributions, not just tempered distributions. The Fourier transform of a test function is an entire function of exponential growth, and the Fourier transforms of distributions are defined by duality. The Fourier transforms of distributions are known as analytic functionals.
You may find an exposition of this topic in the monograph by Gelfand and Shilov.
