Derivation of Fourier Transform of a constant signal I understand that the F.T. of a constant signal is the Dirac. However, I cannot find anywhere showing the derivation or proof for this. I'm trying to do it myself and am getting lost. Can anyone give a worked-out derivation that the Fourier Transform of a constant signal is the Dirac? Thank You for any help!
 A: First, it's trivial that the Fourier transform of the Dirac $\delta$ is a constant function:
$$
\mathcal{F}\{\delta(x)\} = \int_{-\infty}^{\infty} \delta(x) \, e^{-i\xi x} dx
= \left. e^{-i\xi x} \right|_{x=0} = 1.
$$
Then we use the Fourier inversion theorem, saying that if $\mathcal{F}\{f(x)\} = F(\xi)$ then $\mathcal{F}\{F(x)\} = 2\pi \, f(\xi)$:
$$
\mathcal{F}\{1\} = 2\pi\,\delta(\xi).
$$
A: Another approach is to consider the function $e^{-\epsilon x^2/2}.$ Obviously $e^{-\epsilon x^2/2} \to 1$ as $\epsilon \to 0.$ The Fourier transform of $e^{-\epsilon x^2/2}$ is another Gaussian, $C(\epsilon) e^{-\xi^2/(2\epsilon)},$ that tends to $2\pi \, \delta(\xi).$
(Sorry, need to go to bed, so I don't have time to show the calculations.)
A: The derivation is very simple, provided you know what a distribution is. Very briefly: A distribution is a continuous linear functional
$$
L:C_0^\infty(\mathbb{R})\rightarrow\mathbb{C}.
$$
I will not specify what it means for $L$ to be continuous (it's complicated). The Dirac delta distribution is the linear functional
$$
\delta(\varphi):=\varphi(0).
$$
The Fourier transform is defined on a subset of the distributions called tempered distritution. The Fourier transform $\mathcal{F}(L)$ of a (tempered) distribution $L$ is again a (tempered) distribution. It is defined as the linear functional
$$
\mathcal{F}(L)(\varphi):=L(\mathcal{F}(\phi)).
$$
If you want to Fourier transform the constant 1, you first have to identify the constant 1 with a distribution $L_1$. This is done canonically via
$$
L_1(\varphi):=\int_{\mathbb{R}} 1\cdot \varphi(x)\,dx.
$$
Now you can compute the Fourier transform $\mathcal{F}(L_1)$ of $L_1$:
$$
\mathcal{F}(L_1)(\phi) = L_1(\mathcal{F}(\phi)) = 
\int_{\mathbb{R}} 1\cdot \hat\phi(x) dx = 
\int_{\mathbb{R}} e^{2\pi ix\cdot 0}\cdot \hat\phi(x) dx =
\mathcal{F}^{-1}(\hat\phi)(0) = \phi(0) = \delta(\phi).
$$
That is it! We see that the Fourier transform for $L_1$ coincides with the Dirac delta distribution $\delta$. So in the sense of distributions, the the Fourier transform of 1 is the Dirac delta distribution.
A: Yet another solution
Here I first use the formula $\mathcal{F}\{f'(x)\} = i\xi \, \mathcal{F}\{f(x)\}$:
$$
0 
= \mathcal{F}\{0\}
= \mathcal{F}\{\frac{d}{dx}1\}
= i\xi \mathcal{F}\{1\}.
$$
Then I use the fact from distribution theory that the solutions to $x \, u(x) = 0$ are $u(x) = C\,\delta(x),$ where $C$ is a constant:
$$
\mathcal{F}\{1\} = C \, \delta(\xi).
$$
Here we unfortunately don't directly get the value of $C$.
A: Another one...
It's clear that $\chi_{[-R,R]} \to 1$ as $R \to \infty.$ The Fourier transform is $2 \frac{\sin R\xi}{\xi},$ which tends to $2\pi\,\delta(\xi)$ as $R \to \infty.$
