The Fourier transform of Dirac delta is often naively calculated by considering Delta function as a function that makes sense within an integral and by using its fundamental property:
$$
\int_{-\infty}^{+\infty} f(x) \delta(x-x_0)dx = f(x_0)
$$
Then the Fourier transform of Dirac delta function can be evaluated to be $\hat{\delta(p)} = \frac{1}{\sqrt{2\pi}}$ simply by applying such property to the definition of Fourier transform.
However, here I am seeking for a formal proof using theory of distributions. In particolar, we know that the succession $D_n(x) := nD(nx)$ approximates Dirac delta $\delta(x)$, with $D(x) = \frac{1}{\sqrt{\pi}} e^{-x^2}$.
My objective is to derive the same result above discussed using this approach.
The Fourier transform of a Gaussian-like function is known from theory, therefore it is straightforward that:
$$
\mathcal{F} [ D(x) ]
= \mathcal{F} \left[ \frac{e^{-x^2}}{\sqrt{\pi}} \right]
= \frac{e^{-\frac{p^2}{4}}}{\sqrt{2\pi}}
$$
But this doesn't itself give me directly the solution, as there's still the term $e^{-\frac{p^2}{4}}$; how should I proceed to prove my thesis formally?
Best Answer
In the language of distributions, the Dirac delta distribution is the map $\delta$ from the space of test functions (smooth compactly supported functions) to, say $\mathbb{R}$ with the "operation" $(\delta, f) = f(0)$ for every test function $f$.
To figure out the Fourier transform of a distribution, you need to determine the Fourier transform of a test function $f$.
$$\widehat{f}(\xi) = \frac{1}{\sqrt{2\pi}}\int_\mathbb{R} e^{-ix \xi}f(x) \, dx$$
By definition, the Fourier transform of a distribution $\varphi$ is defined by $(\widehat{\varphi},f)=(\varphi,\widehat{f})$ for every test function $f$.
EDIT: As commenters below pointed out, I should say Schwartz function instead of test function and tempered distribution instead of distribution..
Therefore
$$(\widehat{\delta},f) = (\delta,\widehat{f}) = \widehat{f}(0) = \frac{1}{\sqrt{2\pi}}\int_\mathbb{R} e^{-i x \cdot 0} f(x) \, dx =\frac{1}{\sqrt{2\pi}}\int_\mathbb{R} f(x) \, dx = (\frac{1}{\sqrt{2\pi}},f)$$
where the last equality is because the "constant" distribution 1 is regular, i.e., can be represented in integral form. Therefore, as a distribution, $\widehat{\delta} = (2\pi)^{-1/2}$.