Formal derivation of the Fourier transform of Dirac delta using a distribution 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?
 A: 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, $\widehat{\delta} = (2\pi)^{-1/2}$.
A: You forgot to rescale the Gaussian. For $D_n(x)= n D(nx)$, we have that
$$\tag{1}\mathcal F[D_n](p)=\mathcal F[D]\left(\frac p n\right), $$
so, applying the formula mentioned in the question, we see that
$$\tag{2}
\mathcal F[D_n](p)=\frac{ e^{-\frac{p^2}{4n^2}} }{\sqrt{2\pi}}\to \frac{1}{\sqrt{2\pi}}$$
as $n\to \infty$, as expected.
The convergence in (2) is in the sense of tempered distributions. Indeed, letting $\phi$ denote a Schwartz function, we have that
$$
\int_{-\infty}^\infty \frac{ e^{-\frac{p^2}{4n^2}} }{\sqrt{2\pi}}\phi(p)\, dp \to \int_{-\infty}^\infty \frac{ 1 }{\sqrt{2\pi}}\phi(p)\, dp$$
by dominated convergence. The right-hand side is exactly the pairing of the constant function $\frac{1}{\sqrt{2\pi}}$ with the test function $\phi$, which proves that the convergence in (2) holds in the sense of tempered distributions.
Conclusion. This answer contains a rigorous proof that $\mathcal{F}(\delta)=\frac{1}{\sqrt{2\pi}}$ using rescaled Gaussian, as requested.
Remark. We used a Gaussian but that is not necessary. Any integrable $D=D(x)$ such that $\mathcal F[D](0)=1$ will do, by the exact same reasoning.
