Are integral representations of the Dirac Delta formally equivalent to the Dirac Delta distribution? I know that the Dirac Delta "function" $\delta(x)=\begin{cases}\infty&x=0\\0&x\neq0\end{cases},\int_{\Bbb{R}}\delta(x)\,\mathrm{d}x=1$ is supposedly a distribution, not a genuine function, (although I know nothing about distributions and what that really means). I also know that it has various integral representations, of which the one I know best I will present here:
$$\delta(\omega)=\frac{1}{2\pi}\int_{\Bbb{R}}\exp(it\cdot\omega)\,\rm{dt}$$
Formally I assume this integral does not actually exist for $\omega\neq0$, since $\lim_{n\to\pm\infty}\exp(in\cdot\omega)$ does not exist due to oscillating behaviour, so this representation raises some concerns for me. This representation is used to present answers to Fourier transforms, and that integral (which I do not believe exists) is "calculated" and whisked under the rug of Dirac Delta. Is this a correct way to define the distribution?
I am aware that in physics/engineering, it is common to heuristically use $\delta$ like this, with such representations, and treat it as a valid function, since the real world is too complex to treat everything with absolute rigour, but in the world of rigorous mathematics, I ask: is this correct?
For example:
$$\begin{align}\mathcal{F}(\cos)(\omega)&=\frac{1}{\sqrt{2\pi}}\int_{\Bbb{R}}\cos(t)\exp(it\cdot\omega)\,\mathrm{dt}\\&=\frac{1}{2\sqrt{2\pi}}\int_{\Bbb{R}}\exp(it(\omega+1))+\exp(it(\omega-1))\,\mathrm{dt}\\&\overset{\color{red}*}=\sqrt{\pi/2}\cdot(\delta(\omega+1)+\delta(\omega-1))\end{align}$$
Which correctly suggests that cosine consists of pure tones at the radial frequencies $\omega=\pm1$ (although the multiplication by $\sqrt{\pi/2}$ means what, exactly?)
Is this, namely the step denoted by $\color{red}*$, formal (in the context of distributions)? Or is this just some standard physicist's heuristic?
Many thanks! Note that I really do not know any distribution theory, but am just very curious at where the line between rigorous Dirac/Fourier and heuristic Dirac/Fourier is drawn!
 A: The "definition" you give of $\delta$ is not mathematically rigorous. The correct definition is that $\langle \delta, \phi \rangle = \phi(0)$ for all $\phi\in C^\infty_c(\mathbb{R})$ or $\phi\in \mathcal{S}(\mathbb{R})$ when working with Fourier transforms. Here $\langle u, \phi \rangle$ is a pairing of a distribution $u$ with a test function $\phi,$ returning a number. It's a bit similar to inner products, but the objects are of different types. You can think of it as the integral $\int_{\mathbb{R}} u(x)\,\phi(x)\,dx.$
An ordinary function $f$ is considered as a distribution by
$\langle f, \phi \rangle := \int f(x) \, \phi(x) \, dx.$
The formula $\delta(\omega)=\frac{1}{2\pi}\int_{\mathbb{R}} e^{i\omega t}\,dt$ is also not rigorous, but I would say that it is at least somewhat better as it can be interpreted within the theory of distributions as the inverse Fourier transform of constant function $\mathbf{1}(t) = 1.$
Fourier transforms of distributions are defined by moving the transform to the test function: $\langle \mathcal{F}u, \varphi \rangle = \langle u, \mathcal{F}\varphi \rangle.$ From this and the definition of $\delta$ above we get
$$
\langle \mathcal{F}\delta, \phi \rangle
= \langle \delta, \mathcal{F}\phi \rangle
= \mathcal{F}\phi(0)
= \left. \int \phi(t) e^{-i\omega t} \, dt \right|_{\omega=0}
= \int \phi(t) \, dt
= \langle \mathbf{1}, \phi \rangle.
$$
Now,
$$
\langle \int_{\mathbb{R}} e^{i\omega t} dt, \phi(\omega) \rangle
= \langle \mathcal{F}\mathbf{1}(-\omega), \phi(\omega) \rangle
= \langle \mathbf{1}(-\omega), \mathcal{F}\phi(\omega) \rangle
= \langle \mathbf{1}(\omega), \mathcal{F}\phi(-\omega) \rangle
\\
= \langle \mathcal{F}\delta(\omega), \mathcal{F}\phi(-\omega) \rangle
= \langle \delta(\omega), \mathcal{F}\mathcal{F}\phi(-\omega) \rangle
= \langle \delta(\omega), 2\pi\phi(\omega) \rangle
= \langle 2\pi\delta(\omega), \phi(\omega) \rangle
,
$$
for every test function $\phi$ so $\int_{\mathbb{R}} e^{i\omega t} dt=2\pi\delta(\omega)$ where the left hand side has been interpreted as $\mathcal{F}\mathbf{1}(-\omega).$
Thus, the integral itself is not rigorous, but there is a rigorous interpretation of it.
