Fourier Transform leading to $\delta$: How does the Integration work? So it is well-known that the complex exponential
$$f(t) = e^{i\omega_0t}$$
has Fourier transform
$$F(\omega) = 2\pi \delta(\omega-\omega_0) \ .$$
The transformation integral
$$F(\omega) = \int_\mathbb{R} f(t) e^{-i\omega t} dt \ ,$$
gives
$$2\pi \delta(\omega-\omega_0) = \int_\mathbb{R} e^{i(\omega_0-\omega)t} dt \ .
\tag{1}\label{eq:a} $$
It's clear to me that
$$F(\omega_0) = \int_\mathbb{R} 1 \ dt$$
is infinite and I fully understand that $F(\omega)$ is concentrated into a peak at $\omega_0$ from the spectral-intuition-side-of-things. But what has boggled me ever since is how to argue that 
$$\int_\mathbb{R} e^{iat} dt
\tag{2}\label{eq:b} $$
vanishes for $a \neq 0$. Sure, intuitively $e^{iat}$ has zero mean, but technically the above integral doesn't even exist.
So my questions are:


*

*What is a clean way of showing \eqref{eq:a}, especially the integral vanishing for $\omega \neq \omega_0$, without going deep into harmonic analysis, distribution theory, etc.? Since this topic is taught in many engineering undergraduate programs, this should be possible, right?

*How does Fourier theory manage to give meaning to the divergent integral \eqref{eq:b}?

*I am aware that $\delta$ requires special care because it is a distribution and not a function. Is this the cause of all the trouble?

 A: 

*Yes. You can call it trouble, and you can call it convenience.


In fact, for a distribution $u$, the Fourier transform is not defined as for nice functions, but rather,
$$
\mathcal F(u)(\phi)=u(\mathcal F\phi),
$$
i.e. the Fourier transform of the distribution $u$, when acting on a test function is defined to be the distribution we get when $u$ is acting on the (ordinary) Fourier transform of the test function $\phi$.
It is practical to write things like $\int f(x)\delta_{x_0}\,dx=f(x_0)$ and so on, since many laws of calculation agree with the ones for integrals, but one should be careful, since that integral is not an integral in the ordinary sense. It might be better, in that particular case, to write $\delta_{x_0}(f)=f(x_0)$ and avoid integrals if one is confused about them.
A: Yes, the $δ$ being a distribution and not a function is the problem. To be pedantic, one does not even define $δ(x)$ for $x∈\Bbb R$, i.e. we refrain from allowing the evaluation of $δ$ at $x$. Instead, we define $\delta$ based on how it 'integrates against functions'. 
This is motivated by the fact that for (some of) the purposes of fourier analysis, it is not $g(x)$ we are concerned about, but rather $⟨g, · ⟩$,  how it acts when you integrate against a (nice) function,
$$⟨g,f⟩ := ∫_ℝ  g(x) f(x) \ dx$$
So we make the definition in analogy,
$$⟨ δ, f⟩ := \delta(f) := f(0)$$
Which completely circumvents trying to integrate $δ$.
We also define the Fourier transform by analogy with the situation with nice functions,
$$⟨\mathcal{F}δ , f⟩ := ⟨ δ, \mathcal{F}f⟩ $$
With this definition one can prove that $\mathcal{F}\delta = 1$, 
$$⟨\mathcal{F}δ , f⟩ = ⟨ δ, \mathcal{F}f⟩ = \mathcal{F}f(0) = ∫f(x) \ dx = ⟨ 1,f ⟩ $$
Similarly, $\mathcal{F}1=δ$, which in a sense, answers 2.
But we can abusively write $δ(x) = \lim f_n(x)$ for (e.g) the rescaled normal densities that 'tend to 'δ' (there is a precise definition of this), and in this sense we do have $δ(x) = ∞ \Bbb 1_{x}$.
If you know measure theory, it turns out that signed measures give rise to distributions in a natural way, and the $δ$ distribution is also given by a measure, the dirac delta measure,
$$ δ(A) = \begin{cases} 1 & 0∈ A \\ 0 & 0\not\in{} A\end{cases}$$
and in this sense, we can actually write this as a (Lebesgue) integral,
$$⟨δ,f⟩ = ∫_ℝ f(x) dδ(x) $$
