There are some nice regular distributions that are defined as follow: $$ e_k(\phi)=\int \phi(x) e^{ikx} dx$$
I have two questions:
The first question I doubt it can be "easily" answered but I'll ask anyway: what does these distributions represent? I mean, the dirac-delta distribution can be understood as a spike at $0$ (or more precisely, associate $\phi (0)$ to $\phi$), it is easy to visualize but these exponential distributions look much more complex!
The second question (more rigorous):
How to prove this equality $$ \delta = \frac{1}{2\pi} \int e_k dk \tag{0}$$
Thank you very much !
The distribution $e_k$ is just the distribution associated to the function $$ f(x) = e^{ikx}. $$ So it represents a complex valued function with norm one and oscillating with period $2\pi/k$. And this is the role of distributions, to extend the notion of functions. Every locally integrable function $f$ can be identified with a distribution acting on smooth test functions $\varphi$ by setting (where I denote $\langle f,\varphi\rangle$ the action of the distribution $f$ on $\varphi$, that you denoted $f(\varphi)$, to avoid the ambiguity with $f(x)$) $$ \langle f,\varphi\rangle := \int_{\Bbb R} f(x)\,\varphi(x)\,\mathrm d x. $$ The advantage of distributions is that it is a class more wide than functions, and for example there are no functions that verify the property of the Dirac delta $\delta_0$, defined as a distribution by $$ \langle \delta_0,\varphi\rangle := \varphi(0). $$ Two distributions $T$ and $S$ are therefore equal if for any nice function $\varphi$ $$ \langle S,\varphi\rangle = \langle T,\varphi\rangle $$ The equality you indicate is false if interpreted in the classical theory of functions and integrals. However, there is a close interpretation in terms of distributions. For a distribution $T$, one defines its Fourier transform by setting $$ \langle \widehat{T},\varphi\rangle := \langle T,\widehat\varphi\rangle, $$ for any nice function $\varphi$ (Schwartz function more precisely). Here $\widehat\varphi$ represents the classical integral formula for the Fourier transform $$ \widehat\varphi(x) = \int_{\Bbb R} e^{-2i\pi yx}\,\varphi(y)\,\mathrm d y. $$ But even if I use the same notation, this might not be true for $\widehat{T}$. In the case when $T$ is a nice function and one can replace the brackets by integrals, this is compatible with the usual property of the Fourier transform, and in this case, $\widehat{T}$ is also given by the usual integral. In particular, an immediate property is the fact that the Fourier transform of the Dirac delta is given by $$ \langle \widehat{\delta_0},\varphi\rangle := \langle \delta_0,\widehat\varphi\rangle = \widehat\varphi(0) = \int_{\Bbb R} \varphi\,\mathrm d x = \langle 1,\varphi\rangle. $$ That is, in the sense of distributions $\widehat{\delta_0} = 1$. Now the property you seek follows from the Fourier inversion theorem, which also works for distributions, yielding $$\tag{1}\label{1} \widehat{1} = \delta_0. $$
Remark: In terms of integrals and functions, looking at the action on a test function of each of the distributions involved in the above identity, yields $\langle 1,\widehat{\varphi}\rangle = \langle\widehat{1},\varphi\rangle = \langle\delta_0,\varphi\rangle$, which can be written $$\tag{2}\label{2} \int_{\Bbb R} \widehat{\varphi} = \varphi(0) $$ that follows from the usual Fourier inversion theorem. This can be written $$\tag{3}\label{3} \int_{\Bbb R} \int_{\Bbb R} e^{-2i\pi yx}\,\varphi(y)\,\mathrm d y\,\mathrm d x = \varphi(0). $$ If it was allowed to use Fubini (but this is not allowed in this case) then we could write $\langle\delta_0,\varphi\rangle = \varphi(0) = \int_{\Bbb R} \varphi(y) \int_{\Bbb R} e^{-2i\pi yx}\,\,\mathrm d x \,\mathrm d y = \langle \int_{\Bbb R} e^{-2i\pi yx}\,\,\mathrm d x,\varphi\rangle$, and so $\delta_0 = \int_{\Bbb R} e^{-2i\pi yx}\,\,\mathrm d x$, but instead, this last integral expression is not well defined. It is however common, in particular in the physics literature, to use the notation $\int_{\Bbb R} e^{-2i\pi yx}\,\,\mathrm d x$ to denote $\widehat{1}$, giving rise to the formula you mention. It should however rather be interpreted as one the equations \eqref{1}, \eqref{2} or \eqref{3}. This use of simplifying notations is good as long as one knows what he is doing (which is the case of most of the professional physicists and mathematicians) but can be dangerous for students or people that do not use these concepts often.