Identifying $\int_{-\infty}^\infty e^{i k x} dx$ as Dirac delta distribution

Question

The expression $\int_{-\infty}^\infty e^{i k x} dx$ is sometimes identified as the Dirac delta function. This identification is said "formal" or "symbolic", and some physics texts say that the theory of distribution makes this identification rigorous. So I read some texts on distributions and I think now I understand what delta function is as a distribution. However, I still cannot give any meaninig on the expression as a distribution. Is it possible to give a well defined meaning as a mathematical object to the expression?

Answer 1

The way to give meaning to an expression as a distribution is to see its effect on a test function. If $\phi(x)$ is a test function (a member of the Schwartz space $\mathscr S(\mathbb R)$), then the action of the formal expression $T = \int_{-\infty}^\infty e^{ikx}\; dx$ on $\phi$ is obtained by integration (and switching the order of the integrals): $$ T(\phi) = \int_{-\infty}^\infty dx \int_{-\infty}^\infty dk \; \phi(k) e^{ikx} = \int_{-\infty}^\infty dx \; \hat{\phi}(-x/(2\pi)) = 2 \pi \phi(0)$$

Answer 2

$$ \langle \int_{-R}^{R}e^{ikx}dk,\varphi\rangle=\int_{-\infty}^{\infty}\varphi(x)\int_{-R}^{R}e^{ikx}dkdx \\ = \int_{-R}^{R}\int_{-\infty}^{\infty}\varphi(x)e^{ikx}dx dk $$ As $R\rightarrow\infty$ the above tends to $2\pi\varphi(0)$. So the truncated integral converges in the distribution sense to a scalar times the Delta function at $0$: $$ \int_{-R}^{R}e^{ikx}dk \rightarrow 2\pi\delta_{0}. $$ The improper integral converges in the dual. It's an element of $\mathscr{S}^*$.