Fourier Representation of Dirac's Delta Function This question is related to this other question on Phys.SE.
In quantum mechanics is often useful to use the following statement:
$$\int_{-\infty}^\infty dx\, e^{ikx} = 2\pi \delta(k)$$
where $\delta(k)$ is intended to represent Dirac's Delta Function. I would like to understand this statement, or at least know a justification of it, rather than blindly apply this result. From what I currently understand about this topic the equation above should be the Fourier representation of the Dirac's Delta Function, however I don't see how to prove it. Furthermore, since the Delta Function is not even a function, this statement appears to me as really strange.
Keep in mind that I am no expert on this topic, and an elementary explanation is what I seek. I would prefer a proof suited for an undergraduate student rather than a really rigorous and complex one.
 A: I'll put a rigorous explanation first, then a loosey-goosey one afterwards.
This is all rooted in distribution theory. I'll work in $\mathbb{R}^n$ and use the convention that the Fourier transform has a $(2\pi)^{-n/2}$ out front (making it unitary), as well as the more standard sign. That is, $$\mathcal{F}f(\xi)=(2\pi)^{-n/2}\int\limits_{\mathbb{R}^n}f(x)e^{-ix\cdot\xi}\, dx$$
The Dirac delta is an example of a tempered distribution, a continuous linear functional on the Schwartz space. We can define the Fourier transform by duality: $$\langle\mathcal{F} u,\varphi\rangle=\langle u,\mathcal{F}\varphi\rangle$$ for $u\in\mathcal{S}'$ and $\varphi\in\mathcal{S}.$ Here, $\langle \cdot,\cdot\rangle$ denotes the distributional pairing. In particular, the Fourier inversion formula still holds. So, for $u=\delta,$
$$\langle\mathcal{F}\delta, \varphi\rangle=\langle\delta,\mathcal{F}\varphi\rangle=\mathcal{F}\varphi(0)=\langle (2\pi)^{-n/2},\varphi\rangle\implies \mathcal{F}\delta=(2\pi)^{-n/2}.$$ Now, the inversion formula gives that $$(2\pi)^{n/2}\delta=\mathcal{F}1,$$ and $\mathcal{F}1$ "equals" $$(2\pi)^{-n/2}\int\limits_{\mathbb{R}^n}e^{-ix\cdot \xi}\, dx$$ (sign in the exponential doesn't matter here). This is what you wrote if $n=1$.
Since you also wanted a less rigorous answer, this is how you might see it done in physics books:
Loosely, $$\mathcal{F}\delta(\xi)=(2\pi)^{-1/2}\int\limits_{-\infty}^\infty \delta(x)e^{-ix\xi}\, dx=(2\pi)^{-1/2}e^{-ix\xi}|_{x=0}=(2\pi)^{-1/2},$$ so "Fourier inversion" gives
$$\delta(x)=(2\pi)^{-1/2}\int\limits_{-\infty}^\infty \mathcal{F}\delta(\xi)e^{ix\xi}\, d\xi=(2\pi)^{-1}\int\limits_{-\infty}^\infty e^{ix\xi}\, d\xi.$$
Of course, these formal calculations are made rigorous by doing what I original wrote.
A: Let the Fourier transform of a function $f$ be
$$
\mathcal{F}\{f(x)\} = \int_{-\infty}^{\infty} f(x) \, e^{-ikx} \, dx.
$$
Then the Fourier transform of the Dirac delta-function (well, actually it's not a function, but the calculations work anyways) is
$$
\mathcal{F}\{\delta(x)\} = \int_{-\infty}^{\infty} \delta(x) \, e^{-ikx} \, dx = 1.
$$
According to the Fourier inversion theorem, if $\mathcal{F}\{f(x)\} = F(k)$ then $\mathcal{F}\{F(x)\} = 2\pi\,f(-k).$ Applying this, we get
$$
\int_{-\infty}^{\infty} e^{-ikx} \, dx
= \int_{-\infty}^{\infty} 1(x) \, e^{-ikx} \, dx
= \mathcal{F}\{1(x)\}
= 2\pi\,\delta(k)
.
$$
By symmetry we also have
$$
\int_{-\infty}^{\infty} e^{ikx} \, dx
= 2\pi\,\delta(k)
.
$$
A: I thought that it might be instructive to present a way forward that uses a regularization of the Dirac Delta.  To that end we proceed.


PRELIMINARIES:
Let $\displaystyle \delta_L(k)=\frac1{2\pi}\int_{-L}^Le^{ikx}\,dx$.  Then, we can write
$$\delta_L(k)=\begin{cases}\frac{\sin(kL)}{\pi k}&,k\ne0\\\\\frac L\pi&,k=0\tag1\end{cases}$$
The function  $\delta_L(k)$ has the following properties:

*

*For each $L$, $\delta_L(k)$ is an analytic function of $k$.

*$\lim_{L\to \infty} \delta_L(0)= \infty$

*$\left|\int_{-\infty}^x \delta_L(k')\,dk'\right|$ is uniformly bounded.

*$\lim_{L\to \infty}\int_{-\infty}^{k}\delta_L(k')\,dk'=u(k)$, where $u$
is the unit step funciton.

*For each $L>0$, $\int_{-\infty}^\infty \delta_L(k)\,dk=1$
While heuristically $\delta_L(k)$ "approximates" a Dirac Delta when $L$ is "large," the limit of $\delta_L(k)$ fails to exist.  However, if we interpret this limit in the distributional sense, then $\lim_{L\to\infty}\delta_L(k)\sim\delta(k)$.  We will now show that this is indeed that case.


ANALYSIS:
Let $\phi(k)\in S$ where $S$ is the Schwarz Space of functions.
We will now evaluate the limit
$$\begin{align}
\lim_{L\to \infty}\int_{-\infty}^\infty \delta_L(k)\phi(k)\,dk=\lim_{L\to \infty}\int_{-\infty}^\infty \frac{\sin(kL)}{\pi k}\phi(k)\,dk\tag1
\end{align}$$
Integrating by parts the integral on the right-hand side of $(1)$ with $u=\phi(k)$ and $v=\int_{-\infty}^{kL}\frac{\sin(x)}{\pi x}\,dx$ reveals
$$\begin{align}
\lim_{L\to \infty}\int_{-\infty}^\infty \delta_L(k)\phi(k)\,dk&=-\lim_{L\to \infty}\int_{-\infty}^\infty \phi'(k)\int_{-\infty}^{kL}\frac{\sin(x)}{\pi x}\,dx\,dk\tag2
\end{align}$$
Using Property 3 in the Preliminaries section, there exists a number $C$ such that $\left|\phi'(k)\int_{-\infty}^{kL}\frac{\sin(x)}{x}\,dx\right|\le C\,|\phi'(k)|$.  Inasmuch as $C|\phi'(k)|$ is  integrable, the Dominated Convergence Theorem guarantees that
$$\begin{align}
\lim_{L\to \infty}\int_{-\infty}^\infty \delta_L(k)\phi(k)\,dk&=-\lim_{L\to \infty}\int_{-\infty}^\infty \phi'(k)\int_{-\infty}^{kL}\frac{\sin(x)}{\pi x}\,dx\,dk\tag3\\\\
&=-\int_{-\infty}^\infty \phi'(k)\lim_{L\to \infty}\left(\int_{-\infty}^{kL}\frac{\sin(x)}{\pi x}\,dx\right)\,dk\\\\
&=- \int_{-\infty}^\infty \phi'(k)\underbrace{u(k)}_{\text{Unit Step}}\,dx\\\\
&=-\int_0^\infty \phi'(k)\,dk\\\\
&=\phi(0)
\end{align}$$
Therefore, in the sense of distributions as given by $(3)$, we assert that $\lim_{L\to\infty}\delta_L(k)\sim \delta(k)$ whereby rescaling yields the distributional limit
$$\bbox[5px,border:2px solid #C0A000]{\lim_{L\to \infty}\int_{-L}^Le^{ikx}\,dx\sim 2\pi \delta(k)}$$
as was to be shown!
