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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.

  • $\begingroup$ Hi! It's been a while. I hope you're staying safe and healthy during the pandemic. If you would, please let me know how I can improve my answer. I really want to give you the best answer I can. And feel free to up vote an answer as you see fit. ;-) $\endgroup$
    – Mark Viola
    Feb 28, 2021 at 20:01

3 Answers 3


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.

  • $\begingroup$ By $\langle\cdot,\cdot\rangle$, do you mean the usual Hilbert-space inner product on $\mathrm L^2(\Bbb R)$, using the fact that the Schwartz space $\mathrm S(\Bbb R)$ is a subspace of $\mathrm L^2(\Bbb R)$ ? $\endgroup$ Sep 4, 2020 at 13:49
  • $\begingroup$ Good comment, I meant the distributional pairing! It coincides with the real $L^2$-pairing if the distribution's action is integration (which is the case for e.g. $L^p$ functions). $\endgroup$
    – cmk
    Sep 4, 2020 at 13:49
  • $\begingroup$ Also, while the Schwartz space is a subspace of $L^2$ (as a vector space), but we given it a different topology. The Schwartz space is not complete with respect to the standard $L^2$ topology (not closed, it's dense!), so we like to give it something else. In particular, we give it a Frechet space topology generated by semi-norms $\|\varphi\|_{\alpha,\beta}=\sup_x|x^\alpha D^\beta \varphi|.$ $\endgroup$
    – cmk
    Sep 4, 2020 at 13:59
  • $\begingroup$ The condition $\langle\mathcal{F} u,\varphi\rangle=\langle u,\mathcal{F}\varphi\rangle$ is that the Fourier transform is self-adjoint. How do you know this is true? Did you mean to write $\langle\mathcal{F^{\dagger}} u,\varphi\rangle=\langle u,\mathcal{F}\varphi\rangle = \langle\mathcal{F^{-1}} u,\varphi\rangle?$ If so, how do you know what form the inverse Fourier transform takes/how do you know it is unitary? $\endgroup$
    – Jbag1212
    Oct 27, 2021 at 14:11
  • $\begingroup$ @Jbag1212 in my post, the angled brackets denote the duality/distributional pairing, not the complex $L^2$ inner product. $\endgroup$
    – cmk
    Nov 2, 2021 at 13:51

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 $$ \int_{-\infty}^{\infty} e^{ikx} \, dx = 2\pi\,\delta(k) . $$

  • $\begingroup$ If $\int \exp(ikx)\,dx = \int \exp(-ikx)\,dx = 2\pi \delta(k)$, then I get $k = \pi n/x, n \in \mathbb{Z}$. As $x$ is a continuous real variable, this means that the two expressions for $\delta(k)$ is true for any $k \in \mathbb{R}$. What do you think about this conclusion? $\endgroup$
    – rainman
    Oct 28, 2021 at 8:20
  • $\begingroup$ @rainman. I don't understand it. It looks like you have asserted $e^{i2kx}=1.$ But why? Do you think that $\int_{-\infty}^{\infty} \exp(ikx)\,dx = \int_{-\infty}^{\infty} \exp(-ikx)\,dx$ requires $\exp(ikx)=\exp(-ikx)$? It doesn't. $\endgroup$
    – md2perpe
    Oct 28, 2021 at 8:27
  • $\begingroup$ Yes, I thought that I requires $\exp(ikx) = \exp(-ikx)$. Could you please explain why this is not the case? $\endgroup$
    – rainman
    Oct 28, 2021 at 8:29
  • $\begingroup$ @rainman. Let's look at another example: let $f(x)=1$ when $0<x<1$ and $f(x)=0$ otherwise, and let $g(x)=1$ when $-1<x<0$ and $g(x)=0$ otherwise. Then $\int_{-\infty}^{\infty} f(x) \, dx = 1 = \int_{-\infty}^{\infty} g(x) \, dx$ although $f(x)$ and $g(x)$ are not equal. Two functions can have the same integral over $\mathbb{R}$ without being equal. $\endgroup$
    – md2perpe
    Oct 28, 2021 at 8:37
  • $\begingroup$ @rainman. For any integrable function $f$ one has $\int_{-\infty}^{\infty} f(-x) \, dx = \int_{-\infty}^{\infty} f(x) \, dx$. That doesn't require $f(x)=f(-x).$ $\endgroup$
    – md2perpe
    Oct 28, 2021 at 8:41

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.


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:

  1. For each $L$, $\delta_L(k)$ is an analytic function of $k$.
  2. $\lim_{L\to \infty} \delta_L(0)= \infty$
  3. $\left|\int_{-\infty}^x \delta_L(k')\,dk'\right|$ is uniformly bounded.
  4. $\lim_{L\to \infty}\int_{-\infty}^{k}\delta_L(k')\,dk'=u(k)$, where $u$ is the unit step funciton.
  5. 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.


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!

  • $\begingroup$ @Noumeno Hi! It's been a while. I hope you're staying safe and healthy during the pandemic. If you would, please let me know how I can improve my answer. I really want to give you the best answer I can. And feel free to up vote an answer as you see fit. ;-) $\endgroup$
    – Mark Viola
    Feb 28, 2021 at 20:00
  • $\begingroup$ The most rigorous answer (+1) $\endgroup$
    – Svyatoslav
    Oct 27, 2021 at 15:46
  • $\begingroup$ @svyatoslav Thank you! Much appreciated. $\endgroup$
    – Mark Viola
    Oct 27, 2021 at 19:00

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