The defining property of the Dirac delta function is that for any function $f(x)$: $$\int^{\infty}_{-\infty}f(x)\delta(x-a)dx=f(a)$$

The Dirac delta function can be expressed as: $$\delta(x-a) = \frac{1}{2\pi}\int^{\infty}_{-\infty}e^{ip(x-a)}dp$$

How can we show that the latter expression possesses the property defined by the first equation? I.e. is there a proof of the following:

$$\frac{1}{2\pi} \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} f(x)e^{ip(x-a)}dp dx=f(a)$$

  • 1
    $\begingroup$ Your "defining property" is not correct. That equation holds only when $f$ is continuous at $a$. $\endgroup$
    – GEdgar
    Jun 20, 2023 at 12:00

1 Answer 1


First note that the Dirac is a distribibution, not a function. Suppose $f$ is an absolutely integrable and piecewise continous function, then, using the Fourier transfom $\mathcal{F}$, the inversion property :

$$f(a) = \mathcal{F}^{-1}\mathcal{F}(f)(x) \\ = \frac{1}{2\pi}\int e^{-ipa}\int e^{ips}f(s)ds dp$$

With Fubini theorem, you have that :

$$f(a) = \frac{1}{2\pi}\int \int e^{ip(s-a)}f(s)ds dp$$

  • $\begingroup$ Perhaps state what properties $f$ must have for this to work. $\endgroup$
    – GEdgar
    Jun 20, 2023 at 12:03
  • $\begingroup$ Yes you are right, I wanted to make it "unformal". I edit my post. $\endgroup$
    – NancyBoy
    Jun 20, 2023 at 12:08
  • $\begingroup$ @Gaetano But this gives a exp(ip*(a-x)) term instead of exp(ip*(x-a)). We can let p -> (-p), but then dp -> (-dp), so the condition from my question would still not be satisfied. $\endgroup$
    – pll04
    Jun 20, 2023 at 13:42
  • $\begingroup$ I edited my post, the sign in the Fourier transform is just a convention (you just need to keep the opposite when inversion). $\endgroup$
    – NancyBoy
    Jun 20, 2023 at 13:48

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