0

The following formula is often used: $$\int_{-\infty}^{+\infty}e^{ikx}dk = \delta(x).$$

This is equivalent to $$\int_{-\infty}^{+\infty}\left( \int_{-\infty}^{+\infty}e^{ikx}f(x)dx\right)dk = 2 \pi f(0).$$

How is the second equality proved? When I try to google the question, many answers make use of the "inverse Fourier transform." However, in order to show that the inverse Fourier transform takes the form that it does, the preceding equality must be used, so this is circular.

How can the second equality be proved (without using the inverse Fourier transform)? I should add that $f(x)$ goes to zero "sufficiently fast" (square integrable), and so the integral in parentheses in the second equality is a real number.

Jbag1212
  • 1,698
  • What sens do you give to the first equality? – blamethelag Oct 26 '21 at 21:50
  • @blamethelag The sense of the first equality is given entirely by the second equality – Jbag1212 Oct 26 '21 at 21:52
  • the formula cannot be proven because is false. The integral$\int_{-\infty}^{+\infty}e^{itx}dt$ doesn't exists, for any chosen $x$. –  Oct 26 '21 at 22:03
  • @Masacroso Although the first formula cannot be "proven" - the second formula most definitely can assuming $f(x)$ goes to zero sufficiently fast. – Jbag1212 Oct 26 '21 at 22:12
  • This is usually done by introducing a regularization using the standard Gaussian (the reason for this choice is that these functions are nice in both time and frequency, after all they are their own Fourier transform), to justify formal manipulations (i.e. the rigorous way of obtaining your second identity from the first by Fubini's theorem) and then taking a limit. Of course this is done for Schwartz functions $f$ (smooth, rapidly-decreasing). – Jose27 Oct 26 '21 at 23:01
  • 1
    https://math.stackexchange.com/questions/3814073/fourier-representation-of-diracs-delta-function/3814112#3814112 – cmk Oct 27 '21 at 03:40

1 Answers1

1

First of all let me clarify that this is defined for test functions only, which are the ones which vanish outside some compact (closed and bounded) interval of $\mathbb{R}$.

First note that what we want to prove is that $\lim_{\lambda\to \infty} \int_{-\lambda} ^\lambda e^{ikx} dk=\delta (x) $. Now, the proof is as follows:

$ \lim_{\lambda\to \infty}\int_{-\infty} ^\infty \int_{-\lambda} ^\lambda e^{ikx}\phi(x) dkdx = \lim_{\lambda\to \infty}\int_{-\infty} ^\infty 2\frac{\sin(\lambda x)}{x}\phi(x) dx $

Where I have trivially evaluated the integral on $k$.

Now it is only necessary to prove that $\frac{\sin(\lambda x)}{x}=\pi\delta(x) $. To do that, just do the change $\lambda=\frac{1}{\epsilon}$ (and $\epsilon \to 0$) and do the change of variable in the integral $\frac{x}{\epsilon}=u$ so that you get the integral of $\frac{\sin(u)}{u}\phi(\epsilon u) $. Then you can pull the limit inside the integral (since the integrand is bounded by $\frac{\sin(u)}{u}\max(\phi(\epsilon u))$ which is integrable and therefore you can use the theorem of dominated convergence), giving $\phi(0)\int_{-\infty} ^\infty\frac{\sin x}{x}dx=\pi\phi(0). \ \ \square$

PS: $ \int_{-\infty} ^\infty e^{ikx} dk=2\pi\delta (x) $ , you didn't put that $2\pi$ in your question.

Sami Calvo
  • 11
  • 1