# Fourier transforms: divergent integrals?

I am studying different integral transform methods, and I am confused on why saying things such as $$\mathcal{F}^{-1} = \delta(x)$$ is valid? If you actually plug this in, $$\mathcal{F}^{-1} = \frac{1}{2\pi}\int_{-\infty}^{\infty}e^{ikx}dx$$ this does not converge for any $${x}$$ at all. It diverges everywhere. I understand the principle value of the integral when $${x\neq 0}$$ is $$0$$, but I don't know why it's valid to just call it $${\delta(x)}$$ based on it's principle value.

• Probably wrong domain. It's not on R but on hmm $[-\pi,\pi]$ or something like that Mar 31, 2021 at 17:22

The Fourier Transform is a Tempered Distribution (See THIS). Hence, we interpret the object $$\mathscr{F^{-1}}\{1\}=\int_{-\infty}^\infty (1) e^{ikx}\,dx$$ as such. That is to say, the object is not an integral.

Rather, the Fourier Transform acts on a function $$\phi\in \mathbb{S}$$, where $$\mathbb{S}$$ is the Schwartz space of functions. Here, the Fourier transform of $$1$$ acts on $$\phi\in \mathbb{S}$$ as follows:

\begin{align} \langle \mathscr{F^{-1}}\{1\}, \phi\rangle &=\langle 1,\mathscr{F^{-1}}\{\phi\}\rangle\\\\ &=\int_{-\infty}^\infty (1)\frac1{2\pi}\int_{-\infty}^\infty \phi(x)e^{ikx}\,dx\,dk\\\\ &=\phi(0)\tag1 \end{align}

where $$(1)$$ is due to the Inverse Transform Theorem.

Therefore, we identify the object $$\mathscr{F^{-1}}\{1\}$$ as the Dirac Delta distribution, $$\delta(k)$$, which has the property such that for any $$\phi \in \mathbb{S}$$ we have

$$\langle \delta,\phi\rangle =\phi(0)$$

• I knew you'd post here! Mar 31, 2021 at 18:31
• Thank you! I'm not familiar with distributions, I am actually studying the transforms as part of a class which is essentially geared towards calculation - but just to check I have understood correctly, you are saying that the interpretation of ${\mathcal{F}^{-1}}$ is not actually an integral in this particular case? I really do appreciate it! Mar 31, 2021 at 19:03
• You're welcome. My pleasure. And Yes. If it were imterpretes as an integral, then it would be an integral that fails to exist. You have it now. Mar 31, 2021 at 19:10

Some initial remarks:

@Mark Viola considers the Fourier transform defined on the space of tempered distributions which contains not only copies of nice functions (integrable functions for example) but also finite measures, and many other things.

The $$\delta$$ "function" can also been considered as a finite Borel measure in the line, the one that assign mass $$1$$ to $$\{0\}$$ and zero mass to any other set.

I will restrict the Fourier transform to the set of measures of finite variation ($$|\mu|(\mathbb{R})<\infty$$), which contains integrable functions (think of $$f\mapsto \mu_f:=f\,dx$$). The Fourier transform there is defined as

$$\widehat{\mu}(t):=\int e^{-ixt}\mu(dx)$$

With that in mind, for the measure $$\delta(A)=1$$ if $$0\in A$$ and $$\delta(A)=0$$ otherwise, one has $$\widehat{\delta}(t)=\int e^{-ixt}\delta(dx)=e^{i0t}=1$$

• It turns out that in the space of finite measures, the map $$\mu\mapsto\widehat{\mu}$$ is also injective.

• There is criteria (Bochner's theorem) that say when a continuous function $$g$$ is the Fourier transform of a (positive) finite measure the inverse Fourier transfer, and threre is also an inversion formula, which is typically study in courses of probability - think of characteristic functions of probability distributions.

• In this contexts, the inverse Fourier transform of $$g(t)\equiv1$$ is the measure $$\delta(dx)$$.