# Are integral representations of the Dirac Delta formally equivalent to the Dirac Delta distribution?

I know that the Dirac Delta "function" $$\delta(x)=\begin{cases}\infty&x=0\\0&x\neq0\end{cases},\int_{\Bbb{R}}\delta(x)\,\mathrm{d}x=1$$ is supposedly a distribution, not a genuine function, (although I know nothing about distributions and what that really means). I also know that it has various integral representations, of which the one I know best I will present here:

$$\delta(\omega)=\frac{1}{2\pi}\int_{\Bbb{R}}\exp(it\cdot\omega)\,\rm{dt}$$

Formally I assume this integral does not actually exist for $$\omega\neq0$$, since $$\lim_{n\to\pm\infty}\exp(in\cdot\omega)$$ does not exist due to oscillating behaviour, so this representation raises some concerns for me. This representation is used to present answers to Fourier transforms, and that integral (which I do not believe exists) is "calculated" and whisked under the rug of Dirac Delta. Is this a correct way to define the distribution?

I am aware that in physics/engineering, it is common to heuristically use $$\delta$$ like this, with such representations, and treat it as a valid function, since the real world is too complex to treat everything with absolute rigour, but in the world of rigorous mathematics, I ask: is this correct?

For example:

\begin{align}\mathcal{F}(\cos)(\omega)&=\frac{1}{\sqrt{2\pi}}\int_{\Bbb{R}}\cos(t)\exp(it\cdot\omega)\,\mathrm{dt}\\&=\frac{1}{2\sqrt{2\pi}}\int_{\Bbb{R}}\exp(it(\omega+1))+\exp(it(\omega-1))\,\mathrm{dt}\\&\overset{\color{red}*}=\sqrt{\pi/2}\cdot(\delta(\omega+1)+\delta(\omega-1))\end{align}

Which correctly suggests that cosine consists of pure tones at the radial frequencies $$\omega=\pm1$$ (although the multiplication by $$\sqrt{\pi/2}$$ means what, exactly?)

Is this, namely the step denoted by $$\color{red}*$$, formal (in the context of distributions)? Or is this just some standard physicist's heuristic?

Many thanks! Note that I really do not know any distribution theory, but am just very curious at where the line between rigorous Dirac/Fourier and heuristic Dirac/Fourier is drawn!

The "definition" you give of $$\delta$$ is not mathematically rigorous. The correct definition is that $$\langle \delta, \phi \rangle = \phi(0)$$ for all $$\phi\in C^\infty_c(\mathbb{R})$$ or $$\phi\in \mathcal{S}(\mathbb{R})$$ when working with Fourier transforms. Here $$\langle u, \phi \rangle$$ is a pairing of a distribution $$u$$ with a test function $$\phi,$$ returning a number. It's a bit similar to inner products, but the objects are of different types. You can think of it as the integral $$\int_{\mathbb{R}} u(x)\,\phi(x)\,dx.$$

An ordinary function $$f$$ is considered as a distribution by $$\langle f, \phi \rangle := \int f(x) \, \phi(x) \, dx.$$

The formula $$\delta(\omega)=\frac{1}{2\pi}\int_{\mathbb{R}} e^{i\omega t}\,dt$$ is also not rigorous, but I would say that it is at least somewhat better as it can be interpreted within the theory of distributions as the inverse Fourier transform of constant function $$\mathbf{1}(t) = 1.$$

Fourier transforms of distributions are defined by moving the transform to the test function: $$\langle \mathcal{F}u, \varphi \rangle = \langle u, \mathcal{F}\varphi \rangle.$$ From this and the definition of $$\delta$$ above we get $$\langle \mathcal{F}\delta, \phi \rangle = \langle \delta, \mathcal{F}\phi \rangle = \mathcal{F}\phi(0) = \left. \int \phi(t) e^{-i\omega t} \, dt \right|_{\omega=0} = \int \phi(t) \, dt = \langle \mathbf{1}, \phi \rangle.$$ Now, $$\langle \int_{\mathbb{R}} e^{i\omega t} dt, \phi(\omega) \rangle = \langle \mathcal{F}\mathbf{1}(-\omega), \phi(\omega) \rangle = \langle \mathbf{1}(-\omega), \mathcal{F}\phi(\omega) \rangle = \langle \mathbf{1}(\omega), \mathcal{F}\phi(-\omega) \rangle \\ = \langle \mathcal{F}\delta(\omega), \mathcal{F}\phi(-\omega) \rangle = \langle \delta(\omega), \mathcal{F}\mathcal{F}\phi(-\omega) \rangle = \langle \delta(\omega), 2\pi\phi(\omega) \rangle = \langle 2\pi\delta(\omega), \phi(\omega) \rangle ,$$ for every test function $$\phi$$ so $$\int_{\mathbb{R}} e^{i\omega t} dt=2\pi\delta(\omega)$$ where the left hand side has been interpreted as $$\mathcal{F}\mathbf{1}(-\omega).$$

Thus, the integral itself is not rigorous, but there is a rigorous interpretation of it.

• Thank you for your answer. What is $\mathcal{S}(\Bbb{R})$? And what about the Fourier transform of cosine that I showed? Was that rigorous/unrigorous? What would be more correct? Sep 2, 2021 at 14:17
• $\mathcal{S}(\mathbb{R})$ is the Schwartz space. It's closed under Fourier transforms. Sep 2, 2021 at 14:20
• The Fourier transform of $\cos$ can be made rigorous (although the integral you have is not), and the result is $\delta(\omega-1)+\delta(\omega+1)$ times some factor. Sep 2, 2021 at 14:22
• Thank you. So the heuristic Fourier transform I took has the correct answer but a non-rigorous method? Sep 2, 2021 at 14:23
• If the distribution is given by an $L^1$ function then the transpose is given by the exact same integral. Try yourself to show that $\int \mathcal{F}f(\omega) \, g(\omega)\,d\omega=\int f(\omega)\,\mathcal{F}g(\omega)\,d\omega$ when $f,g$ are nice. On a distribution the integral isn't defined and instead the Fourier transform is defined by moving the transform to the test function. This is based on the previous formula. Sep 2, 2021 at 14:44