The Dirac delta function can be defined the following way:

$$\delta(x) = \frac{1}{2\pi}\int_{-\infty}^\infty e^{itx}\, dt={\displaystyle {\begin{cases}\frac1{2\pi}\int_{-\infty}^\infty dt,&{\text{if }}x=0\\0,&{\text{if }}x\ne 0.\end{cases}}}$$

The later formula formally follows from the Fourier transform definition.

We also know that $\int_{-\infty}^\infty f(x)\delta(x)dx=f(0)$

It seems that if the expression under integral has a singularity at a point, the integral may gain wonderful changes there. Some other examples:

$$\delta''(x)= -\frac{1}{2\pi}\int_{-\infty}^\infty t^2 e^{itx}\, dt={\begin{cases}-\frac1{2\pi}\int_{-\infty}^\infty t^2 dt,&{\text{if }}x=0\\0,&{\text{if }}x\ne 0.\end{cases}}$$

And $\int_{-\infty}^\infty f(x)\delta''(x)dx=f''(0)$

Another example $(x\in\mathbb{R})$: $$\delta(ai+x)= \frac{1}{2\pi}\int_{-\infty}^\infty e^{-at} e^{itx}\, dt={\begin{cases}\frac1{2\pi}\int_{-\infty}^\infty e^{-at} dt,&{\text{if }}x=0\\0,&{\text{if }}x\ne 0.\end{cases}}$$

And we know that $\int_{-\infty}^{\infty}f(x)\delta(ai+x)dx=f(-ai)$

So, given a divergent integral $S$, an expression $$\phi(x)={\begin{cases}S,&{\text{if }}x=0\\0,&{\text{if }}x\ne 0.\end{cases}}$$

What can we say about the functional $\int_{-\infty}^\infty f(x)\phi(x)dx$? Can we, knowing the functional, find the corresponding divergent integral $S$?

It seems to me, the expressions under the Fourier transform are similar to the algebra of linear operators.


2 Answers 2


Distributions don't work this way. $\delta$ and its derivatives are not functions, that's not what $\delta(x) = \frac{1}{2\pi}\int_{-\infty}^\infty e^{itx}\, dt$ means.

What it means is that $$\delta(x) =\lim_{T\to \infty} \frac{1}{2\pi}\int_{-T}^T e^{itx}\, dt$$ where the limit is in the sense of distributions, which means that for all $\phi \in C^\infty_c(\Bbb{R})$ $$\phi(0)=\langle \delta,\phi \rangle=\lim_{T\to \infty} \int_{-\infty}^\infty \phi(x)(\frac{1}{2\pi}\int_{-T}^T e^{itx}\, dt) dx$$


$$\phi^{(k)}(0)=\langle (-1)^k\delta^{(k)},\phi \rangle=\lim_{T\to \infty} \int_{-\infty}^\infty \phi(x)(\frac{1}{2\pi}\int_{-T}^T (-it)^k e^{itx}\, dt) dx$$

For $a\in \Bbb{C}$ the convergence of $\delta(x-a) =\lim_{T\to \infty} \frac{1}{2\pi}\int_{-T}^T e^{-i at} e^{itx}\, dt$ in the sense of analytic functionals follows the same idea.

  • $\begingroup$ this is more a question about divergent integrals, I use $\delta$ as a symbol for the integral, not as a distribution. $\endgroup$
    – Anixx
    Mar 2, 2021 at 13:32
  • 2
    $\begingroup$ There are no divergent integrals. $\endgroup$
    – reuns
    Mar 2, 2021 at 13:34
  • $\begingroup$ What interests me is that any divergent integral, germ, singularity corresponds to a linear functional, and also to a linear operator. You make a function singular at a point, and the integral of that function jumps at that point in an interesting way. $\endgroup$
    – Anixx
    Mar 2, 2021 at 13:35
  • 2
    $\begingroup$ Learn distributions first. $\endgroup$
    – reuns
    Mar 2, 2021 at 13:35
  • $\begingroup$ I treat divergent integrals like numbers that extend real line. In that case, delta function is a function, but those infinite numbers behave in an interesting way under integrals. $\endgroup$
    – Anixx
    Mar 2, 2021 at 13:39

Following up on @reuns' points:

It is possible to reasonably think about distributions as "generalized functions", but then one must give up "point-wise values" as a description of "functions". Yes, there is still a notion of localization of distributions, and a notion of "wavefront set" and/or "smooth set", and so on, so it is sometimes possible to talk about pointwise values on certain sets. But not altogether.

Similarly, Fourier transform maps (and such) are indeed linear, and continuous in various topologies. But the integral definition does not literally apply outside $L^1$ functions. Either extend by continuity, or define by duality.

So, again, I don't think it's about "divergent integrals" per se, but about literal definitions of operators (etc) via integrals which fail to converge even when the operator has a reasonable extension. But those extensions very often require giving up tooo-classical pointwise senses of both functions and integrals.

  • $\begingroup$ Still, it is interesting how we can make a correspondence between functionals in a general sense and divergent integrals. I never thought there can be such bijection. Any divergent integral in fact tells a story on what it will do to the integral of a function which has a singularity, corresponding to that divergent integral (for instance, the imaginary part of logarithm jumps at zero because the singularity there is similar to $\mp i \pi \delta(0)$ ). $\endgroup$
    – Anixx
    Mar 2, 2021 at 20:05

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