When the following property of integrals holds?

Question

Natural integral is the distinguished antiderivative of a function that can be understood as an analytic continuation of consecutive derivatives of a function towards $-1$th order. It is defined as

$$f^{(-1)}(x)=\sum_{m=0}^{\infty} (-1)^m \sum_{k=0}^m\binom mk(-1)^{m-k}f^{(k)}(x)$$

or

$$f^{(-1)}(x)=\frac{i}{2\pi}\int_{-\infty}^{+\infty} e^{- i \omega x}\omega^{-1} \int_{-\infty}^{+\infty}f(t)e^{i\omega t}dt \, d\omega = i\mathcal{F}^{-1}\left(x^{-1} \mathcal{F}(f(x))\right)$$

I would like to see the verification of the conjectured property of natural integral that

$$\int_{-\infty}^0f(x)dx-\int_{0}^\infty f(x)dx=2f^{(-1)}(0 )$$

Where if the integrals diverge they are taken as their regularized values and if $f^{(-1)}(x)$ has a pole at zero it should be taken as Cauchy principal value.

In other words, I wonder whether the following definition of natural integral is equivalent to the previous ones:

$$f^{(-1)}(x)=\int_0^x f(t) \, dt+\frac{1}{2} \left(\int_{-\infty }^0 f(t) \, dt-\int_0^{\infty } f(t) \, dt\right)$$

Note that $0$ in these 3 integrals can be replaced by any constant as the value does not depend on it.

If the property does not hold for all functions, in what cases does it hold?

Answer 1

It seems linked to the fact that the Fourier transform of the inverse function. More specifically, assuming that $x^{-1}$ is the distribution called the principal value of $1/x$, then its Fourier transform is $\frac{i}{2}\,\mathrm{sign}(x)$.

This gives exactly $$f^{(-1)}(0) = i\,\mathcal{F}_x^{-1}(x^{-1}\mathcal{F}(f))(0) = i\,\int_{\mathbb{R}} x^{-1}\mathcal{F}(f) = i\,\int_{\mathbb{R}} \mathcal{F}(x^{-1})f = \frac{1}{2}\left( \int_{\mathbb{R}_-} f - \int_{\mathbb{R}_+} f\right).$$

(To be more rigorous, one should perhaps replace the integrals in the intermediate steps by duality products, and verify that $x^{-1}$ defines a continuous functional over $\mathcal{F}(L^1)$, which would imply that the formula is valid as soon as $f$ is integrable.)