1
$\begingroup$

Natural integral (as we will define it) is a distinguished antiderivative of a function that can be understood as interpolation of the sequence of consecutive derivatives to the $-1$. It has a naturally defined integration constant. While it is possible to define natural integral in various ways, it all boils down to the following property:

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

where the integrals in the brackets should be understood in the sense of regularization, if they diverge. The $0$ is not important here and can be replaced by any point, this will not affect the answer.

That said, I wonder, if such natural antiderivative ever appears in physical applications?

UPDATE

As @Qmechanic pointed out, it appears in many areas of physics. I prefer examples from classical mechanics or elementary quantum theory.

$\endgroup$
0

1 Answer 1

1
$\begingroup$

The natural antiderivative $$f^{(-1)}(x)~:=~\frac{1}{2}\int_{\mathbb{R}}\!\mathrm{d}x^{\prime} ~{\rm sgn}(x\!-\!x^{\prime})~f(x^{\prime})$$ is the most symmetric choice of integration constant. The kernel ${\rm sgn}(x\!-\!x^{\prime})$ appears all over physics. It is the Fourier transform of ${\rm PV}\frac{1}{k}$ up to a multiplicative constant. Examples:

  1. The symplectic potential $\frac{1}{2}z^I\omega_{IJ}\dot{z}^J$ in the Lagrangian leads to the Feynman propagator/Greens function $$G^{IJ}_F(t\!-\!t^{\prime})~=~\frac{1}{2}\omega^{IJ} {\rm sgn}(t\!-\!t^{\prime}).$$ See also my related Phys.SE answer here.

  2. The Poisson commutation relations $\{\phi(x),\phi(y)\}=\frac{1}{2}{\rm sgn}(x\!-\!x^{\prime})$ for a self-dual boson field becomes $$\{\Phi[f],\Phi[g]\}=\frac{1}{2}\int_{\mathbb{R}}\!\mathrm{d}x \int_{\mathbb{R}}\!\mathrm{d}x^{\prime}~{\rm sgn}(x\!-\!x^{\prime})f(x) g(x^{\prime})$$ in terms of test functions $f,g$.

$\endgroup$
2
  • $\begingroup$ I wonder why the mathematicians are not aware of this. when it comes to differintegral definitions or generalizing the polygamma function to the negative orders, all kinds of definitions pop out. $\endgroup$
    – Anixx
    Commented May 17, 2020 at 12:39
  • $\begingroup$ By the way, does it have relation to Berezin integral, where essentially also integration constant is fixed? $\endgroup$
    – Anixx
    Commented Nov 11, 2023 at 7:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.