1
$\begingroup$

The calculation of the area of the $\mathbb{R}^2$ plane depends on filtering used. I think, the most natural filtering is along the radius in polar coordinates:

$$S_{\mathbb{R}^2}=\int_0^\infty 2\pi r dr=2\pi\left(\frac{\tau^2}2+\frac1{24}\right)=\pi\tau^2+\frac\pi{12}$$

where $\tau=\int_0^\infty dx$.

The regularized value of this area is $0$. On the other hand, the area of a disk with radius $\tau$ (equal to the length of the real semi-axis) is $S_c=\pi\tau^2$, and its regularized value is $-\frac\pi{12}$.

Thus, $S_{\mathbb{R}^2}-S_c=\frac\pi{12}$. I wonder, where this area difference comes from? Does it originate from the fact that the plane should not be considered a disk of infinite radius? Or it is some glitch of integration technique?

$\endgroup$
3
  • 6
    $\begingroup$ How on earth can it be meaningful to attach any value to this integral except $\infty$? $\endgroup$ – David Loeffler Apr 19 '20 at 8:39
  • 3
    $\begingroup$ How do you get the integral equal to $2\pi\left(\frac{\tau^2}2+\frac1{24}\right)$? $\endgroup$ – Wojowu Apr 19 '20 at 11:38
  • 2
    $\begingroup$ @Wojowu the formula appears in this post of OP, I got lost before arriving to it so I cannot comment further on it. $\endgroup$ – Dabed Apr 19 '20 at 23:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.