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?

  • 7
    $\begingroup$ How on earth can it be meaningful to attach any value to this integral except $\infty$? $\endgroup$ Apr 19, 2020 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, 2020 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, 2020 at 23:58
  • $\begingroup$ @Anixx: You ask quite a few questions about your own theory of divergent integrals, which surely very few people are qualified to answer. Is there a good reference where your theory is developed in full rigour and detail, so that it might be possible for mathematicians other than you to answer your questions? $\endgroup$
    – Ben McKay
    Jan 15, 2022 at 17:22
  • $\begingroup$ @BenMcKay for now, I think, there is no theory because I found that the proposed approach was non-natural definition. It seems, the Levi-Civita type of construction is more natural. On the other hand, I am looking at whether one can be merged/embedded into the other. If you want an outdated text, I can provide a reference... $\endgroup$
    – Anixx
    Jan 15, 2022 at 17:38

1 Answer 1


Apparently, the discrepancy comes from my use of non-natural definition of multiplication of divergent integrals. When using a more natural and intuitive Levi-Civita field kind of construction, the multiplication gives

$\int_0^\infty dx\cdot \int_0^\infty dx=\omega^2=2\int_0^\infty x dx$.

So, the areas in both cases are $\pi\omega^2$, where $\omega=\int_0^\infty dx$.


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