Are the shapes of the $\mathbb{R}^2$ plane and a disk of infinite radius different? Or otherwise, why their areas differ by $\frac\pi{12}$? [closed]

Question

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?

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 $\omega=\int_0^\infty dx$.