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?



Are the shapes of the $\mathbb{R}^2$ plane and a disk of infinite radius different? Or otherwise, why their areas differ by $\frac\pi{12}$?
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](https://mathoverflow.net/q/413873/10059), 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 $\omega=\int\_0^\infty dx$.


