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

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?

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