# What gets broken if we assign a value to the square of Delta function?

Particularly, what gets broken if $$\pi^2\delta^2(x)=2i\pi\delta'(x)-\frac1{12}$$?

• How do you motivate that value? – md2perpe Sep 30 at 16:46
• Bear in mind $\delta$ doesn't have a value as such either; what we have is values for $\int_{\Bbb R}f(x)dx$ ($f(0)$) provided $f$ is a Schwartz function, or values for nascent delta functions $\delta_n(x)=\frac1n\delta_1\left(\frac{x}{n}\right)$ for an even PDF $\delta_1$ satisfying $\delta_1(0)\ne0$, from which we get an $n\to\infty$ pointwise limit of $\infty$ if $x=0$ or $0$ otherwise. Similarly, $\lim_{n\to\infty}\delta_n^2(x)$ is $\infty$ if $x=0$ or $0$ otherwise, but $$\int_{\Bbb R}\delta^2(x)f(x)dx=\delta(0)f(0)$$is probably not what you want either. – J.G. Sep 30 at 19:57
• @md2perpe look here: mathoverflow.net/a/342651/10059 After you read that, basically $\int_0^\infty x dx=\int_0^\infty \frac13dx=\frac{\tau^2}2+\frac1{24}=i\pi\delta'(0)$ And $\tau=\pi\delta(0)$. The values of delta function at zero are taken formally, it is just that the Fourier transform becomes the respective divergent integrals at zero. – Anixx Oct 1 at 4:49
• @J.G. your expression is very interesting. – Anixx Oct 1 at 4:52

Benci and Baglini provide a more modern writeup here with an exact list of conditions. So no matter how you want to define $$\delta^2$$, of of these will break. Many different suggestions have been made on what the "proper" generalization should be. (for example Colombeau Algebra) However from my experience with the topic one doesn't come across any "satisfying" answer.