Dirac delta distribution in $1D$

Question

We know that $$\frac{d^2}{dx^2}\left(\frac{1}{|x|}\right) =-4\pi\delta(x)$$ where $\delta(x)$ is Dirac delta distribution. $$\Rightarrow \lim_{x\to 0}\frac{d^2}{dx^2}\left(\frac{1}{|x|}\right) =-\infty$$ $$\frac{d^2}{dx^2}\left(\frac{1}{\sqrt{x^2 + a^2}}\right) =\frac{3 x^2}{(a^2 + x^2)^{5/2}}-\frac{1}{(a^2 + x^2)^{3/2}}$$ $$\lim_{a\to 0}\frac{d^2}{dx^2}\left(\frac{1}{\sqrt{x^2 + a^2}}\right) \approx\frac{3x^2}{|x|^5}-\frac{1}{|x|^3}=\frac{2}{|x|^3}$$ $$\lim_{x\to 0}\lim_{a\to 0}\frac{d^2}{dx^2}\left(\frac{1}{\sqrt{x^2 + a^2}}\right) \approx\color{red}{+\infty}$$ Shouldn't the red part be $-\infty$? Where did something go wrong?

Answer 1

In 1D, the fundamental solution is given by the absolute value.

$$ {d^2\over dx^2} |x| = {d\over dx} ( -{\bf 1}_{]-\infty,0[}+{\bf 1}_{]0,\infty[}) = 2\delta(x). $$

The expression you give at the start of your question is related to the laplacian of the radial function $(x,y,z) \mapsto {1\over \sqrt{x^2+y^2+z^2}}$ in 3D. You can use the following formula to compute it outside the origin. Setting $r = \sqrt{x^2+y^2+z^2}$ and for $f : [0,\infty[ \rightarrow {\bf R}$,

$$\Delta f(r) = {1\over r^2}{\partial\over \partial r} \Bigl(r^2 {\partial f \over \partial r}\Bigr).$$

Answer 2

Writing @coudy's answer more clearly. Not rigorous but intuitively correct.

Let $f=\frac{d}{dx}|x|$. Then $f(x>0)=1$ and $f(x<0)=-1$ then \begin{align} f(\epsilon)-f(-\epsilon)&=2=\int_{-\epsilon}^{\epsilon}2\delta(x)dx\\ \implies \frac{df}{dx}&=\frac{d^2|x|}{dx^2}=2\delta(x)\\ \implies \frac{d^2}{dx^2}\left(\frac{|x|}{2}\right)&=\delta(x) \end{align}