Does anybody know if such function exists? As I understand it, the function $$\frac{1}{x^2}$$ itself could be used as a delta function if it had no 1/0 infinity.
That is why I'm in a search of an "inverse square law" delta function that has maybe a shape like parabola in the range (-1; 1). Knowing its integral expression would be useful, too.
Maybe the Cauchy distribution $p(x,\sigma)$ fits the bill: $$ p(x,\sigma) = \frac{\sigma/\pi}{\sigma^2+x^2} = \frac{1/(\pi\,\sigma)}{1+(x/\sigma)^2} \qquad \mbox{with} \quad \sigma > 0$$ Where: $$\lim_{\sigma\rightarrow 0}\, p(x,\sigma) = \delta(x)$$ Check that $\,p(x)\,$ is normed (: substitute $t=x/\sigma$): $$ \int_{-\infty}^{+\infty} \frac{1/(\pi\,\sigma)}{1+(x/\sigma)^2}\,dx = \frac{1}{\pi} \int_{-\infty}^{+\infty} \frac{dt}{1+t^2} = \frac{1}{\pi} \left[\, \arctan(t)\, \right]_{-\infty}^{+\infty} = \pi/\pi = 1 $$ And: $$ \lim_{x\rightarrow\pm\infty} \frac{\sigma/\pi}{\sigma^2+x^2} = 0 \qquad ; \qquad \lim_{\sigma\rightarrow 0} \, \frac{\sigma/\pi}{\sigma^2+x^2} = \left\{ \begin{array}{ll} \infty & \mbox{if} \quad x = 0 \\ 0 & \mbox{if} \quad x \ne 0 \end{array} \right.$$
