# Representation of the Dirac delta function

The Dirac delta function appears in the Sokhotsky formula, $$\text{Im}\lim_{\epsilon\to 0^+} \frac{1}{x-i\epsilon} = \pi\delta(x),$$ to be understood in the integral sense $$\text{Im}\lim_{\epsilon\to 0^+} \int \frac{f(y)}{y-x-i\epsilon}dy=\pi f(x),$$ for a real valued function $$f(x)$$.

I stumbled on an identity that has a similar flavour, $$\lim_{\epsilon\to 0^+}\int_x^b \frac{\epsilon f(y)}{(y-x)^{1-\epsilon}} dy=f(x).\label{1}\tag{\ast}$$ The upper limit $$b>x$$ of the integral is arbitrary, one may send it to infinity if $$f(x)$$ has compact support. A corollary is $$\lim_{\epsilon\to 0^+}\int_a^b \frac{\epsilon f(x)}{[(b-x)(x-a)]^{1-\epsilon}}\,dx=\frac{f(a)+f(b)}{b-a}.$$

All of this can be interpreted as a delta function representation in terms of the unit step function $$\theta(x)$$, $$\lim_{\epsilon\to 0^+} \frac{\epsilon\theta(x)}{x^{1-\epsilon}}=\delta(x),\tag{\ast\ast}$$ acting on compactly supported functions.

Q: One can readily check the formula \eqref{1} for polynomial functions $$f(x)$$. Is there a more comprehensive derivation? Is this representation of the delta function known?

As usual in such examples, there is no need to integrate against a test function. One can simply use the fact that if a sequence (or net) of distributions converges in the distributional sense, then so does the one obtained by differentiating term by term. In particular, this applies when the sequence consists of functions which converge in pretty well any sensible classical sense, e.g., locally $$L^1$$ as in the case in point, that of the functions which are defined as $$x^{\epsilon}$$ on the positive real axis, and $$0$$ elsewhere. They converge to the Heaviside function and we can differentiate to obtain the required result.

Most of the examples of $$\delta$$-sequences in the literature can be verified in this way: consider the terms’ primitives and show that they converge to the Heaviside function. The result then follows as above.

The first example (Sokhotsky) in the question can be proved in one line, after integrating $$\dfrac{\epsilon}{x^2+\epsilon^2}.$$


This follows easily by writing $$\int_x^b=\int_x^{x+\de}+\int_{x+\de}^b$$ for small $$\de>0$$.

In your case, just take $$K_\ep(z)=\ep z^{\ep-1}$$ for real $$z>0$$.

Another example of an appropriate family of kernels -- converging to the delta function as $$\ep\downarrow0$$ -- would be $$K_\ep(z)=e^{-z/\ep}/\ep$$ for real $$z>0$$. More generally, one can take $$K_\ep(z)=k(z/\ep)/\ep$$ for real $$z>0$$, where $$k\colon(0,\infty)\to\mathbb R$$ is any nonnegative measurable function such that $$\int_0^\infty k(t)\,dt=1$$.

Your transformation (*) is related to the Mellin transform (with $$\epsilon=s$$). In particular, your result is obtained as the limit $$s\to0^+$$ of Ramanujan's master theorem $$\frac{\sin(\pi s)}{\pi}\int_0^\infty x^{s-1}\left(\,\lambda(0) - x\,\lambda(1) + x^2\,\lambda(2) -\,\cdots\,\right) dx = \,\lambda(-s)\,.$$

Here is a (quick) derivation that $$(\ast\ast)$$ is the Dirac delta.

I notice $$\frac{\epsilon\theta(x)}{x^{1-\epsilon}}= \frac{d}{dx}(x^\epsilon)\theta(x)\,.$$ so that for $$\epsilon>0$$ $$\int_{\mathbb R} d x\; \frac{d}{dx}(x^\epsilon)\theta(x) f(x)=\int_0^{\infty} \frac{d}{dx}(x^\epsilon) f(x)=0 -\int_0^\infty x^\epsilon f'(x)$$ where I assumed $$f$$ is smooth and of compact support.

I cannot be bothered to find some function that dominates $$x^\epsilon f'(x)$$. Anyway, given one exists, you move the $$\lim_{\epsilon \to 0^+}$$ inside the integral and pick up $$f(0)$$.