Generalization of Dirac delta identity

Question

The Dirac delta distribution obeys the following identity in $\mathbb{R}$

$$\lim_{\epsilon\to 0}\dfrac{1}{\pi}\dfrac{\epsilon}{\epsilon^2+x^2}=\delta(x)\tag1.$$

I know how to prove this using the Sokhotski–Plemelj theorem, namely $$\lim_{\epsilon\to 0}\dfrac{1}{x\pm i\epsilon}=\mp i\pi\delta(x)+{\cal P}\dfrac{1}{x},\tag2$$

where $\cal P$ means the principal value. Indeed subtracting the formula for the two signs it follows that

$$\lim_{\epsilon \to 0}\dfrac{1}{\pi}\left[\dfrac{\epsilon}{x^2+\epsilon^2}\right]=\dfrac{1}{2\pi i}\lim_{\epsilon \to 0}\left[\dfrac{1}{x-i\epsilon}-\dfrac{1}{x+i\epsilon}\right]=\delta(x)\tag{3}.$$

All that said, I'm interested in a generalization. I want to understand the small $\epsilon$ expansion of

$$\left(\dfrac{\epsilon}{\epsilon^2+\|x\|^2}\right)^a\tag{4}$$

where $a\in \mathbb{C}$ and $x\in \mathbb{R}^n$. For example this often appears in some Physics papers when discussing the wave equation in hyperbolic spaces. It is often claimed that one has an expansion of the form $$\left(\dfrac{\epsilon}{\epsilon^2+\|x\|^2}\right)^a=\pi^{n/2}\dfrac{\Gamma\left(a-\frac{n}{2}\right)}{\Gamma(a)}\epsilon^{n-a}\delta^{(n)}(x)+\dfrac{\epsilon^a}{\|x\|^{2a}}+\cdots\tag{5}$$

where the dots denote subleading terms in the small $\epsilon$ expansion. Setting $a=1$ and $n=1$ and taking $\epsilon\to 0$ we recover (1). I would like to understand how to prove (5) and how to identify the order of the small $\epsilon$ corrections. I also want to understand whether some restriction on $a$ has to be made, another reason I want to fully understand the proof of this result. How can this be shown?

Answer 1

For distribution theory proofs I always find it helpful to simply start by definition. Take test function $\phi\in C_0^\infty(\Bbb{R}^n)$ and consider

$$\tilde{f_\epsilon}(\phi)\equiv \widetilde{\left(\frac{\epsilon}{\epsilon^2+x^2}\right)^a}(\phi) \equiv\int_{\Bbb{R}^n}\left(\frac{\epsilon}{\epsilon^2+x^2}\right)^a\phi(x)\:dx$$

Simply put, the first non-delta contribution to the expansion can be found by considering $\phi$ with $0\notin\operatorname{Supp}\phi$

$$\tilde{f_\epsilon}(\phi) = \epsilon^a \int_{\Bbb{R}}\frac{1}{|x|^{2a}}\left(1-\frac{\epsilon^2}{x^2}+\cdots\right)^a\phi(x)\:dx \approx \widetilde{\left(\frac{\epsilon}{x^2}\right)^a}(\phi) + \widetilde{O\left(\left(\frac{\epsilon}{x^2}\right)^{a+2}\right)}(\phi)$$

as $\epsilon\to 0^+$. But now consider $\phi$ with $0\in\operatorname{Supp}\phi$ and use the substitution $x = \epsilon y$:

$$\tilde{f_\epsilon}(\phi) = \int_{\Bbb{R}^n}\left(\frac{\epsilon}{\epsilon^2+\epsilon^2y^2}\right)^a\phi(\epsilon y)\:\epsilon^ndy = \epsilon^{n-a}\int_{\Bbb{R}^n}\frac{\phi(\epsilon y)}{(1+y^2)^a}dy $$

Then by dominated convergence take the limit of

$$\lim_{\epsilon\to0^+} \epsilon^{a-n}\tilde{f_\epsilon}(\phi) = \int_{\Bbb{R}^n}\frac{\phi(0)}{(1+y^2)^a}dy = \operatorname{Area}\left(S^{n-1}\right)\cdot \int_0^\infty \frac{r^{n-1}}{(1+r^2)^a}dr \cdot \delta(\phi)$$

by converting to $n$-dimensional spherical coordinates. The constant out front can be computed to be the constant used in the approximation formula by plugging the appropriate formulas for the area of an $n-1$-sphere and the value of that integral into the above. More interestingly, we can answer your question about the restrictions on $a$. Namely, we have that

$$\operatorname{Re}\{a\} > \frac{n}{2}$$

by $p$-test. Therefore we have proven

$$\tilde{f_\epsilon}(\phi) \approx C(n,a)\delta(\phi) + \widetilde{\left(\frac{\epsilon}{x^2}\right)^a}(\phi) + \cdots$$

or less rigorously using the convention of representing deltas as functions we have

$$f_\epsilon(x) \approx C(n,a)\delta(x) + \left(\frac{\epsilon}{x^2}\right)^a + \cdots$$

A couple notes on notation:

I use $x^2$ to unambigiously mean the dot product $x\cdot x$ for $x\in\Bbb{R}^n$. But when appropriate, I do use single bars for the norm and reserved the use of double bars for norms on functions if any had appeared. Lastly, I'm not a fan of the physics notation $\delta^{(n)}(x)$ to mean the $n$-dimensional Dirac delta (I stared at this formula in confusion for quite some time because it looks exactly like the notation for the $n$th derivative of delta), so I simply write $\delta(x)$ to mean the unrigorous function representation of the more rigorously defined linear functional that performs $\delta(\phi) \equiv \phi(0)$ regardless of the dimension of the $0$. Most of these notational quirks are standard for mathematician-oriented texts on distribution theory.

Answer 2

The OP question can be extended to much more general phenomenon.

Let $f$ be a continuous function absolutely integrable on $\mathbb{R}^n$ so that $$\int\limits_{\mathbb{R}^n}f(x)\,dx =1$$ Then $$\lim_{\varepsilon\to 0^+} \varepsilon^{-n} f(\varepsilon^{-1}x)\to \delta(x)\quad (*)$$ In the OP example we have $n=1$ and $f(x)=\pi^{-1}(1+x^2)^{-1}.$

The equality $(*)$ means $$\int\limits_{\mathbb{R}^n} \varepsilon^{-n} f(\varepsilon^{-1}x)g(x)\,dx =g(0),\qquad g\in C_b(\mathbb{R}^n)\quad (**)$$ where $C_b(\mathbb{R}^n)$ denotes the set of all continuous bounded functions. The proof of $(**)$ follows. $$\int\limits_{\mathbb{R}^n} \varepsilon^{-n} f(\varepsilon^{-1}x)g(x)\,dx-g(0)\underset{x=\varepsilon y}{=} \int\limits_{\mathbb{R}^n} f(y)g(\varepsilon y)\,dy-g(0)\\ = \int\limits_{\mathbb{R}^n} f(y)[g(\varepsilon y)-g(0)]\,dy=\int\limits_{\|y\|\le N} f(y)[g(\varepsilon y)-g(0)]\,dy+\int\limits_{\|y\|> N} f(y)[g(\varepsilon y)-g(0)]\,dy$$ Next $$\left |\int\limits_{\|y\|> N} f(y)[g(\varepsilon y)-g(0)]\,dy\right |\le 2\max |g(x)|\int\limits_{\|y\|>N}|f(y)|\,dy$$ For given $\delta>0$ there is $N$ so that $$2\max |g(x)|\int\limits_{\|y\|>N}|f(y)|\,dy<\delta$$ Next $$\left | \int\limits_{\|y\|\le N} f(y)[g(\varepsilon y)-g(0)]\,dy\right |\le \max_{\|u\|\le N\varepsilon}|g((u)-g(0)|\int\limits_{\|u\|>N}|f(u)\,du\le \max_{\|u\|\le N\varepsilon}|g((u)-g(0)|\,\|f\|_1 $$ By the continuity of $g$ at $0$ there is $\varepsilon_0>0$ such that $$ \max_{\|u\|\le N\varepsilon_0}|g((u)-g(0)|\,\|f\|_1<\delta$$ In particular for $0<\varepsilon\le \varepsilon_0$ we have $$ \max_{\|u\|\le N\varepsilon}|g((u)-g(0)|<\delta$$ Summarizing the calculations we get $$ \left |\int\limits_{\mathbb{R}^n} \varepsilon^{-n} f(\varepsilon^{-1}x)g(x)\,dx-g(0)\right|<2\delta,\quad 0<\varepsilon\le \varepsilon_0$$

The example in OP is $$f(x)=c_a(1+\|x\|)^{-a}$$ where $c_a$ is the normalizing constant so that the integral of $f$ is equal $1.$ The integrability assumption requires ${\rm Re}\,a>n/2.$