The text I use is the one by R.F. Hoskins and J.Sousa Pinto
Distribution, Ultradistributions and other Generalised Functions because it covers a very large number of topics in very short space.
I'm going to give a proof sketch of the Fisher's result
x^{-r} \cdot \delta^{(r-1)} = (-1)^{r} \frac{(r-1)!}{2(2r-1)!}\delta^{(2r-1)}(x).
The result is published in
Proc Camb Phil Soc, 72, pp 201-204. I going to assume you know the basics, namely that the space of test functions ##\mathcal{D}## is a dense subset of ##\mathcal{D}'##.
The (classical) Fisher product is an example of what we call
sequential products. The ideal is simple: for any two Schwartz distributions ##\mu## and ##\nu##, we find sequences of smooth functions ##\mu_n \to \mu## and ##\nu_n \to \nu## and define ##\mu \cdot \nu = \lim_{n\to\infty} \mu_n \nu_n##. This ultimately is a losing battle. The more constraints you place on your sequences the more functions you can multiply, but the properties of your product get steadily worse (such as no distributive law).
We start by choosing a smooth function ##\rho(x)## satisfying
- ##\rho## is non-negative,
- the area under the curve over the reals is 1,
- ##\rho(x) = 0## for all ##-1 \leq x \leq 1##,
- ##\rho(-x) = \rho(x)## for all x,
- ##\rho^{(r)}(x)## has only ##r## changes of sign for ##r=1,2,3,\ldots##.
The sequence ##\rho_n(x) = n\rho(nx)## is called a
symmetric model sequence and we set ##\mu_n = \mu * \rho_n## (where * is the convolution of distributions). Such a ##\rho## exists, namely the bump function,
\rho(x) = A exp(1/(x^2-1)) for ##|x|\leq1## and zero elsewhere. Here A is just a normalization constant. It should be clear that ##\rho_n \to \delta##.
The singular distribution ##x^{-r}## is defined as
x^{-r} = \frac{(-1)^r}{(r-1)!}D^r \log|x|
and our sequence
(x^{-r})_n = \frac{(-1)^{r-1}}{(r-1)!} \int_{-1/n}^{1/n} \rho^{(r)}_{n}(t) \log|x-t|\, dt.
Now we define ##\mathcal{I}f(x) := \int_{-\infty}^x f(t) dt## then
F_n(x) := \mathcal{I}^{2r-1}[(x^{-r})_n\rho^{(r-1)}_n(x)] = \frac{1}{(2r-2)!}\int_{-1/n}^{x}(t^{-r})_n \rho^{(r-1)}_n(x-t)^{2r-2}dt.
It can be shown that
\int_{-1/n}^{1/n}(x^{-r})_n \rho_n^{(r-1)}(x)x^{m}dx = \frac{(-1)^{r+1}}{2}(r-1)!
when ##m=2r-1## and zero for ##m=0,1,\ldots, 2r-2##.
Putting the results together you get
\int_{1/n}^{-1/n}(t^{-r})_n \rho_n^{(r-1)}(t) (1/n-t)^{2r-2}dt =0
and using ##\rho_n^{(r-1)}(x) = 0## for ##|x| \geq 1/n##; you get ##\mathcal{I}^{2r-1}[(x^{-r})_n\rho_n^{(r-1)}(x)] = 0## for ##|x| \geq 1/n## as well. The basically means that the support is converging to ##\{0\}##.
Moving on. ##\rho^{(r)}## has only r changes of sign; therefore ##(x^{-r})_n## has only r changes of sign, therefore the ##(2r-1)##th primitive is either always non-negative or always non-positive. And finally,
##\int_{-1/n}^{1/n}\mathcal{I}^{2r-1}[(x^{-r})_n\rho_n^{(r-1)}(x)]dx = (-1)^r\frac{(r-1)!}{2(2r-1)!}.##
Hence ##F_n## converges distributionally to ##(-1)^r\frac{(r-1)!}{2(2r-1)!} \delta(x)##. Differentiate (2r-1) times to get the result. Setting r=1, we get ##x^{-1}\cdot\delta(x) = -\frac{1}{2}\delta'(x)## as required.
