And still Cambridge mathematicians (not Dirac) often misunderstand Heaviside, disadvantaging their students.
A common claim that has been often repeated is that the fractional calculus (FC) as envisioned by Euler and Heaviside doesn't obey the law of exponents
$D^{\alpha}D^{\beta} =D^{\alpha+\beta},$
and, specifically, differentiation $D$ and integration $D^{-1}$ do not commute, and consequently, neither do they obey the law of exponents. An example is given on the webpage Fractional Calculus III by Beardon:
$$D^{-1}D \; e^x = \int_0^x e^t dt = e^x-1 \neq DD^{-1} e^x = D(e^x-1) = e^x = D^0e^x = D^{-1+1}e^x.$$
However, introducing the Heaviside step function $H(x)$,
$$D^{-1}D \; H(x) e^x =H(x) \int_0^x (\delta(t) +e^t) dt =H(x)( 1+ e^x-1) =H(x) e^x = D^0H(x) e^x .$$
The Euler-Heaviside FC can be framed a number of ways to ensure the fundamental operator action is interpreted as
$$D^{\alpha}D^{\beta} H(x) \frac{x^{\gamma}}{\gamma!} =D^{\beta} D^{\alpha} H(x) \frac{x^{\gamma}}{\gamma!} =D^{\alpha+\beta}H(x) \frac{x^{\gamma}}{\gamma!} = H(x) \frac{x^{\gamma-\alpha-\beta}}{(\gamma-\alpha-\beta)!}$$
for $\alpha,\beta,$ and $\gamma$ any real numbers.
In one interpretation (e.g., see Gelfand and Shilov's Generalized Functions Vol. I),
$$H(x) \frac{x^{-n-1}}{(-n-1)!}= D^n \delta(x)= \delta^{(n)}(x)$$
such that, under a finite part construction or other analytic continuation,
$$D^{n} H(x)\frac{x^{\alpha}}{\alpha!} =H(x) \int_0^x \frac{(x-t)^{-n-1}}{(-n-1)!} \frac{t^{\alpha}}{\alpha!}dt = H(x)\oint_{|z-x|=x} \frac{n!}{(z-x)^{n+1}}\frac{z^{\alpha}}{\alpha!}dz =H(x) \frac{x^{\alpha-n}}{(\alpha-n)!}$$
and, more generally,
$$D^{\beta} H(x)\frac{x^{\alpha}}{\alpha!} = H(x)\int_{-\infty}^\infty H(x-t) \frac{(x-t)^{-\beta-1}}{(-\beta-1)!} H(t)\frac{t^{\alpha}}{\alpha!}dt$$
$$= H(x)\int_0^x \frac{(x-t)^{-\beta-1}}{(-\beta-1)!} \frac{t^{\alpha}}{\alpha!}dt = H(x)\oint_{|z-x|=x} \frac{\beta!}{(z-x)^{\beta+1}}\frac{z^{\alpha}}{\alpha!}dz =H(x) \frac{x^{\alpha-n}}{(\alpha-n)!}.$$
Then the Euler-Heaviside FC gives
$$D^{\frac{1}{2}} H(x)\frac{x^{\frac{-1}{2}}}{(\frac{-1}{2})!} = H(x)\frac{x^{-1}}{(-1)!} = \delta(x)$$
and
$$D^{\frac{1}{2}}D^{\frac{1}{2}} H(x)\frac{x^{\frac{-1}{2}}}{(\frac{-1}{2})!}$$$$ = D^{\frac{1}{2}} H(x)\frac{x^{-1}}{(-1)!} = H(x)\frac{x^{\frac{-3}{2}}}{(\frac{-3}{2})!} = D H(x)\frac{x^{\frac{-1}{2}}}{(\frac{-1}{2})!},$$
so, in this case,
$$D^{\frac{1}{2}}D^{\frac{1}{2}} = D$$
whereas Beardon concludes that
$$D^{\frac{1}{2}}x^{\frac{-1}{2}} = 0,$$
implying that the law of exponents is violated since then
$$D^{\frac{1}{2}} D^{\frac{1}{2}}x^{\frac{-1}{2}} = D^{\frac{1}{2}}0 = 0 \neq D x^{\frac{-1}{2}} =\frac{-1}{2}x^{\frac{-3}{2}} .$$
(If I recall correctly from my readings a couple or so decades ago, H. Jeffreys in his oft referenced book Operational Methods In Mathematical Physics (1927) imposed the same misleading interpretations as Beardon.)
If the FC is constructed using an infinitesimal generator, the analytic continuations can be dodged, or a Pochhammer contour integral can be invoked for generalizing the beta function integral. These constructions are consistent with Laplace- and Mellin-transform approaches over the domains of common convergence and analytic continuation of the reps, with Pincherle's axiomatic treatment of a canonical FC, and with the calculus of Appell Sheffer polynomial sequences.