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It's a well know fact that $$d(x^a)=ax^{a-1}dx\; \text{ for }\;a\in \Bbb{R}$$ It's a less well-known, but easily provable, fact that this generalizes to $$d^n(x^a)=\frac{a!}{(a-n)!}x^{a-n}\; \text{ for }\;n\in \Bbb{N}$$ Using fractional calculus, we can generalize this to $$D^n(x^a)=\frac{\Gamma(a+1)}{\Gamma(a-n+1)}x^{a-n}\;\text{ for }\;n\in \Bbb{R}^+$$ (for $\Bbb{R}^-$, and thus antiderivatives, and thus indefinite integrals, we need the slightly different $(-1)^{-n}\frac{\Gamma(n-a)}{\Gamma(-a)}x^{(a+n)}$).

There are just two problems: $\int(x^{-1})dx=\ln(x)+C$, which isn't a power function and stays that way under further (integral) integration, and $d(x^0)=0(x^{-1})$, which is technically a power function, but stops further (integral) differentiation cold with that 0.

Now, these wouldn't normally be problems (in fact, there are very good reasons the integral of the reciprocal function is the natural logarithm, not the least of which because the derivative of the constant function is 0). The problem isn't that the inconsistency is there - it's that I don't know how it behaves.

Are the fractional integrals of the reciprocal function logarithmic or power? (or both? or neither?) If both, where is the line drawn? Or is it "both" in the sense of multi-valued? What about the fractional derivatives of the constant function? Are they identically zero or do they wiggle around between the integral values?

And what about the square root? It's first derivative is on the order of the inverse square-root, while its three-halves derivative is on the order of the reciprocal, implying that the half-integral of the reciprocal should be on the order of the inverse square-root, but it apparently isn't.

And what about other non-integer powers going through this zone? How do they avoid getting zeroed by the constant function or getting swept up in the logarithm? Do the fractional "differintegrals" of the power functions depend on where in the chain you start?

Edit: It would seem, in some naive sense, that it is the case that fractional derivatives depend on where in the chain you start, if only because constants of integration are themselves integrated in subsequent reintegrations. So, better question: Do the constants of differintegration, which must be normalized by initial conditions, allow the differintegration chain to pass through the interval $a \in [-1, 0]$? And if so, how?

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  • $\begingroup$ The important bit is in bold. The other questions can be considered background. Or extra credit, if you're feeling ambitious. $\endgroup$ – No Name Mar 19 '18 at 15:37
  • $\begingroup$ The important bit is still in bold. The old important bit(s) are in italics. The rest of my previous comment still stands. $\endgroup$ – No Name Mar 22 '18 at 6:23

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