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It is known that the left-side Weyl-Marchaud fractional derivative $$ D^\mu f(x) = \frac{\mu}{\Gamma(1-\mu)}\int_0^\infty \frac{f(x)-f(x-t)}{t^{\mu+1}}dt $$ for continuous and integrable $f(x)$ on $\mathbb{R}$.

I wonder if we can find a corresponding fractional integral operator $I^\mu$ such that $$ I^\mu D^\mu f(x) = D^\mu I^\mu f(x) = f(x).$$

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  • $\begingroup$ I do not think "continuous and integrable" is sufficient for the integral in the definition of $D^\mu f(x)$ to converge. Then, as far as I remember, given appropriate assumptions on $f$ and $f'$, the fractional derivative in your question is the same as the usual Riemann–Liouville fractional derivative, is it not? $\endgroup$ – Mateusz Kwaśnicki Oct 6 at 21:35

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