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What would be the fractional derivative of any order 'b' of the function

$ (a-x) $ ?

My guess is: $$ \frac{d^{s}}{dx^{s}}(a-x)^{-1}= \frac{\Gamma(s+1)}{(a-x)^{s+1}} $$

Is this correct?

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  • $\begingroup$ I don't think that your guess is correct. The corresponding Rieman-Liouville fractional integral involves a Gauss hypergeometric function: hypergeometric2F1(1,1;1-s;x/a)/(a*(x^s)*Gamma(1-s)) $\endgroup$ – JJacquelin Jan 12 '14 at 11:21
  • $\begingroup$ Your guess should be correct if the definition of the fractional derivatives was based on the Weyl's transform instead of the Rieman-Liouville transform, which should be not the usual definition. $\endgroup$ – JJacquelin Jan 12 '14 at 11:37
  • $\begingroup$ Function 1-x or 1/(a-x)? $\endgroup$ – Anixx Dec 18 '14 at 14:22
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It is not sadly. The more correct answer is much more complicated, and is given as so:

$$\frac{d^s}{dx^s}\frac1{a-x}=\frac{d^s}{dx^s}\frac{-1}{x-a}=\frac{-(x-a)^{-1-s}\left[\ln(x-a)-\gamma-\psi^{(0)}(-s)\right]}{\Gamma(-s)}$$

And for $s\in\mathbb N$,

$$=\lim_{k\to s}\frac{d^s}{dx^s}\frac{-1}{x-a}=\frac{-(x-a)^{-1-k}\left[\ln(x-a)-\gamma-\psi^{(0)}(-k)\right]}{\Gamma(-k)}$$

where $\gamma$ is the Euler-Mascheroni constant and $\psi^{(0)}$ is the digamma function

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