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?

  • $\begingroup$ 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

It is not sadly. The more correct answer is much more complicated, and is given as so:


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

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.