# A question about fractional derivatives

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?

• 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)) – JJacquelin Jan 12 '14 at 11:21
• 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. – JJacquelin Jan 12 '14 at 11:37
• Function 1-x or 1/(a-x)? – Anixx Dec 18 '14 at 14:22

## 1 Answer

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