Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

According to Fractional calculus, we know that $$(J^\alpha f) ( x ) = { 1 \over \Gamma ( \alpha ) } \int_0^x (x-t)^{\alpha-1} f(t) \; dt$$

It's in real analysis, but what about in complex analysis? As we know, if $\alpha$ is not a integer, then $(x-t)^{\alpha-1}$ may returns more than one values.

So my question is can we find a method to let fractional calculus working on complex field?

share|cite|improve this question
up vote -4 down vote accepted

Its easy. Just do the integral and then just replace your real x with complex z.

share|cite|improve this answer
I don't think it is as easy as it seems. Note that $2^{\frac{1}{3}}$ has only 1 value in real, but has 3 value in complex. So this may cause the equation to be not well-defined. – Popopo Sep 18 '12 at 2:08
But that problem gets fixed automaticly by analytic continuation , since analytic garantees that you picked the correct branch. – mick Sep 29 '12 at 18:37
Er...I'm not quite clear, need it work at Riemann surface? – Popopo Sep 30 '12 at 1:59
@Popopo : Yes that is one way of looking at it. – mick Sep 30 '12 at 17:11
Okay, finally you are right. $(x-t)^{1-\alpha}$ denote its principal value $|x-t|^{1-\alpha}e^{i(1-\alpha)arg(t)}$ – Popopo Oct 6 '12 at 1:37

There are a lot of definitions of fractional derivatives. If you want an (almost complete) answer for your question, then try the book Samko, Kilbas, Marichev, Fractional integrals and derivatives: theory and applications (1993). Specially, Ch 4, § 22 Fractional Integrals and Derivatives in the Complex Plane . "We emphasize that any work with definitions requires precision aimed to single out a branch of the multivalued function. It is usually achieved by means of a cut which goes from the branching point to infinity or by fixing $\arg(t - z)$ in one or another way. Different choices of a cut, which fixes the branch of the function $(t - z)^{1+\alpha}$ , and of the curve, gives different values of $f^{(\alpha)}(z)$ in general.

share|cite|improve this answer
Although you are correct that there are a lot of definitions of fractional derivatives , the OP has already given one kind. That is to say that there different definitions for REAL fractional derivatives and the OP has already picked a common one. So whatever branch is chosen for the real corresponds to the complex as long as both branches agree. – mick Oct 2 '12 at 21:12
Many thanks to your book. – Popopo Oct 6 '12 at 1:23

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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