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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?

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2 Answers 2

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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.

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    $\begingroup$ 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. $\endgroup$
    – mick
    Oct 2, 2012 at 21:12
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    $\begingroup$ Many thanks to your book. $\endgroup$
    – Popopo
    Oct 6, 2012 at 1:23
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    $\begingroup$ Thank you for providing the quote. This book has been out of print for some time. $\endgroup$
    – Joel
    May 23, 2017 at 19:27
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Its easy. Just do the integral and then just replace your real x with complex z.

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  • $\begingroup$ 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. $\endgroup$
    – Popopo
    Sep 18, 2012 at 2:08
  • $\begingroup$ But that problem gets fixed automaticly by analytic continuation , since analytic garantees that you picked the correct branch. $\endgroup$
    – mick
    Sep 29, 2012 at 18:37
  • $\begingroup$ Er...I'm not quite clear, need it work at Riemann surface? $\endgroup$
    – Popopo
    Sep 30, 2012 at 1:59
  • $\begingroup$ @Popopo : Yes that is one way of looking at it. $\endgroup$
    – mick
    Sep 30, 2012 at 17:11
  • $\begingroup$ Okay, finally you are right. $(x-t)^{1-\alpha}$ denote its principal value $|x-t|^{1-\alpha}e^{i(1-\alpha)arg(t)}$ $\endgroup$
    – Popopo
    Oct 6, 2012 at 1:37

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