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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?

2 Answers 2

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

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    Okay, finally you are right. $(x-t)^{1-\alpha}$ denote its principal value $|x-t|^{1-\alpha}e^{i(1-\alpha)arg(t)}$2012-10-06
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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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    Thank you for providing the quote. This book has been out of print for some time.2017-05-23