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The derivative of a function $f(x)$ is the limit of the quotient $$\lim_{\Delta x\to 0}\frac{f(x+\Delta x)-f(x)}{\Delta x}$$ A formula defining the fractional derivative of the same function is for example: $$\frac{d^{\frac{1}{2}}}{d(x-a)^{\frac{1}{2}}}f(x)=\frac{1}{\sqrt{(\pi)}}\frac{d}{dx}\int_{a}^{x}\frac{f(\tau)}{\sqrt{(x-\tau})}d\tau$$ My question is: is it possible to define an imaginary derivative $$\frac{d^i}{dx^{i}}f(x)$$ of the function $f(x)$? Tanks.

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A google search found complex order mentioned here: and here: – GEdgar Apr 18 '12 at 15:50
E.R. Love "Fractional derivatives of imaginary order." J. London Math. Soc. (2) 3 1971 241–259. – Bill Cook Apr 18 '12 at 15:59
up vote 4 down vote accepted

Let $\alpha$ be a complex number with $\mathrm{Re}(\alpha) \geq 0$. Then the Riemann-Liouville fractional derivative of order $\alpha$ of a function $f(t)$ is given by $$_a D^\alpha_t f(t) = \frac{1}{\Gamma(n - \alpha)} \frac{d^n}{dx^n} \int_a^x \frac{f(\tau)}{(t - \tau)^{n-\alpha-1}} ~d\tau,$$ where $n = [\mathrm{Re}(\alpha)] + 1$ and $[\mathrm{Re}(\alpha)]$ is the integer part of $\mathrm{Re}(\alpha)$. Note that this definition indeed makes sense for complex $\alpha$ with positive real part, and in particular, $$_a D^i_t f(t) = \frac{1}{\Gamma(1 - i)} \frac{d}{dx} \int_a^x \frac{f(\tau)}{(t - \tau)^{-i}} ~d\tau.$$

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There is also the Caputo definition, which performs the differentiation first and the integration last... – J. M. Apr 18 '12 at 16:09
I don't know if the Caputo would work for complex orders: it is not analytic in general. In particular if $f(t) = t$, then its Caputo derivative as a function of the differentiation order as a real number is eventually constant, but not constant over the entire line, thus cannot be analytic. – mike4ty4 May 18 '15 at 0:11

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