Ytukyg

I would like to find the imaginary-order derivative of a function (let's just focus on a simple function for now). There is the Riemann-Liouville fractional-derivative: $$ _{a}D^{i}_{t} f(t) = frac{1}{Gamma(1-i)} frac{d}{dt} int_{a}^t frac{f(tau)}{(t-tau)^{-i}} dtau$$

For a constant function $ f(t) = c, c in mathbb{R}$, this definition will result in something that has $0^{i}$, which is undefined (see my other post: What is $0^{i}$?)

$$ _{a}D^{i}_{t} f(t) = frac{1}{Gamma(1-i)} frac{d}{dt} int_{a}^t (t-tau)^{i} dtau = frac{1}{Gamma(1-i)} frac{d}{dt} frac{1}{1+i}(0^{1+i} - (t-a)^{1+i}) $$

Is there another way the imaginary-order derivative can be found, without getting an undefined result?

There is a better formula, Fourier transform based differintegral:

$$f^{(s)}(x)=frac{1}{2pi}int_{-infty}^{+infty} e^{- i omega x}(-i omega)^s int_{-infty}^{+infty}f(t)e^{iomega t}dt , domega$$

This stems from the nature of Fourier transform: multiplying the image function by $(-iomega)$ corresponds to taking derivative from the original.

Putting here $i$ instead of $s$ will give you the i-th order derivative of sine:

$$(sin x)^{(i)}=isinhleft(fracpi 2-ixright)$$

There is also another formula based on Newton series which may be useful in some cases:

$$f^{(s)}(x)=sum_{m=0}^{infty} binom {s}m sum_{k=0}^mbinom mk(-1)^{m-k}f^{(k)}(x)$$

but for sine it diverges. When converges, it coincides with the prevuous formula.

That said, there is a general expression for s-th derivative of sine:

$$(sin x)^{(s)}=sinleft(frac{pi s}2+xright)$$

When $s=i$ this coincides with the prevuously obtained result.

Complex exponentiation involves multi-valued functions. To use this formula, we should choose a branch of the complex logarithm that is defined on the open segment $(0,t-a)$. Let's assume, for simplicity, that $t-a$ is a positive real. Then we have:

$$int_a^t (t-tau)^i dtau = int_a^t e^{i log(t-tau)}dtau = lim_{x to t}frac{1}{1+i}[ e^{(1+i)log(t-tau)}]mid_a^x =$$

$$lim_{xto t} frac{1}{1+i} (e^{(1+i)log(t-x)} - e^{(1+i)log(t-a)}) = -frac{1}{1+i} e^{(1+i)log(t-a)} = -frac{1}{1+i} (t-a)^{1+i}$$

Here the limit is taken from the right, and the last exponentiation is understood to depend on the choice of logarithm.

The step that gets around this $0^i$ business is $lim_{xto t} e^{(1+i)log(t-x)} = 0$, which follows from the fact that $1+i$ has positive real part.

As I pointed out in my answer to your other question, we can arrive at this result in a more ad hoc way by interpreting $0^i$ as a complex number with undefined argument but finite magnitude. Then $0^{1+i} = 0^1 cdot 0^i = 0cdot 0^i = 0$.