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

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 $0^{i}$?)

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

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

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  • $\begingroup$ I haven't looked closely at the details, but $0^{1+i}$ is much easier to assign a value to than $0^i$. If you look at my answer on your other post, I think that $0^z = 0$ is safe to write when $z$ has positive real part. But in general, complex exponentiation is multi-valued, so if you want to use Riemann-Liouville you'll need to decide exactly what you mean by $(t-\tau)^i$. $\endgroup$ Jan 12, 2015 at 20:07

2 Answers 2

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There is a better formula, Fourier transform based differintegral:

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

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

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

$$(\sin x)^{(i)}=i\sinh\left(\frac\pi 2-ix\right)$$

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

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

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:

$$(\sin x)^{(s)}=\sin\left(\frac{\pi s}2+x\right)$$

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

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – davidlowryduda
    Jan 14, 2015 at 5:02
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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:

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

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

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 $\lim_{x\to t} e^{(1+i)\log(t-x)} = 0$, 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 $0^{1+i} = 0^1 \cdot 0^i = 0\cdot 0^i = 0$.

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