Imaginary-Order Derivative

Solution 1:

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.