The falling and rising factorial are defined by

$$z^\underline n:=\prod_{k=0}^{n-1}(z-k),\quad z^\overline n:=\prod_{k=0}^{n-1}(z+k),\quad z\in\Bbb C,n\in\Bbb N_{\ge 0}\tag{1}$$

In first place $(1)$ seems to be equivalent to

$$z^\underline n=\lim_{w\to z}\frac{w!}{(w-n)!},\quad z^\overline n=\lim_{w\to z}\frac{(w+n-1)!}{(w-1)!}\tag{2}$$

where $z!:=\Gamma(z+1)$ (at least for most cases, not sure about it full equivalence). Then from $(2)$ a natural extension seems to be

$$z^\underline\alpha:=\lim_{w\to z}\frac{w!}{(w-\alpha)!},\quad z^\overline\alpha:=\lim_{w\to z}\frac{(w+\alpha-1)!}{(w-1)!},\quad z,\alpha\in\Bbb C\tag{3}$$

where fortunately the important functional equations

$$\quad z^{\underline\alpha}=(z-\alpha+1)^\overline\alpha,\quad z^\overline\alpha=(z+\alpha-1)^\underline\alpha\\ z^{\underline {-\alpha}}=\frac1{(z+1)^\overline \alpha},\quad z^{\overline{-\alpha}}=\frac1{(z-1)^\underline \alpha},\quad z,\alpha\in\Bbb C\tag{4}$$

seems to hold (at least in most cases).

My problem comes when I tried to go further generalizing the already generalized factorial power defined by

$$z^{[n,\beta]}:=\prod_{k=0}^{n-1}(z+\beta k)=\lim_{w\to z}\beta^n\frac{(w/\beta+n-1)!}{(w/\beta-1)!}=\beta^n (z/\beta)^\overline n,\quad z,\beta\in\Bbb C\tag{5}$$


$$z^{[\alpha,\beta]}:=\beta^\alpha (z/\beta)^\overline\alpha,\quad z,\alpha,\beta\in\Bbb C\tag{6}$$

The first problem that I found with $(6)$ is that it is supposed that $z^{[\alpha,-1]}$ must be equivalent to $z^\underline\alpha$, however with $(6)$ I get that

$$z^{[\alpha,-1]}=(-1)^\alpha (-z)^{\overline\alpha}=e^{i\pi\alpha}(-z+\alpha-1)^\underline\alpha\tag{7}$$

Then I tried to solve the equation

$$e^{i\pi\alpha}(-z+\alpha-1)^\underline\alpha=z^\underline\alpha\iff e^{i\pi\alpha}\lim_{w\to z}\frac{(-w+\alpha-1)!}{(-w-1)!}=\lim_{w\to z}\frac{w!}{(w-\alpha)!}\tag{8}$$

For $z\notin\Bbb Z$ and $z+\alpha\notin\Bbb Z$ we can ignore the limit process, then using the following identity

$$z!\,(-z)!=\frac{\pi z}{\sin(\pi z)},\quad z\notin\Bbb Z\tag{9}$$

I had written $(8)$ as

$$e^{i\pi\alpha}\frac{\pi(z-\alpha)}{\sin(\pi(z-\alpha))}\cdot\frac1{\alpha-z}=\frac{\pi z}{\sin(\pi z)}\cdot\frac1{-z}\implies\frac{\sin(\pi(z-\alpha))}{\sin(\pi z)}=e^{i\pi\alpha}\\\implies e^{i\pi(z-\alpha)}-e^{-i\pi(z-\alpha)}=e^{i\pi(z+\alpha)}-e^{-i\pi(z-\alpha)}\\\implies e^{i\pi(z-\alpha)}=e^{i\pi(z+\alpha)}\implies e^{i2\pi\alpha}=1\tag{10}$$

And from $(10)$ we can see that $(6)$ doesnt work as intended (however is clear that $(5)$ works correctly, because $e^{i2\pi n}=1$ for $n\in\Bbb N$).

Question: there is a known way to extend $(5)$ to complex $n$?

WIP: a possible solution could be stick to $(2)$ and $(6)$ and redefine $(3)$.


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