Is this another version of the gamma function?

Question

I know that $\Gamma \left( x \right) $ is the unique function on $x \in (0, \infty)$ such that

$f \left( 1 \right) =1$

$f(x+1)=xf(x)$

${\frac {d^{2}}{d{x}^{2}}}ln(f \left( x \right))>0$

However define the set of functions $g(x)=\Gamma(x){e^{2\pi inx}}\:for\:n\in \Bbb Z$ and $i=\sqrt{-1}$, then

$g(1)=\Gamma(1)e^{2\pi in}=1$

$g(x+1)=\Gamma(x+1){e^{2\pi in(x+1)}}=x\Gamma(x){e^{2\pi inx}}{e^{2\pi in}}=x\Gamma(x){e^{2\pi inx}}=xg(x)$

${\frac {d^{2}}{d{x}^{2}}}ln(g \left( x \right))={\frac {d^{2}}{d{x}^{2}}}(ln(\Gamma(x))+2\pi i n x)={\frac {d^2}{dx^2}}ln(\Gamma(x))>0$

so therefore $g(x)=\Gamma(x)$ for $x\in(0,\infty)$. However, this is obviously not true, but it satisfies all the conditions of the Bohr-Mollerup theorem. Why is this? Is this some other version of the Gamma function?

If anyone's curious, I got this idea from this post

How did you conclude that $ {\frac {d^2}{dx^2}}\ln(\Gamma(x))>0 $? — Mhenni Benghorbal, Jul 26 '14 at 11:12
Your $g$ is not real-valued on $\mathbb{R}$. The Bohr-Møllerup theorem deals with functions $f\colon (0,\infty) \to (0,\infty)$. (Yes, that is sorely lacking on the wikipedia page.) — Daniel Fischer, Jul 26 '14 at 11:28
@PaulyB: I asked you two questions but you have not answered them yet. — Mhenni Benghorbal, Jul 26 '14 at 11:31
@MhenniBenghorbal: for the first question the imaginary numbers are not ordered, so it's meaning less to say $2\pi ni > 0$ and the answer to the second question is a property of gamma function being log-convex — Pauly B, Jul 26 '14 at 11:36
@DanielFischer Thank you! Wow that is a real important fact... I'm surprised it neglects that on the page — Pauly B, Jul 26 '14 at 11:39

score 3 · Accepted Answer · answered Jul 26 '14 at 13:02

The very anticlimactic answer here is, that the Bohr-Mollerup-Theorem only includes real-valued functions, see here for example. Otherwise, for general complex-valued functions, you also would have problems to evaluate the condition $\frac{d^2}{dx^2}\ln f(x)>0$.

Is this another version of the gamma function?

1 Answers1