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

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


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

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

  • $\begingroup$ What about $2\pi i n $? $\endgroup$ Jul 26, 2014 at 10:53
  • $\begingroup$ Is $2\pi n i >0,\quad i=\sqrt{-1} $. $\endgroup$ Jul 26, 2014 at 11:02
  • $\begingroup$ edited now; now says $i=\sqrt{-1}$ $\endgroup$
    – Pauly B
    Jul 26, 2014 at 11:09
  • 2
    $\begingroup$ 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.) $\endgroup$ Jul 26, 2014 at 11:28
  • 1
    $\begingroup$ @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 $\endgroup$
    – Pauly B
    Jul 26, 2014 at 11:36

1 Answer 1


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 $\frac{d^2}{dx^2}\ln f(x)>0$.


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