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Various questions on MO concerning the "surprise" occurrence of the gamma function in the functional equation of the Riemann zeta function got me wondering whether the Gamma function alone suffice for an analytic formulation of the fundamental theorem of arithmetic.

Pondering this lead me to consider the product

$$(x+1)\prod_n \Gamma\left(1+\frac{x}{n}\right)^{\mu(n)}$$

whose convergence is apparently equivalent to the prime number theorem. (Fix $x\not=0,-1$, apply $\ln$ and linearize to get a series that eventually behaves like $1 - 1/2 - 1/3 -1/5 + 1/6 - 1/7 +\cdots$.) The limit appears to equal $e^x$.

Of course my motivation lay in using Mobius-type inclusion-exclusion to eliminate one by one all the poles.

My question: does this expression occur in the literature?

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Wait--fundamental theorem of arithmetic, or prime number theorem? – Daniel Litt Dec 25 '10 at 8:30
David, the function that you consider is quite interesting. Have you considered that it is based on the product definition of the the $\Gamma(x)$ and the limit form of $e^x$? You also might want to consider the multiplication theorem found at Some modification of this theorem might help. If none of this helps then you might want to prove that the function has no roots. Please get back to us on this. Your function is interesting and I haven't seen it in the literature, so I would advise that you definitely keep working on it :) – backstoreality Dec 25 '10 at 14:07
What is $\mu (n)$? – Anixx Dec 25 '10 at 15:22
@Daniel and Anixx $\mu(n)$ is the Mobius function: -1 to the number of prime factors of a square-free number, 0 for numbers divisible by a square. – David Feldman Dec 25 '10 at 18:14
Thanks Joe! Yes the idea is that the poles of log gamma line up like the positive integers themselves, so one can get multiplicative identities from combinatorial identities. The multiplication theorems you pointed me to were in my mind, and best understood I think in terms of expressing the sequence of poles of log gamma in two different ways - and here I'm just playing a more complicated version of the game. After a night's sleep, I believe I understand the $e^x$. For $x$ a large integer one plays inclusion-exclusion with the factors of $x!$. In the end one needs $d/dx 1/\zeta(z)$ at $z=1.$ – David Feldman Dec 25 '10 at 20:03

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