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The primorial $p_{n}\#$ is given by the product $p_n\# = \prod_{k=1}^n p_k$ (where $p_{k}$ is the $k$th prime) -- is there a natural (a la the gamma function $\Gamma(z)$) way of interpolating it for arguments not necessarily a natural number? (or in $\mathbb{C}$?)

I tried starting with the following definition of the gamma function:

$$\Gamma(z) = \lim_{n \to \infty} \frac{n! \; n^z}{z \; (z+1)\cdots(z+n)} = \frac{1}{z} \prod_{n=1}^\infty \frac{\left(1+\frac{1}{n}\right)^z}{1+\frac{z}{n}}$$

My first thought was to modify the Pochhammer symbol in the denominator:

$$\Gamma_{?}(z) = \lim_{n \to \infty} \frac{p_{n}\# \; p_{n}^z}{z \; (z+p_{1})\cdots(z+p_{n})}$$

But this clearly doesn't work, because the primes aren't regularly spaced.

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Maybe first you should decide what's "natural" about the Gamma function. –  Gerry Myerson Feb 6 '12 at 10:08
    
Or consider what might a function that is the generating function for the primorial (or its reciprocal) might look like, and then apply Cauchy's differentiation formula. –  J. M. Feb 6 '12 at 10:16
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There's a bunch of computer code entitled "Variants of primorial and lcmultorial extended to be continuous over the positive reals" at phodd.net/gnu-bc/code/orialc.bc - maybe it will mean more to others than it means to me. –  Gerry Myerson Jul 24 '12 at 22:52
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1 Answer 1

up vote 5 down vote accepted

Take the log of $p_n\# = \prod_{k=1}^n p_k$ to get $$ \log p_n\# = \sum_{k=1}^n \log p_n, $$ where you recognize the first Chebyshev function $\theta(n)$, which has an asymptotic behaviour of $\theta(n)\sim n$. Write the sum as integral and use $$ \int_2^x f(t) d(\pi(t))=f(t)\pi(t)\biggr|_{2}^{x}+\int_{2}^{x}f'(t)\pi(t)dt. $$ from here to get: $$ \begin{eqnarray} \sum_{k=1}^n \log p_n &=& \int_2^n \log k\; d\pi(k)\\ &=& \log(k)\pi(k)\biggr|_{2}^{n}+\int_{2}^{n}\frac1k \pi(k)dk. \end{eqnarray} $$ Now, put in your favorite representation for $\pi(x)$, like $ \pi(x) = \operatorname{R}(x^1) - \sum_{\rho}\operatorname{R}(x^{\rho}) , $ with $ \operatorname{R}(x) = \sum_{n=1}^{\infty} \frac{ \mu (n)}{n} \operatorname{li}(x^{1/n})\;$ and $\rho$ running over all the zeros of $\zeta$, to get $\log p_n\#\;$.

Choose, for example, the approximation $\pi(n)\sim \frac{n}{\log n}$, then you get $$ \log p_n\# \sim \log(k)\frac{k}{\log k}\biggr|_{2}^{n}+\int_{2}^{n}\frac1k \frac{k}{\log k}dk = (n-1)+\text{Li}(x) \;. \tag{$*$} $$ Exponentiate $(*)$ and you are done...

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