Interpolating the primorial $p_{n}\#$ 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.
 A: 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... 
