Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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
There's a bunch of computer code entitled "Variants of primorial and lcmultorial extended to be continuous over the positive reals" at - maybe it will mean more to others than it means to me. – Gerry Myerson Jul 24 '12 at 22:52
up vote 6 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...

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.