What would this evaluate to?

Question

What would this evaluate to? How can you find an analytic continuation? What are the zeroes of this function? And lastly, can we use the zeroes of this function to evaluate the integer counting function?

$$f(s)=\prod_{n=1}^{\infty} \frac {1}{1-\frac{1}{n^s}}$$

(Edit: the product starts at n=2)

Now, we all know the Riemann Zeta function: $$\zeta(s)=\sum_{n=1}^\infty \frac {1}{n^s}$$ And its Euler Product expansion. $$\zeta(s)=\prod_{p \in \Bbb P}^{\infty} \frac {1}{1-\frac{1}{p^s}}$$

Now I am interested in the first function, and whether it contains information on the integer counting function. What exactly is this function? Is there some sort of analytic continuation for it.

I believe that if we could maybe "decode" this function. And find the zeroes. We could gain a better understanding of the Integer counting function (In the case that this function is even related to the integer counting function).

In the end and if this is actually related to the integer counting function, we might be able to get a better understanding of the prime counting function.

So if anyone can evaluate this, it would help me greatly, as this question has been on my mind for several days now and I find myself unable to solve it.

Answer 1

You meant $$f(s)=\prod_{n\ge 2}\frac1{1-n^{-s}}$$ $$\log f(s)=\sum_{k\ge 1} \frac{\zeta(ks)-1}{k}$$ It provides the meromorphic continuation to $\Re(s) > 0$. There is no simpler expression.

The coefficients of $$f(s)=\sum_n a_n n^{-s}$$ have some chance to appear elsewhere, $a_n$ is the number of ways to write $n$ as a product of integers $\ge 2$ where the order doesn't count. When the order counts we obtain $\frac1{1-(\zeta(s)-1)}$. Relating the two is an interesting problem because the Riemann hypothesis can be interpreted as a relation between $\prod_{n\ge 2} \frac1{1+n^{-s}}$ and $\frac1{1+(\zeta(s)-1)}$.

Answer 2

(too long for a comment...)

Let's consider $\;\displaystyle f(s):=\prod_{n=2}^{\infty} \frac {1}{1-\frac{1}{n^s}}\;$ (starting with $n=1\,$ is not so interesting!).

Then $\;\displaystyle I(s):=\log(f(s))=\sum_{k=1}^\infty \frac {\zeta(sk)-1}{k},\;\quad \text{for}\;\Re(s)> 0\quad$ (+1 reuns).
This function was illustrated and extended with many more details in this answer (with additional integral re-writings by the OP and myself. A rather interesting subject btw!). $$-$$ Another track possibly of interest : the closed forms for $\,f(s)\,$ for integers values of $s\ge 2$ :

$$f(s)=\begin{cases} s\prod_{j=1}^{s-1}\Gamma\left(-e^{\pi i\, (j+j/s)}\right) & \text{for $s$ odd}\\ \large{\frac{s(\pi i)^{(s/2)-1}}{\large{\prod_{j=1}^{(s/2)-1}\sin\left(\pi\,e^{2\pi i\,j/s}\right)}}} & \text{for $s$ even} \end{cases}$$ $$\text{(I replaced the somewhat ambiguous notation $(-1)^{\alpha}$})$$ \begin{array} {l|c} s&f(s)\\ \hline 2&2\\ 3&\large{\frac{3\,\pi}{\cosh(\pi\sqrt{3}/2)}}\\ 4&\large{\frac{4\,\pi}{\sinh(\pi)}}\\ \end{array}

This was provided by D. W. Cantrell to MathWorld ($(20)$ to $(26)$) and results from a more general formula in Prudnikov's "Integral and series $1$" (equation $20.$ page $754$ for integers $\,s\ge 2$) :

$$f_x(s):=\prod_{n=1}^{\infty} \frac {\large{1}}{\large{1-\left(\frac xn\right)^s}}=-x^s\prod_{j=0}^{s-1}\;\Gamma\left(-e^{2\pi i\,j/s}\,x\right)$$ (here $\,n\,$ starts at $1$ and limits have to be considered for integer values of $\,x$, note that $x$ can be complex)