I bet the product $$ \prod_{n=2}^\infty\frac 1 {1-n^{-s}}, $$ which is convergent for ${\rm Re}(s)>1$, has been studied before. Can it be analytically extended across the line ${\rm Re}(s)=1$? If so, is there a functional equation?
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2 Answers
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Yes, the inverse of it, $\prod_{n=2}^{\infty} \frac{n^s-1}{n^s}$
has been studied in connection with the so-called $k$-almost prime zeta functions of the second kind.
It can be evaluated to
$$
\prod_{n=2}^{\infty} \frac{n^s-1}{n^s}=\prod_{n=1}^{\infty} \frac{(n+1)^s-1}{(n+1)^s}=
\frac{1}{\prod_{j=1}^{s-1}\Gamma (2-e^{2\pi ij/s})}.
$$
Then one can use the properties for the $\Gamma$-function.
A reference is the paper by Richard J Mathar, on HARDY-LITTLEWOOD CONSTANTS EMBEDDED INTO
INFINITE PRODUCTS OVER ALL POSITIVE INTEGERS, and the references therein. The connection with the $k$-almost prime zeta function is given by
$$
\prod_{n=2}^{\infty}\left(1-\frac{1}{n^s}\right)=\prod_{k=1}^{\infty}\frac{1}{\zeta_k(s)}
.$$
Edit: I apologize, but $s$ is assumed to be an integer above. But the question has been discussed already, see Values where infinite products of primes and composites are equal.
answered Jun 12, 2013 at 7:53
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$\begingroup$ @Dietrich: thanks for the answer. However, this is for integral $s$ only. I would be interested in complex $s$ and analytic continuation. $\endgroup$– user1688Jun 12, 2013 at 8:01
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$\begingroup$ @Anton: yes, I am sorry. I found the above discussion on your product, called $T(s)$, but I am not sure if this gives you an answer, i.e., how to extend $T(s)$ meromorphically to $\mathbb{C}$. $\endgroup$ Jun 12, 2013 at 9:11
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Yes, It can be analytically extended across the line Re$(s)=1$. Taking the logarithm of the function and using the series expansion of $\log(1+x)$ gives you the sum
$$ \sum_{k=1}^\infty \frac{\zeta(ks)-1} k $$ which has a meromorphic continuation up to the line Re$(s)=0$, with poles at $s=1/k$. There is some speculation on the other page on the problem Values where infinite products of primes and composites are equal whether the continuation should go beyond the line Re$(s)=0$ but I agree with Greg Martin's comment that this is not clear. However you asked about the line Re$(s)=1$ and that follows from the above. Note though that the logarithm of the function has meromorphic continuation and this means that your function does not have meromorphic continuation since the singularity at Re$(s)=1$ will not be a pole. I doubt that the function has a functional equation.
