Assume $s,a \in \mathbb{C}, a \pm in \ne 0$.

The following infinite product nicely converges and can be expressed in a closed form:

$$\displaystyle \prod_{n=1}^\infty \left(1- \frac{s}{a+i n} \right) \left(1- \frac{s}{{a- i n}} \right) = {\frac {a\sinh \left( \pi \left( a-s \right) \right) }{ \left( a-s \right) \sinh \left( \pi a \right) }}$$

The individual factors however diverge, so I tried to exchange the sub factors for each $n$ and found that:

$$\displaystyle \prod_{n=1}^\infty \left(1- \frac{s}{a+ (-1)^n i n} \right)$$


$$\displaystyle \prod_{n=1}^\infty \left(1- \frac{s}{a+ (-1)^{n+1} i n} \right)$$

actually do (slowly but surely) converge and when multiplied together yield the closed form above.

Is there a closed form for these two individual factors?

EDIT (and follow up question):

Many thanks to Carlo for answering the question so quickly.

A (maybe too) provocative follow up question deals with a similar product that also has a closed form (when assuming RH is true):

$$\prod_{n=1}^\infty \left(1- \frac{s}{a + i \Im(\rho_n)} \right) \left(1- \frac{s}{{a - i \Im(\rho_n)}} \right) = \frac{\xi_{rie}(\frac12 - a+s)}{\xi_{rie}(\frac12 - a)}$$

so runs through the non-trivial zeros $\rho_n$ with:$\xi_{rie}(s)= \frac12 s(s-1) \pi^{-\frac{s}{2}} \Gamma(\frac{s}{2}) \zeta(s)$

Again, the individual factors diverge, but after exchanging the sub factors for each $n$, I again found that:

$$\displaystyle \prod_{n=1}^\infty \left(1- \frac{s}{a+ (-1)^n i \Im(\rho_n)} \right)$$


$$\displaystyle \prod_{n=1}^\infty \left(1- \frac{s}{a+ (-1)^{n+1} i \Im(\rho_n)} \right)$$

actually do converge.

A closed form for these would obviously be quite spectacular... Could it exist?


In another context, I asked about the similarity between these products here: closed forms infinite products


2 Answers 2


$\displaystyle \prod_{n=1}^\infty \left(1- \frac{s}{a+ (-1)^n i n} \right)=\frac{\Gamma \left(\frac{1}{2}+\frac{i a}{2}\right) \Gamma \left(1-\frac{i a}{2}\right)}{\Gamma \left(\frac{1}{2}-\frac{1}{2} i (s-a)\right) \Gamma \left(1+\frac{1}{2} i (s-a)\right)}$

$\displaystyle \prod_{n=1}^\infty \left(1- \frac{s}{a+ (-1)^{n+1} i n} \right)=\frac{\Gamma \left(1+\frac{i a}{2}\right) \Gamma \left(\frac{1}{2}-\frac{i a}{2}\right)}{\Gamma \left(1-\frac{1}{2} i (s-a)\right) \Gamma \left(\frac{1}{2}+\frac{1}{2} i (s-a)\right)}$

  • $\begingroup$ Brilliant. Many thanks, Carlo! I have followed up with a (very) provocative second question, that might not have an answer like this, but maybe it does... $\endgroup$
    – Agno
    Commented Mar 28, 2013 at 23:04

I did explore the second question a bit further and now have a follow up question.

Assuming the RH, the following product:

$$\displaystyle \displaystyle \prod_{n=1}^\infty \left(1- \frac{s}{a+ (-1)^n i \Im(\rho_n)} \right) \left(1- \frac{s}{a+ (-1)^{n+1} i \Im(\rho_n)} \right) = \frac{\xi_{rie}(\frac12 - a+s)}{\xi_{rie}(\frac12 - a)}$$

runs through the alternating non-trivial zeros $\rho_n$, with $\xi_{rie}(s)= \frac12 s(s-1) \pi^{-\frac{s}{2}} \Gamma(\frac{s}{2}) \zeta(s)$.

Based on $\sinh(s)$ in the original question being factored into closed forms containing $\Gamma(s)$ and $\Gamma(1-s)$, I more and more believe that similar closed form factors could exist for $\frac{\xi_{rie}(\frac12 - a+s)}{\xi_{rie}(\frac12 - a)}$.

Assume $a=\frac12$. This makes the function to be split into two factors:

$$s(s-1) \pi^{-\frac{s}{2}} \Gamma(\frac{s}{2}) \zeta(s)$$

For $s$ and $s-1$ factoring seems trivial, and also for $\Gamma(\frac{s}{2})$ and $\pi^{-\frac{s}{2}}$, further factoring might be done by for instance using the function: $\Gamma(\frac{s}{2}) = \Gamma(\frac{s}{4})\Gamma(\frac{s}{4}+\frac12)\pi^{-\frac12}2^{\frac{s}{2}-1}$.

The tricky part lies in how to factor $\zeta(s)$ into two complementary functions that each generate alternating non-trivial zeros; i.e. for the first function: $\frac12+14.13...i$, $\frac12-21.02...i$, $\frac12+25.01...i$, etc.

Such a function would still require $\zeta(s)$ itself, since it is the only known function that induces $\rho_n$'s, but it now also requires additional information on each subsequent zero about whether it should be generated or be 'suppressed'. Such information does not seem to exist in the real or the imaginary parts of $\zeta(s)$ nor does it in its derivatives. However I did find that when defining $Ξ(t) = \xi_{rie}(s)$ when $s=\frac12 + ti$:

$\displaystyle \frac{|\Xi'(t)|}{\Xi'(t)}$ alternates between $-1$ and $1$ at each subsequent $\rho_n$.

And this brought me to the following factors for $\zeta(s)$, that do what they should do, but I do find very ugly:

$$\displaystyle \zeta\left(1 \pm \frac{|\Xi'(t)|}{2\Xi'(t)}+it \right)$$

There must be a smarter way to produce the alternating $\rho_n$'s and I'd appreciate any help/steers you have.



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