I found that the following infinite product with $\mu = a +n b i$ and a,b real, $s \in \mathbb{C}$:

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

can be expressed in a closed form (with poles at $a,b = 0$ or $a=1$ and when $a=s$):

$$\dfrac{\left( {a}^{2}-a \right)} {\left( {a} ^{2}-a+s-{s}^{2} \right)} \dfrac{\Gamma \left( {\frac {-ia}{b}} \right) \Gamma \left( {\frac {-i \left( a-1 \right) }{b}} \right)}{\Gamma \left( {\frac {-i \left( a-s \right) }{b}} \right) \Gamma \left( {\frac {-i \left( a+s-1 \right) }{b}} \right)}$$

When $a=\frac12$ this could be further reduced to (poles at $s=\frac12$ and $b=0$):

$$ \dfrac{1}{(2s-1)} \dfrac{\sinh \left( {\frac { \left( 2s-1 \right) \pi }{2b}} \right)} { \sinh \left({\frac {\pi }{2b}} \right)}$$

Encouraged by this result, my wish was to use it to find new hints about the Hadamard product:

$$\displaystyle \prod_\rho \left(1- \frac{s}{\rho} \right) \left(1- \frac{s}{1-\rho} \right) = \dfrac{2(s-1)\Gamma(1+\frac{s}{2}){\zeta(s)}}{ \pi^{\frac{s}{2}}}$$

but it is quite obviously an impossible task to transform the linear element $nb$ into to very random imaginary parts of the $\rho$s. However, it still triggered a follow up question:

With $\rho = \sigma + ti$ and $t,x$ real, the following product:

$$Had(s,x):=\displaystyle \prod_\rho \left(1- \frac{s}{\sigma + xti} \right) \left(1- \frac{s}{1-(\sigma + xti)} \right)$$

allows for "scaling" of the imaginary parts of the $\rho$s.

Since $Had(s,1)$ has a closed form and is entire, does this imply that the (linearly) scaled $\Im(\rho_n)$ must also induce entire functions and have closed forms (possibly related to $\zeta(s)$ and assuming RH is true)?

Edit: Extra question:

To take it a step further: similar for the infinite products with $n$ above, could the known closed form:

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

just be the 'reduced' version for $\sigma=\frac12$ and be extended with $\sigma$ and $x$ to express $Had(s,x,\sigma)$?


I wrote a program to calculate $Had(s,x)$ by using the first 2 mln $\rho$s from Andrew Odlyzko's table, however when calibrating the results with the known $Had(2,1) =\dfrac{\pi}{3}$, I found that the accuracy is limited to 5 decimals max. (i.e. too few to link it to known constants). Are there any larger $\rho$-files available on the web?

  • $\begingroup$ For more zeta zeros check out lmfdb.org/zeros/zeta There are over 36 million there. $\endgroup$
    – ncr
    Dec 2, 2012 at 3:56
  • $\begingroup$ For real $x$ and RH the product should be purely real on the critical line because of cancellation, right? Wouldn't this allow expressing $\Re \zeta(z) / \Im \zeta(z)$ in "elementary" functions for some $z$? $\endgroup$
    – joro
    Mar 25, 2013 at 15:56

4 Answers 4


I assume Riemann hypothesis on all this answer.
I want a closed form for

$$f(s,x):= \prod_{n=1}^\infty\Bigl(1-\frac{s}{\frac12+i x\gamma_n}\Bigr) \Bigl(1-\frac{s}{\frac12-i x\gamma_n}\Bigr).$$ Of course $\gamma_n$ runs here over the ordinates of the zeros of $\zeta(s)$ but only those with $\gamma>0$.

We know that $$\Xi(t)=\Xi(0)\prod_\gamma\Bigl(1-\frac{t^2}{\gamma^2}\Bigr).\qquad (1)$$ Therefore $$\Xi(t/x)=\Xi(0)\prod_\gamma\Bigl(1-\frac{t^2}{x^2\gamma^2}\Bigr).$$ Substitute here $s=\frac12+it$ then $it=s-\frac12$ $$\Xi(t/x)=\Xi(0)\prod_\gamma\Bigl(1+\frac{(s-\frac12)^2}{x^2\gamma^2}\Bigr)=$$ $$= \Xi(0)\prod_\gamma\Bigl(\frac{(s-\frac12-ix\gamma)(s-1/2+ix\gamma)}{x^2\gamma^2}\Bigr).$$ Now we call $\rho=\frac12+ix\gamma$ and we get $$\Xi(t/x)=\Xi(0) \prod_\gamma\Bigl(\frac{\rho(1-\rho)}{x^2\gamma_n^2}\Bigr)\cdot \prod_\gamma\Bigl(1-\frac{s}{\rho}\Bigr)\Bigl(1-\frac{s}{1-\rho}\Bigr).$$ By (1), this is equal to $$\Xi(t/x)=\Xi(0) \prod_\gamma\Bigl(\frac{\frac14+x^2\gamma^2}{x^2\gamma^2}\Bigr)\cdot \prod_\gamma\Bigl(1-\frac{s}{\rho}\Bigr)\Bigl(1-\frac{s}{1-\rho}\Bigr)=$$ $$= \Xi(i/2x)\prod_\gamma\Bigl(1-\frac{s}{\rho}\Bigr)\Bigl(1-\frac{s}{1-\rho}\Bigr)$$

Therefore we have $$\prod_\gamma\Bigl(1-\frac{s}{\rho}\Bigr)\Bigl(1-\frac{s}{1-\rho}\Bigr)= \frac{\Xi(t/x)}{\Xi(i/2x)}$$

By definition we have

$$\Xi(t)=\frac{s(s-1)}{2}\pi^{-s/2}\Gamma(s/2)\zeta(s), \qquad \text{if} \quad s=\frac12+it.$$

I shall continue in other answer because the TeX do not runs well


... continue the above answer

Therefore $$\prod_\gamma\Bigl(1-\frac{s}{\rho}\Bigr)\Bigl(1-\frac{s}{1-\rho}\Bigr)= \frac{\frac14+\frac{t^2}{x^2}}{\frac14-\frac{1}{4x^2}}\frac{\pi^{-\frac{it}{2x}}} {\pi^{\frac{1}{4x}}}\frac{\Gamma(\frac14+\frac{it}{2x})} {\Gamma(\frac14-\frac{1}{4x})}\frac{\zeta(\frac12+i\frac{t}{x})} {\zeta(\frac12-\frac{1}{2x})},$$ or equivalently $$\prod_\gamma\Bigl(1-\frac{s}{\rho}\Bigr)\Bigl(1-\frac{s}{1-\rho}\Bigr)=$$ $$ \frac{x^2-4(s-\frac12)^2}{x^2-1}\pi^{-s/2x} \frac{\Gamma(\frac{1}{4}-\frac{1}{4x}+\frac{s}{2x})} {\Gamma(\frac14-\frac{1}{4x})}\frac{\zeta(\frac12+\frac{s}{x}-\frac{1}{2x})} {\zeta(\frac12-\frac{1}{2x})}.$$ If my computation are correct.

  • $\begingroup$ If my computations are correct, I do not think the result depends on Riemann hypothesis. Really my computation does not have assumed it. I have not used my gamma's are real. $\endgroup$
    – juan
    Dec 2, 2012 at 11:31
  • $\begingroup$ Many thank Juan. Checked your formula against the numbers I have computed with 'brute force' and it all looks correct! The structure of your closed form also looks very similar to the ones for the infinite products with $n$ and therefore might be easily expandable to $Had(s,x, \sigma)$. Could the RH just be that $Had(s,1, \frac12) = \dfrac{2(s-1)\Gamma(1+\frac{s}{2}){\zeta(s)}}{ \pi^{\frac{s}{2}}}$ (i.e. reduced form of a more general product formula)? $\endgroup$
    – Agno
    Dec 2, 2012 at 11:45
  • $\begingroup$ @Juan. It is indeed easy to expand and: $$Had(s,x, \sigma)=\frac{x^2-4(s-\sigma)^2}{x^2-1}\pi^{-s/2x} \frac{\Gamma(\sigma^2-\frac{1}{4x}+\frac{s}{2x})} {\Gamma(\sigma^2-\frac{1}{4x})}\frac{\zeta(\sigma+\frac{s}{x}-\frac{1}{2x})} {\zeta(\sigma-\frac{1}{2x})}$$ For $x \rightarrow 1$ it reduces to the known form for $\sigma=\frac12$. What does this mean? $\endgroup$
    – Agno
    Dec 2, 2012 at 12:31
  • $\begingroup$ I do not think this is the correct formula, $\zeta(\sigma+\frac{s}{x}-\frac{1}{2x})$ has zeros at $\frac{1+x}{2}-\sigma x+ix\gamma$ where do you want $Had(s,x,\sigma)$ have its zeros? $\endgroup$
    – juan
    Dec 2, 2012 at 12:45
  • $\begingroup$ Juan, you're right. I was way too quick. This is not the correct formula $Had(s,x,\sigma)$. $\endgroup$
    – Agno
    Dec 2, 2012 at 14:33

Have not given up yet on whether or not there exists a closed form for:

$$Had(s, \sigma, x):=\displaystyle \prod_\rho \left(1- \frac{s}{\sigma + xti} \right) \left(1- \frac{s}{1-(\sigma + xti)} \right)$$

that, as Juan proved above, reduces to (assuming RH):

$$Had(s,\frac12,x):=\frac{x^2-4(s-\frac12)^2}{x^2-1}\pi^{-s/2x} \frac{\Gamma(\frac{1}{4}-\frac{1}{4x}+\frac{s}{2x})} {\Gamma(\frac14-\frac{1}{4x})}\frac{\zeta(\frac12+\frac{s}{x}-\frac{1}{2x})} {\zeta(\frac12-\frac{1}{2x})}$$

and for $x=1$, further reduces to the Hadamard product:

$$Had(s, \frac12, 1):=\dfrac{2(s-1)\Gamma(1+\frac{s}{2}){\zeta(s)}}{ \pi^{\frac{s}{2}}}$$

Assuming RH, a closed form for $Had(s, \sigma, x)$ requires:

  • $Had(0, \sigma, x)=1$ and $Had(1, \sigma, x)=1$.
  • $Had(s, \sigma, x)= Had(s, 1-\sigma, x)$
  • $Had(\frac12, \sigma, x)$ is the function's minimum.
  • $Had(s, \sigma, x)$ to reduce to the closed forms for $Had(s,\frac12,x)$ and $Had(s, \frac12, 1)$
  • the Zeta function's non-trivial zeros to be the 'source' for all (horizontally shifted) zeros.
  • the function to be entire (all poles annihilated by zeros).

The following function does meet all the criteria, except for the second:

$$\displaystyle {\frac {{x}^{2}-4 \left( \sigma-s \right) ^{2}}{{x}^{2}-4 \left( 2s\sigma- s-\sigma \right) ^{2}}}{\pi }^{{\frac {s \left( \sigma-1 \right) }{x}}} \dfrac{\Gamma \left( \dfrac{\frac12-{\frac {\sigma}{x}}+{\frac {s}{x}}}{2}\right)}{\Gamma \left( \dfrac{\frac12-{\frac {\sigma}{x}}+{\frac {s(2\sigma-1)}{x}}}{2}\right)} \dfrac{\zeta \left( \frac12-{\frac {\sigma}{x}}+{\frac {s}{x}} \right)}{\zeta \left( \frac12-{\frac {\sigma}{x}}+{\frac {s(2\sigma-1)}{x}} \right)}$$

My 'brute force' infinite product calculations (based on the first 2 mln zeros) show that the shapes of the curves are close (but not equal), however the results for $Had(s, \sigma, x)$ and $Had(s, 1-\sigma, x)$ differ (slightly, yet consistently) from each other.

Could there be any way to improve this?


About database of zeros. mpmath can compute the zeta zeros to arbitrary precision and sage has optional package containing a lot of zeros (though IIRC with not much precision).

You are asking about products over zeros with scaled imaginary parts, but I suppose such sums are much easier, including finding closed form assuming RH (computing such sums without RH will be interesting to me).

André Voros's More Zeta Functions for the Riemann Zeros explains how to compute:

$$Z_1(\sigma,v) = \sum_{k=1}^\infty (\tau_k^2+v)^{-\sigma} $$

$$Z_2(s,x) = \sum_{k=1}^\infty (x-\rho)^{-s} $$

where {$\rho$} = { $\frac12 \pm i \tau_k$ } k=1,2,... = {the Riemann zeros}

Assume RH (this means $\tau_k \in \mathbb{R}$) and suppose you want to compute $\sum_\rho \frac{1}{1/2+ 2 i \tau_k}$.

Grouping $\rho, 1-\rho$ gives $\sum_\rho \frac{1/4}{\tau_k^2+1/16}$ which is directly computable by $Z_1(1,1/16)$.

Modulo errors the last sum is:

$$1/12\,{\frac {-16\,\zeta \left( 3/4 \right) +3\,\Psi \left( 3/8 \right) \zeta \left( 3/4 \right) -3\,\ln \left( \pi \right) \zeta \left( 3/4 \right) +6\,\zeta' \left(3/4 \right) }{\zeta \left( 3/ 4 \right) }}$$

The same approach works for other scalings.


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