Can the $\Xi(t)$ function be extended this way?

$$\xi(s) = \xi(0) \prod_{\rho} \left(1- \frac{s}{\rho} \right) \left(1- \frac{s}{1-\rho} \right)$$

where $\xi(z) = \frac12 z(z-1) \pi^{-\frac{z}{2}} \Gamma(\frac{z}{2}) \zeta(z)$ and $\rho = \sigma + \gamma i$ is a non-trivial zero of $\zeta(s)$ (i.e. $\gamma_n$ is the imaginary part of the n-th $\rho)$).

Riemann had already conjectured this in 1859 and since he needed an 'always real' function for his next thought steps, he assumed $s=\frac12 + t i$ and defined the function $\Xi(t)=\xi(s)$ that has been proven to be equal to:

$$\Xi(t)= \Xi(0)\prod_\gamma\left(1-\frac{t^2}{\gamma^2}\right)$$

It is known that $\xi(s)=\xi(1-s)$ and $\xi(s)=\overline{\xi(\overline{s})}$ and this implies that $\xi(s)\xi(\overline{s})$ must be real. Also known is that when there is a non-trivial zero $\rho$ lying off the critical line, then also $1-\rho, \overline{\rho}, \overline{1-\rho}$ must be zeros and their product will always be real:

$$\displaystyle\prod_\gamma\left(1-\frac{s}{\sigma+i \gamma}\right) \left(1-\frac{s}{\sigma-i \gamma}\right)\left(1-\frac{s}{1-(\sigma+i \gamma)}\right) \left(1-\frac{s}{\overline{1-(\sigma+i \gamma)}}\right)$$

This brought me to the following conjecture. Suppose $s=a+t i$ and $\Xi_a(t)=\xi(s)$ so $\Xi_a(0)=\xi(a)$, then the following product holds and is always real:

$$\Xi_a(t)\Xi_a(-t) = \Xi_a(0)^2\prod_\gamma \left(1-\frac{t^2}{\gamma^2}\right)\left(1-\frac{(-t)^2}{\gamma^2}\right)$$

or simpler:

$$\Xi_a(t)\Xi_a(-t) = \left(\xi(a)\prod_\gamma \left(1-\frac{t^2}{\gamma^2}\right)\right)^2$$

Numerical tests indicate that the conjecture is correct, but I am obvioulsy keen to find a proof (I do realise this is a way too big of a question to ask here!). Does anybody have a link that explains how $\Xi(t)= \Xi(0)\prod_\gamma\left(1-\frac{t^2}{\gamma^2}\right)$ has been derived? Also appreciate any other thoughts/steers on how to further progress this conjecture.

Thanks!

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This seems to be fit way more for overflow than for SE, though I'm happy to see it here. – DonAntonio Jan 6 '13 at 15:52
Mangoldt has $\xi(t)= \xi(0)\cdot \prod(1 -t^2/\alpha_{\nu^2}).$ Are you saying $\Xi$ is some other function? – daniel Jan 6 '13 at 16:25
Your $\gamma$ is Mangoldt's $\alpha.$ He cites Hadamard for this relation so it must be in one of Hadamard's better-known papers? – daniel Jan 6 '13 at 16:41
Indeed Daniel. I guess that's the formula I am after. Keen to find the proof of it in Hadamard's papers! P.S. There is quite a bit of confusion on the use upper and lower case xi. Riemann confusingly used $\xi(t)$ for $\Xi(t)$ where the latter has become the convention. – Agno Jan 6 '13 at 16:45
@Agno. I would not think that Riemann is confusing. Riemann is the first one who defined ($s=1/2+it$): $\xi(t) := \frac12 s(s-1) \pi^{-\frac{s}{2}} \Gamma(\frac{s}{2}) \zeta(s)$. Later on someone defined $\Xi(t):=\xi(s) := \frac12 s(s-1) \pi^{-\frac{s}{2}} \Gamma(\frac{s}{2}) \zeta(s)$. – mike Oct 22 '15 at 0:37

Hoping to redeem my previous incorrect answer, I offer the following. In an article about Riemann's paper in which he discusses the problem of Riemann mentioned in Edwards' book, Fresan gives your relation on page 26 and says:

"La première preuve rigoureuse de cette identité est dûe à Hadamard, qui l’inclut dans son article “Étude sur les propriétés des fonctions entières et en particulier d’une fonction considérée par Riemann” (1893)." [Full cite at linked PDF.] ["The first rigorous proof of this identity is due to Hadamard, who includes it in his article...etc."]

Note that he has retained Riemann's notation, which employs $\xi$ for $\Xi.$

So this would be the paper of Hadamard in which the proof appears.

Glancing at the paper of Fresan again, I wonder if his calculation on page 26 is not responsive to your underlying question?

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Daniel. I have Edwards' book on my lap and open at page 20. It only shows the product formula for $\xi(s)$ i.e. the lower case version. I am keen to find out how the the product for the $\Xi(t)$, i.e. upper case version assuming $s=\frac12 +t i$, has been derived. – Agno Jan 6 '13 at 16:10
Merci bien, Daniel :-) Dusting off my French. You are right. P26 seems to outline the trick for $s=\frac12+ti$! Now need to check whether the same logic works for $P(a+ti)$ that is equivalent to my $\xi(a+s i)$. – Agno Jan 6 '13 at 18:38
@user46149: Appreciate the up-vote. You should probably wait a bit before accepting because someone may provide a more authoritative answer. I am not an expert but there are many on this site. – daniel Jan 6 '13 at 19:05
Daniel. Yes! This indeed does the trick. Many thanks. I can now prove that when I f.i. assume that all zeros in the infinite product of $\left(\frac{1}{\rho}\right)\left(\frac{1}{\overline{\rho}}\right)$ lie on the line $\rho=\frac13$, that the formula for $\Xi(t)_\frac13$ still perfectly holds :-) Landau's Handbuch is also correct that $\Xi_\frac12(z) = \Xi_\frac12(-z)$ but this is not true for other values of $a$ ! – Agno Jan 6 '13 at 19:09
Much appreciated your help :-) – Agno Jan 6 '13 at 19:11

For completeness' sake and to avoid confusion, I can now confirm that the conjecture (even though the numerical results are very close) must be false. It is easy to see that when:

$$\left(1-\frac{t^2}{\gamma^2}\right)$$

would be independent of $a$, then it could be expressed as:

$$\frac{\Xi(t)}{\Xi(0)}$$

This function has the same zeros as $\zeta(s)$ and therefore its square can not be equal to $\dfrac{\Xi_a(t)\Xi_a(-t)}{\xi(a)^2}$ since the latter doesn't have zeros at those spots (otherwise the RH would also have been immediately proven false).

Back to the drawing board.

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