Take the Hadamard product for $\zeta(s)$:

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

and reshuffle it into:

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

After experimenting with some values for $s$, I found that f.i. $\zeta(3)$ can be simply written as:

$$\zeta(3) = \prod_\rho \left(1- \frac{2}{\rho} \right) \left(1- \frac{2}{1-\rho} \right)\left(1- \frac{3}{\rho} \right) \left(1- \frac{3}{1-\rho} \right)$$

and believe this can be generalized into ($k= 1,2,3...$):

$$\displaystyle \zeta(2k+1) = a[2k+1] \prod_\rho \left(1- \frac{2k}{\rho} \right) \left(1- \frac{2k}{1-\rho} \right)\left(1- \frac{2k+1}{\rho} \right) \left(1- \frac{2k+1}{1-\rho} \right)$$

with $a[3]=1, a[5]=\frac12, a[7]=\frac15, a[9]=\frac{5}{84}, \dots$

1) Is this a known result? Could the $a[2k+1]$ sequence be derived from the Bernoulli numbers?

2) Could it be proven that when solving $\rho$ for one factor of the infinite product, i.e. only equate the 4 subfactors to $\zeta(2k+1)$, that the complex roots always must have $\Re(s)=\frac12$ ?

  • $\begingroup$ See my comment to Greg Martin's response. $\endgroup$
    – GH from MO
    Nov 21, 2012 at 18:54

1 Answer 1


(1) Yes: multiplying your Hadamard products for $\zeta(2k+1)$ and $\zeta(2k)$ and then dividing by the known formula $\zeta(2k) = |B_{2k}|(2\pi)^{2k}/2(2k)!$, we obtain $$ \zeta(2k+1) = \frac1{(2k-1)2k(2k+1)|B_{2k}|} \prod_\rho \bigg( 1-\frac{2k}\rho \bigg) \bigg( 1-\frac{2k}{1-\rho} \bigg) \bigg( 1-\frac{2k+1}\rho \bigg) \bigg( 1-\frac{2k+1}{1-\rho} \bigg). $$

(2) What?

  • $\begingroup$ Thanks Greg. Very clear. Understand your '(2) What?' response, so let me clarify. I know the $\rho_n$ are given and you obviously can't solve all $\rho$s in an infinite product. However, after playing with a number of $\zeta(2k+1)$s, I noted that solving $x$ in $\zeta(2k+1) = \frac1{(2k-1)2k(2k+1)|B_{2k}|} \bigg( 1-\frac{2k}x \bigg) \bigg( 1-\frac{2k}{1-x} \bigg) \bigg( 1-\frac{2k+1}x \bigg) \bigg( 1-\frac{2k+1}{1-x} \bigg).$, always seems to have a set of complex roots with $\Re(x)=\frac12$. Not sure whether this leads anywhere, but can this be proven? $\endgroup$
    – Agno
    Nov 21, 2012 at 9:30
  • $\begingroup$ @Agno: Your observation follows simply from the fact that all the coefficients in your equation are real (hence I don't think it leads anywhere). Indeed, if $x$ is a root, then so is its complex conjugate $\bar x$, whence $\bar x=1-x$, i.e. $x+\bar x=1$, i.e. $\Re(x)=1/2$. If you replace $\zeta(2k+1)$, $B_{2k}$, $2k-1$ etc. by any real numbers in your equation, you still get $\Re(x)=1/2$ for the roots. $\endgroup$
    – GH from MO
    Nov 21, 2012 at 18:53
  • $\begingroup$ @GH. Thanks. Understand your logic. To be 'very picky' though, I don't think it works for 'any real number' in the equation. For instance with $\zeta(3)$ and the factor $\frac1{(2k-1)2k(2k+1)|B_{2k}|} < 0$, there are 4 complex roots (two pairs $x, \bar x$ ) with $\Re(s) \ne \frac12$. We can obviously ignore this situation, since the factor will always be positive. Outcomes can therefore only be 4 real roots or 2 real roots + 2 complex roots (with always $\Re(x)=1/2$). My guess is that the formula above can only have the latter set of $2+2$ roots due to the factor being always $<1$. $\endgroup$
    – Agno
    Nov 22, 2012 at 16:43

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