Using the norm

$\displaystyle | \frac {1}{1-1/p^{s}} |^2 =

\left( \frac {1}{1-1/p^{2a}} \right)

\left(\frac {p^{2a}-1}{p^{2a}-2 p^{a}\cos(b\ln{p})+1} \right) (1)$

valid for any $\displaystyle s = a + ib \in \field{C}$, $\displaystyle p \in \field{N}$, consider the case of $\displaystyle a=Re\left( s \right) > \frac {1}{2}$, and multiplying (1) over all primes p, define a real function of two variables:

$\displaystyle

| M \left( a, b \right) |^2 \equiv \prod^{\infty}_{p=primes} | \frac {1}{1-1/p^{s}} |^2

=\zeta \left( 2a \right) \prod^{\infty}_{p=primes}

\left(\frac {p^{2a}-1}{p^{2a}-2 p^{a}\cos(b\ln{p})+1} \right) (2)$

where $\displaystyle \zeta$ is the Riemann Zeta function.

For $\displaystyle a=Re\left( s \right) > 1$, the right-hand product of (2) converges absolutely and in that range one can equate:

$\displaystyle

\ln \left( \frac {|M \left( a, b \right) |^2 } { \zeta \left( 2a \right) } \right) =

\ln \left( \frac{| \zeta \left( a, ib \right) | ^2 } { \zeta \left( 2a \right) } \right)

=\sum^{\infty}_{p=primes} \ln \left(\frac {p^{2a}-1}{p^{2a}-2 p^{a}\cos(b\ln{p})+1} \right) (3)$

Apply the Laplacian $\displaystyle \nabla^2=\frac{\partial^2 }{\partial a^2} + \frac{\partial^2 }{\partial b^2}$ on each side of (3):

$\displaystyle \nabla^2 \ln \left( \frac{| \zeta \left( a, ib \right) | ^2 } { \zeta \left( 2a \right) } \right)

=\sum^{\infty}_{p=primes} \nabla^2 \ln \left(\frac {p^{2a}-1}{p^{2a}-2 p^{a}\cos(b\ln{p})+1} \right)

= -4 \left( \sum^{\infty}_{p=primes} \left( \frac{ p^{2a} \ln ^2(p)} { (p^{2a}-1)^2 } \right) \right) (4)$

The right-hand side of (4) converges absolutely for $\displaystyle a=Re\left( s \right) > \frac {1}{2}$ without claims as to the convergence of (3) in the same range. Thus the left-hand side of (4) exists, is finite and always negative for all

$\displaystyle s = a + ib \in \field{C}$, $a=Re\left( s \right) > \frac {1}{2}$ which precludes the existence of zeroes of Zeta to the right of the critical line(?)

Can anyone spot an obvious flaw? I have been twisting it in my head for quite a while before posting but can't seem to see what is wrong except it is too simple...The last step in (4) was checked with Mathematica online and is pretty straightforward. Thanks.