## help needed with this Riemann hypothesis argument

Hi, I would be grateful for any comments pointing out gross mistakes in this "proof" as it seems too simple (and short!) to be correct. Did I miss something? Wrong assumptions? Is it common knowledge, and fails for large values of b? I tried to double check each step numerically and it seems correct within machine precision. Thank you.

The reflection functional equation for the Riemann zeta function is
$\zeta \left(1- a-i b \right) = \frac {2}{\left(2 \pi \right)^{a+i b}} \cos(\frac{(a +i b) \pi}{2})\Gamma(a+i b) \zeta \left( a+i b \right) (eq.1)$

where $a, b \in \{R}$ and $a + ib \in \{C}$. For simplicity, restrict what follows to $a, b > 0$, and multiply each side of (1) by its complex conjugate to obtain:
$| \zeta \left(1- a- ib \right) |^2 = \frac {2} {\left(2 \pi \right)^{2a}} (\cos(a\pi) + \cosh(b\pi)) \Gamma(a+i b))\Gamma(a-i b) | \zeta \left( a+ ib \right) |^2 (eq. 2)$

Using $(a+ib)\Gamma(a+i b) = \Gamma(1+a+i b)$ and the Ramanujan identity:

$\frac {\Gamma(a+1)^2} {\Gamma(1+a+i b)\Gamma(1+a-i b)} = \prod^{\infty}_{k=1} \left(1 + \frac {b^2}{(a+k)^2}\right) (eq. 3)$

where the infinite product in (3) converges absolutely for $a, b > 0$, substitute in (2) to obtain:

$\frac {| \zeta \left( a+ ib \right) |^2} {| \zeta \left(1- a- ib \right) |^2} = \frac {\left(2 \pi \right)^{2a} (a^{2}+b^{2})}{2 (\cos(a\pi) + \cosh(b\pi)) \Gamma(a+1)^2 } \prod^{\infty}_{k=1} \left(1 + \frac {b^2}{(a+k)^2}\right) (eq. 4)$

The left-hand side of (4) is identically equal to 1 for all values of b when a= 1/2 (i.e. on the critical line) and for all other values of a, b on the critical strip 0<a<1 that are zeroes of $\zeta \left( a+i b \right)$ since the zeroes are symmetrically distributed about the critical line.

Take the logarithm on each side of (4) and its partial derivative with respect to a,
$\frac {\partial \ log { \frac {| \zeta \left( a+ ib \right) |^2} {| \zeta \left(1- a- ib \right) |^2} } }{\partial a}=2 \ log {(2 \pi) } + \frac{2a} { (a^{2}+b^{2})} + \frac{ \pi \sin(a\pi) } { (\cos(a\pi) + \cosh(b\pi)) }-\frac {\partial \ log { \Gamma(a+1)^2} }{\partial a} -2 \sum^{\infty}_{k=1} \frac {b^2 } {b^2(a+k)+(a+k)^3 } (eq. 5)$

Inside the critical strip 0<a<1, b>0, the first three terms on the right-hand side of (5) are positive, the 4th term is bounded between approximately -1.2 and 0.9, and the infinite sum term is always negative.

For small values of b (between 0 and slightly less than 2 pi (by my estimate)), the positive terms on the RHS of (5) dominate for all 0<a<1 and the derivative $\frac {\partial \ log { \frac {| \zeta \left( a+ ib \right) |^2} {| \zeta \left(1- a- ib \right) |^2} } }{\partial a}$ is uniformly >0 hence $\frac {| \zeta \left( a+ ib \right) |^2} {| \zeta \left(1- a- ib \right) |^2} = 1$ only on the critical line. For larger values of b (slightly more than 2 pi), the negative infinite sum dominates the RHS of equation (5) for all b, and $\frac {| \zeta \left( a+ ib \right) |^2} {| \zeta \left(1- a- ib \right) |^2} = 1$ also only on the critical line.

Hence the above would seem to show that there are no zeroes off the critical line except for a small region near b= 2pi which I did not scrutinize here since others have already shown the RH to be true up to a much greater  lower bound for b.

Thank you.