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Take the well known integral:

$$\displaystyle \pi^\frac{-s}{2}\,\Gamma\left(\frac{s}{2}\right)\, \zeta(s) =\int_1^{\infty} \left({x}^{\frac{s}{2}-1} + {x}^{\frac{-s}{2}-\frac12}\right)\,\psi(x)\, \text{d}x + \frac{1}{s\,(s-1)}$$

with $\displaystyle \psi(x)=\sum_{n=1}^{\infty}e^{-\pi\,n^2\,x}$. Make a simple change and rewrite it into:

$$\displaystyle s\,(s-1)\,\pi^\frac{-s}{2}\,\Gamma\left(\frac{s}{2}\right)\, \zeta(s) - 1 =s\,(s-1)\,\int_1^{\infty} \left({x}^{\frac{s}{2}-1} + {x}^{\frac{-s}{2}-\frac12}\right)\,\psi(x)\, \text{d}x$$

With $\xi(s)=$ Riemann xi-function, the left hand side becomes $2\,\xi(s)-1$ and below are its zeros:

$2\,\xi(s)-1$

The zeros are symmetrical around the line $\Re(s)=\frac12$, however based on the irregular patterns in the imaginary parts of the non-trivial zeros $\rho$, I was actually surprised by their apparent regularity. I therefore have two questions:

1) Can anything be derived about the "tangent shaped" curves on which these zeros seem to reside?

2) The function $2\,\xi(s)-1$ remains entire, so could a Hadamard product of all its zeros ($\mu$) exist for it? If so, this would then yield something of the form:

$$\displaystyle \prod_{n=1}^{\infty} \Bigl(1-\frac{s}{\rho_n}\Bigr)\Bigl(1-\frac{s}{{1-\rho_n}}\Bigr) = \Bigl(z^m \, e^{g(z)}\,\prod_{n=1}^{\infty}\Bigl(1-\frac{s}{\mu_n}\Bigr)\Bigl(1-\frac{s}{{1-\mu_n}}\Bigr)\Bigr) + 1$$

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