We
I assume Riemann hypothesis on all this answer.
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I want a closed form for
$$f(s,x):= \prod_{n=1}^\infty\Bigl(1-\frac{s}{\frac12+i x\gamma_n}\Bigr) \Bigl(1-\frac{s}{\frac12-i x\gamma_n}\Bigr).$$ Of course $\gamma_n$ runs here over the ordinates of the zeros of $\zeta(s)$ but only those with $\gamma>0$.
We know that $$\Xi(t)=\Xi(0)\prod_\gamma\Bigl(1-\frac{t^2}{\gamma^2}\Bigr).\qquad (1)$$ Therefore $$\Xi(t/x)=\Xi(0)\prod_\gamma\Bigl(1-\frac{t^2}{x^2\gamma^2}\Bigr).$$ Substitute here $s=\frac12+it$ then $it=s-\frac12$ $$\Xi(t/x)=\Xi(0)\prod_\gamma\Bigl(1+\frac{(s-\frac12)^2}{x^2\gamma^2}\Bigr)=$$ $$= \Xi(0)\prod_\gamma\Bigl(\frac{(s-\frac12-ix\gamma)(s-1/2+ix\gamma)}{x^2\gamma^2}\Bigr).$$ Now we call $\rho=\frac12+ix\gamma$ and we get $$\Xi(t/x)=\Xi(0) \prod_\gamma\Bigl(\frac{\rho(1-\rho)}{x^2\gamma_n^2}\Bigr)\cdot \prod_\gamma\Bigl(1-\frac{s}{\rho}\Bigr)\Bigl(1-\frac{s}{1-\rho}\Bigr).$$ By (1), this is equal to $$\Xi(t/x)=\Xi(0) \prod_\gamma\Bigl(\frac{\frac14+x^2\gamma^2}{x^2\gamma^2}\Bigr)\cdot \prod_\gamma\Bigl(1-\frac{s}{\rho}\Bigr)\Bigl(1-\frac{s}{1-\rho}\Bigr)=$$ $$= \Xi(i/2x)\prod_\gamma\Bigl(1-\frac{s}{\rho}\Bigr)\Bigl(1-\frac{s}{1-\rho}\Bigr)$$
Therefore we have $$\prod_\gamma\Bigl(1-\frac{s}{\rho}\Bigr)\Bigl(1-\frac{s}{1-\rho}\Bigr)= \frac{\Xi(t/x)}{\Xi(i/2x)}$$
By definition we have
`$$\Xi(t)=\frac{s(s-1)}{2}\pi^{-s/2}\Gamma(s/2)\zeta(s), \qquad \text{if} \quad s=\frac12+it.$$
I shall continue in other answer because the TeX do not runs well
