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juan
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... continue the above answer

Therefore $$\prod_\gamma\Bigl(1-\frac{s}{\rho}\Bigr)\Bigl(1-\frac{s}{1-\rho}\Bigr)= \frac{\frac14+\frac{t^2}{x^2}}{\frac14-\frac{1}{4x^2}}\frac{\pi^{-\frac{it}{2x}}} {\pi^{\frac{1}{4x}}}\frac{\Gamma(\frac14+\frac{it}{2x})} {\Gamma(\frac14-\frac{1}{4x})}\frac{\zeta(\frac12+i\frac{t}{x})} {\zeta(\frac12-\frac{1}{2x})},$$ or equivalently $$\prod_\gamma\Bigl(1-\frac{s}{\rho}\Bigr)\Bigl(1-\frac{s}{1-\rho}\Bigr)=$$ $$ \frac{x^2-4(s-\frac12)^2}{x^2-1}\pi^{-s/2x} \frac{\Gamma(\frac{1}{4}-\frac{1}{4x}+\frac{s}{2x})} {\Gamma(\frac14-\frac{1}{4x})}\frac{\zeta(\frac12+\frac{s}{x}-\frac{1}{2x})} {\zeta(\frac12-\frac{1}{2x})}.$$ If my computation are correct.

continue the above answer

Therefore $$\prod_\gamma\Bigl(1-\frac{s}{\rho}\Bigr)\Bigl(1-\frac{s}{1-\rho}\Bigr)= \frac{\frac14+\frac{t^2}{x^2}}{\frac14-\frac{1}{4x^2}}\frac{\pi^{-\frac{it}{2x}}} {\pi^{\frac{1}{4x}}}\frac{\Gamma(\frac14+\frac{it}{2x})} {\Gamma(\frac14-\frac{1}{4x})}\frac{\zeta(\frac12+i\frac{t}{x})} {\zeta(\frac12-\frac{1}{2x})},$$ or equivalently $$\prod_\gamma\Bigl(1-\frac{s}{\rho}\Bigr)\Bigl(1-\frac{s}{1-\rho}\Bigr)=$$ $$ \frac{x^2-4(s-\frac12)^2}{x^2-1}\pi^{-s/2x} \frac{\Gamma(\frac{1}{4}-\frac{1}{4x}+\frac{s}{2x})} {\Gamma(\frac14-\frac{1}{4x})}\frac{\zeta(\frac12+\frac{s}{x}-\frac{1}{2x})} {\zeta(\frac12-\frac{1}{2x})}.$$ If my computation are correct.

... continue the above answer

Therefore $$\prod_\gamma\Bigl(1-\frac{s}{\rho}\Bigr)\Bigl(1-\frac{s}{1-\rho}\Bigr)= \frac{\frac14+\frac{t^2}{x^2}}{\frac14-\frac{1}{4x^2}}\frac{\pi^{-\frac{it}{2x}}} {\pi^{\frac{1}{4x}}}\frac{\Gamma(\frac14+\frac{it}{2x})} {\Gamma(\frac14-\frac{1}{4x})}\frac{\zeta(\frac12+i\frac{t}{x})} {\zeta(\frac12-\frac{1}{2x})},$$ or equivalently $$\prod_\gamma\Bigl(1-\frac{s}{\rho}\Bigr)\Bigl(1-\frac{s}{1-\rho}\Bigr)=$$ $$ \frac{x^2-4(s-\frac12)^2}{x^2-1}\pi^{-s/2x} \frac{\Gamma(\frac{1}{4}-\frac{1}{4x}+\frac{s}{2x})} {\Gamma(\frac14-\frac{1}{4x})}\frac{\zeta(\frac12+\frac{s}{x}-\frac{1}{2x})} {\zeta(\frac12-\frac{1}{2x})}.$$ If my computation are correct.

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juan
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continue the above answer

Therefore $$\prod_\gamma\Bigl(1-\frac{s}{\rho}\Bigr)\Bigl(1-\frac{s}{1-\rho}\Bigr)= \frac{\frac14+\frac{t^2}{x^2}}{\frac14+\frac{1}{4x^2}}\frac{\pi^{-\frac{it}{2x}}} {\pi^{-\frac{1}{4x}}}\frac{\Gamma(\frac14+\frac{it}{2x})} {\Gamma(\frac14+\frac{i}{4x})}\frac{\zeta(\frac12+i\frac{t}{x})} {\zeta(\frac12+\frac{i}{2x})},$$$$\prod_\gamma\Bigl(1-\frac{s}{\rho}\Bigr)\Bigl(1-\frac{s}{1-\rho}\Bigr)= \frac{\frac14+\frac{t^2}{x^2}}{\frac14-\frac{1}{4x^2}}\frac{\pi^{-\frac{it}{2x}}} {\pi^{\frac{1}{4x}}}\frac{\Gamma(\frac14+\frac{it}{2x})} {\Gamma(\frac14-\frac{1}{4x})}\frac{\zeta(\frac12+i\frac{t}{x})} {\zeta(\frac12-\frac{1}{2x})},$$ or equivalently $$\prod_\gamma\Bigl(1-\frac{s}{\rho}\Bigr)\Bigl(1-\frac{s}{1-\rho}\Bigr)=$$ $$ \frac{x^2-4(s-\frac12)^2}{1+x^2}\pi^{(1-s)/2x} \frac{\Gamma(\frac{1}{4}-\frac{1}{4x}+\frac{s}{2x})} {\Gamma(\frac14+\frac{i}{4x})}\frac{\zeta(\frac12+\frac{s}{x}-\frac{1}{2x})} {\zeta(\frac12+\frac{i}{2x})}.$$$$ \frac{x^2-4(s-\frac12)^2}{x^2-1}\pi^{-s/2x} \frac{\Gamma(\frac{1}{4}-\frac{1}{4x}+\frac{s}{2x})} {\Gamma(\frac14-\frac{1}{4x})}\frac{\zeta(\frac12+\frac{s}{x}-\frac{1}{2x})} {\zeta(\frac12-\frac{1}{2x})}.$$ If my computation are correct.

continue the above answer

Therefore $$\prod_\gamma\Bigl(1-\frac{s}{\rho}\Bigr)\Bigl(1-\frac{s}{1-\rho}\Bigr)= \frac{\frac14+\frac{t^2}{x^2}}{\frac14+\frac{1}{4x^2}}\frac{\pi^{-\frac{it}{2x}}} {\pi^{-\frac{1}{4x}}}\frac{\Gamma(\frac14+\frac{it}{2x})} {\Gamma(\frac14+\frac{i}{4x})}\frac{\zeta(\frac12+i\frac{t}{x})} {\zeta(\frac12+\frac{i}{2x})},$$ or equivalently $$\prod_\gamma\Bigl(1-\frac{s}{\rho}\Bigr)\Bigl(1-\frac{s}{1-\rho}\Bigr)=$$ $$ \frac{x^2-4(s-\frac12)^2}{1+x^2}\pi^{(1-s)/2x} \frac{\Gamma(\frac{1}{4}-\frac{1}{4x}+\frac{s}{2x})} {\Gamma(\frac14+\frac{i}{4x})}\frac{\zeta(\frac12+\frac{s}{x}-\frac{1}{2x})} {\zeta(\frac12+\frac{i}{2x})}.$$ If my computation are correct.

continue the above answer

Therefore $$\prod_\gamma\Bigl(1-\frac{s}{\rho}\Bigr)\Bigl(1-\frac{s}{1-\rho}\Bigr)= \frac{\frac14+\frac{t^2}{x^2}}{\frac14-\frac{1}{4x^2}}\frac{\pi^{-\frac{it}{2x}}} {\pi^{\frac{1}{4x}}}\frac{\Gamma(\frac14+\frac{it}{2x})} {\Gamma(\frac14-\frac{1}{4x})}\frac{\zeta(\frac12+i\frac{t}{x})} {\zeta(\frac12-\frac{1}{2x})},$$ or equivalently $$\prod_\gamma\Bigl(1-\frac{s}{\rho}\Bigr)\Bigl(1-\frac{s}{1-\rho}\Bigr)=$$ $$ \frac{x^2-4(s-\frac12)^2}{x^2-1}\pi^{-s/2x} \frac{\Gamma(\frac{1}{4}-\frac{1}{4x}+\frac{s}{2x})} {\Gamma(\frac14-\frac{1}{4x})}\frac{\zeta(\frac12+\frac{s}{x}-\frac{1}{2x})} {\zeta(\frac12-\frac{1}{2x})}.$$ If my computation are correct.

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juan
juan
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continue the above answer

Therefore $$\prod_\gamma\Bigl(1-\frac{s}{\rho}\Bigr)\Bigl(1-\frac{s}{1-\rho}\Bigr)= \frac{\frac14+\frac{t^2}{x^2}}{\frac14+\frac{1}{4x^2}}\frac{\pi^{-\frac{it}{2x}}} {\pi^{-\frac{1}{4x}}}\frac{\Gamma(\frac14+\frac{it}{2x})} {\Gamma(\frac14+\frac{i}{4x})}\frac{\zeta(\frac12+i\frac{t}{x})} {\zeta(\frac12+\frac{i}{2x})},$$ or equivalently $$\prod_\gamma\Bigl(1-\frac{s}{\rho}\Bigr)\Bigl(1-\frac{s}{1-\rho}\Bigr)=$$ $$ \frac{x^2-4(s-\frac12)^2}{1+x^2}\pi^{(1-s)/2x} \frac{\Gamma(\frac{1}{4}-\frac{1}{4x}+\frac{s}{2x})} {\Gamma(\frac14+\frac{i}{4x})}\frac{\zeta(\frac12+\frac{s}{x}-\frac{1}{2x})} {\zeta(\frac12+\frac{i}{2x})}.$$ If my computation are correct.