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I ask about an idea to prove this formula:

$Γ(1/2-iβ)=((√π)/(√(coshπβ)))exp(-i(2ϑ(β)+βln2π+arctan(tanh(1/2)πβ)))$

where $ϑ(β)$ is the Riemann Siegel function.

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up vote 3 down vote accepted

I know two proof, the first uses $$\cos\frac{\pi s}{2}=\frac{1}{\sqrt{2}}\sqrt{\cosh(\pi t)}\,e^{-i\arctan(\tanh\frac{\pi t}{2})}.\qquad (1)$$ and $$\Gamma(\frac12+i\frac t 2)=|\Gamma(\frac14+i\frac t2)|\,e^{i(\vartheta(t)+\frac t 2\log\pi)},\qquad (2)$$

Since $$\Gamma(z)\Gamma(z+1/2)=2^{1-2z}\sqrt{\pi}\Gamma(2z);\quad \Gamma(z)\Gamma(1-z)=\frac{\pi}{\sin\pi z}$$ we get $$\Gamma(2z)=2^{2z-1}\pi^{-1/2} \Gamma(z)\frac{\pi}{\cos\pi z}\frac{1}{\Gamma(1/2-z)}.$$ We put now $z=\frac14+i\frac t 2$, $t$ real $$ \Gamma(\frac12+it)=2^{-\frac12+it}\frac{\pi^{\frac12}}{\cos\pi(\frac14+i\frac{t}{2})} \frac{\Gamma(\frac14+i\frac{t}{2})}{\Gamma(\frac14-i\frac{t}{2})}. $$ From (1) and (2) we get $$\Gamma(1/2+it)=2^{it}\frac{\pi^{1/2}}{\sqrt{\cosh \pi t}\; e^{-i \arctan\tanh(\pi t/2)}}\pi^{it} e^{2i\vartheta(t)}. $$ so that $$ \Gamma(1/2+it)=\sqrt{\frac{\pi}{\cosh\pi t}}\exp\bigl\{i(2\vartheta(t)+t\log(2\pi)+\arctan\tanh(\pi t/2))\bigr\} $$

The other proof I know uses the functional equation of the zeta function.

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@Juan: Note that all the identities concerning arguments holds only modulo factors of 2π if the argument is being restricted to (−π,π] . How you can deal with arguments figured in this proof. This is a problem since these arguments depend on the variable t . –  Shpigle Dec 16 '12 at 9:38
    
@RH The argument of $\Gamma(s)$ is well defined and harmonic on the plane with a cut along the negative real axis. –  juan Dec 16 '12 at 15:25
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