I ask about an idea to prove this formula:
$Γ(1/2iβ)=((\sqrt{π})/(\sqrt{\coshπβ}))\exp(i(2ϑ(β)+βln2π+\arctan(\tanh(1/2)πβ)))$
where $ϑ(β)$ is the Riemann Siegel function.
I ask about an idea to prove this formula: $Γ(1/2iβ)=((\sqrt{π})/(\sqrt{\coshπβ}))\exp(i(2ϑ(β)+βln2π+\arctan(\tanh(1/2)πβ)))$ where $ϑ(β)$ is the Riemann Siegel function. 


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 $$\Gamma(z)\Gamma(z+1/2)=2^{12z}\sqrt{\pi}\Gamma(2z);\quad \Gamma(z)\Gamma(1z)=\frac{\pi}{\sin\pi z}$$ we get $$\Gamma(2z)=2^{2z1}\pi^{1/2} \Gamma(z)\frac{\pi}{\cos\pi z}\frac{1}{\Gamma(1/2z)}.$$ We put now $z=\frac14+i\frac t 2$, $t$ real $$ \Gamma(\frac12+it)=2^{\frac12+it}\frac{\pi^{\frac12}}{\cos\pi(\frac14+i\frac{t}{2})} \frac{\Gamma(\frac14+i\frac{t}{2})}{\Gamma(\frac14i\frac{t}{2})}. $$ 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. 


I extended juan's proof to a formula for all complex numbers. Consequently, we get the Euler chi function $\chi(z):=\zeta(1z)/\zeta(z)$. (substitution) $\tan^{1}(\tanh(\pi t/2))\longrightarrow\frac{\mathrm{gd}(\pi t)}2$, where $\mathrm{gd}(z)$ is the Gudermannian function. $\sqrt{\frac{\pi}{\cosh(\pi t)}}\longrightarrow\frac{\sqrt{\pi}}{\cosh(\pi t)}\left(\sinh(\frac{\pi t}2)\sin(\frac{\mathrm{gd}(\pi t)}{2})+\cosh(\frac{\pi t}{2})\cos(\frac{\mathrm{gd}(\pi t)}2)\right)$ $t\longrightarrow z$ (simplification) If we covert the above result(trigonometric functions) to exponentials, then we get the following simplified formula. $\Gamma(\frac12+iz)=\frac{\sqrt{\pi}(1+i)(2\pi)^{iz}}{e^{\pi z}+i}e^{\frac{\pi z}2+2i\vartheta(z)}/;z\in\mathbb{C}$ Therefore, $\Gamma(z)=\frac{(2\pi)^z}{1+e^{i\pi z}}e^{i(2\vartheta(\frac{i}2iz)+\frac{\pi z}2)}=\frac{(2\pi)^z}{2\cos(\frac{\pi z}2)}e^{2i\vartheta(\frac{i}2iz)}=\frac{(2\pi)^z}{2\cos(\frac{\pi z}2)}\chi(z)$ 

