Fixing some real $x>0$, there is a result which says that: $$ |\Gamma(ix)| \ = \ \frac{\sqrt{\pi}}{\sqrt{x\sinh(\pi x)}} $$
So this means the following: $$ \Gamma(ix) \ = \ \frac{\sqrt{\pi}}{\sqrt{x\sinh(\pi x)}} e^{i \mathrm{Arg}\left[ \Gamma(ix) \right]} $$
Is there any way to determine $\mathrm{Arg}\left[ \Gamma(ix) \right]$? What properties does this function have? I've included a plot of the function, and there are a bunch of discontinuities for this function.
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \Gamma\pars{\ic x} & = {\Gamma\pars{1 + \ic x} \over \ic x} = -\,{\ic \over x}\int_{0}^{\infty}t^{\ic x}\expo{-t}\dd t = -\,{\ic \over x}\int_{0}^{\infty}\expo{\ic x\ln\pars{t}}\expo{-t}\dd t \\[5mm] & = -\,{\ic \over x}\int_{0}^{\infty} \bracks{\cos\pars{x\ln\pars{t}} + \ic\sin\pars{x\ln\pars{t}}}\expo{-t}\dd t \end{align}
$$ \bbx{\arg\pars{\Gamma\pars{\ic x}} = -\arctan\pars{\ds{\int_{0}^{\infty} \cos\pars{x\ln\pars{t}}\expo{-t}\dd t} \over \ds{\int_{0}^{\infty} \sin\pars{x\ln\pars{t}}\expo{-t}\dd t}}} $$