Is there any way to determine $\mathrm{Arg}[ \Gamma(ix) ]$? 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.

 A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\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}

$$
\bbx{\arg\pars{\Gamma\pars{\ic x}} =
-\arctan\pars{\ds{\int_{0}^{\infty}
\cos\pars{x\ln\pars{t}}\expo{-t}\dd t} \over
\ds{\int_{0}^{\infty}
\sin\pars{x\ln\pars{t}}\expo{-t}\dd t}}}
$$
