Imaginary part of $\log \Gamma$ and when its imarginary argument tends to infinity Define
$$
f(z):=-\gamma z-\operatorname{Log} z+\sum_{k=1}^{\infty}\left ( \frac{z}{k}  -\operatorname{Log}\left ( 1+\frac{z}{k} \right )\right ),
$$
where $\operatorname{Log}z=\ln|z|+i\operatorname{Arg}z$ and $\operatorname{Arg}z\in (-\pi,\pi]$. Then $f$ defines a holomorphic function on $\mathbb{C}^-:=\mathbb{C}\setminus (-\infty,0]$. Since $e^f=\Gamma$, one can write $\log \Gamma:=f$.
Question 1: What is the imaginary part of $\log \Gamma$?
My idea: I take the imarginary part of the series first, but I am unsure if it's true that $\Im \sum^{\infty}(*)=\sum^{\infty}\Im(*)$. Since
$$
\Im\left ( \frac{z}{k}  -\operatorname{Log}\left ( 1+\frac{z}{k} \right )\right )=\frac{\Im z}{k}  -\operatorname{Arg}\left ( 1+\frac{z}{k} \right ),
$$
one has
$$
\Im (\log \Gamma(z))=-\gamma\Im z-\operatorname{Arg}z+\sum_{k=1}^{\infty}\left ( \frac{\Im z}{k}  -\operatorname{Arg}\left ( 1+\frac{z}{k} \right ) \right ).
$$
Is this correct?
Question 2: If I write $z=x+iy$, I am curious to know what happens to $\Im (\log \Gamma(z))$ when $y$ grows from 0 til $\infty$?
I have checked Stirling's formula. The one I have been studying is: For all $\mathbb{C}\setminus (-\infty,0]$, we have
$$
  \Gamma(z)=\sqrt{2\pi}e^{\left ( z-\frac{1}{2} \right )\operatorname{Log}z}e^{-z}e^{\mu(z)};\\
$$
where
$$\mu(z)=\sum_{n=0}^{\infty}\left [ \left ( z+n+\frac{1}{2} \right )\operatorname{Log}\left ( 1+\frac{1}{z+n} \right )-1 \right ].
$$
I really do not know how or where this approximation can be used to answer the Question 2.
 A: $1.$ Note that $\operatorname{Im}(\log \Gamma(z)) =\arg \Gamma(z)$.
$2.$ The general asymptotic expansion of the logarithm of the gamma function is
$$\tag{1}
\log \Gamma (z+h) \sim \left( {z+h - \frac{1}{2}} \right)\log z - z + \frac{1}{2}\log (2\pi ) + \sum\limits_{k = 2}^\infty  {\frac{{B_{k}(h) }}{{k(k - 1)z^{k - 1} }}} ,
$$
as $z\to \infty$ in $|\arg z|\leq \pi -\delta$ ($<\pi$) with any fixed complex $h$. Here $B_k(h)$ denotes the Bernoulli polynomials. Taking $h=x$ and $z=\text{i}y$ we find
$$\tag{2}
\arg \Gamma (x+\text{i}y) \sim y\log y-y +\left(x-\frac{1}{2}\right)\!\frac{\pi }{2}  + \sum\limits_{k = 1}^\infty  {\frac{{( - 1)^k B_{2k}(x) }}{{2k(2k - 1)y^{2k - 1} }}} ,
$$
as $y\to +\infty$ with fixed real $x$. Note that by the Schwarz reflection principle $\arg \Gamma (x-\text{i}y)=-\arg \Gamma (x+\text{i}y)$.
I shall add that there is a Stokes phenomenon happening on the imaginary axis for the logarithm of the Gamma function. This brings in exponentially small corrections to $(1)$, namely
\begin{align*} 
\log \Gamma (z+h) \sim \left( {z+h - \frac{1}{2}} \right)\log z - z + \frac{1}{2}\log (2\pi ) & - \log (1 - {\rm e}^{\pm 2\pi {\rm i}(z + h)} )\\ & + \sum\limits_{k = 2}^\infty  {\frac{{B_{k}(h) }}{{k(k - 1)z^{k - 1} }}} ,
\end{align*}
as $z\to \infty$ in $\frac{\pi}{2}<\pm \arg z \leq \pi -\delta$ ($<\pi$) with any fixed complex $h$. Thus, $(2)$ is most useful when $x\geq 0$, whereas if $x<0$, one may use
$$ 
\arg \Gamma (x+\text{i}y) \sim y\log y-y +\left(x-\frac{1}{2}\right)\!\frac{\pi }{2} +\! \sum\limits_{n = 1}^\infty  {\frac{{\mathrm{e}^{ - 2\pi ny} }}{n}\sin (2\pi nx)} +\! \sum\limits_{k = 1}^\infty  {\frac{{( - 1)^k B_{2k}(x) }}{{2k(2k - 1)y^{2k - 1} }}} ,
$$
as $y\to +\infty$.
