Smoothness of $\frac12[W_0(x)+W_{-1}(x)]$ for real $x<0$ The Lambert W-function, i.e. the multivalued inverse of $z=we^w$, has countably many complex-valued branches $W_k(z)$. The relations between the branches are a bit involved and are summarized here. 
We will consider the behavior of the $k=0,-1$ branches for $x<0$. Using Mathematica, we obtain the following plots of $W_k(x)$ along the negative real axis:

Setting aside the green line for the moment, the two plots give the real and imaginary parts respectively of $W_0(x)$ (blue) and $W_{-1}(x)$ (orange). From this, we see that both branches are real for $x\in (-1/e,0)$. (The $k=0$ branch is additionally real for positive real $x$; no other branches obtain real values along the real line.) This is not surprising, as these branches correspond to the two real-valued inverses of $z=we^w$ along the real line.
What's perhaps surprising, though, is that (according to this plot) $\overline{W_{-1}(x)}=W_{0}(x)$ for all $x\in (-\infty,-1/e)$. (A derivation of this fact may be found in the Q&A linked in the comments below.) From this, we conclude that the average of these two branches, $\frac{1}{2}(W_0(x)+W{-1}(x))$, is real for all negative real $x$. This is the green line plotted above, and from the first plot we further ascertain that $\frac12 (W_0+W_{-1})$ is smooth across the point $x=-1/e$. (Further plotting in Mathematica suggests that $\frac12 (W_0+W_{-1})$ is analytic for all $x\neq 0$ such that $-\pi < \text{arg }x \leq \pi$.)
for all $x<0$ but not holomorphic across the real line.) By contrast, the two branches have square-root branching at $x=-1/e$.
This last property of $\frac{1}{2}(W_0+W_{-1})$ remains mysterious to me, so my question is:

Why is $\frac{1}{2}(W_0(x)+W{-1}(x))$ a smooth function for all real $x<0$?

 A: It’s not a full answer, I only want to show why $\,\displaystyle\frac{W_0(x)+W_{-1}(x)}{2}\,$ is real for $\,x<0\,$ .
We can parameterize $\,\displaystyle W_0(x)=-\ln ((1+\frac{1}{t})^t) \,$ and $\,\displaystyle W_{-1}(x)=-\ln ((1+\frac{1}{t})^{t+1})\,$ 
(where $\,t\,$ is complex) and it follows $\,\displaystyle
\frac{W_0(x)+W_{-1}(x)}{2}=-\ln ((1+\frac{1}{t})^{t+\frac{1}{2}})\,$ .
Using e.g. $\,\displaystyle t:=\frac{1}{e^{-i\alpha}-1}\,$ with real $\,\alpha\,$ we can express the complex line of $\,W_0(x)\,$ by $\,\displaystyle \frac{i\alpha}{e^{-i\alpha}-1}\,$ 
and $\,W_{-1}(x)\,$ by $\,\displaystyle \frac{-i\alpha}{e^{i\alpha}-1}\,$ so that we see $\,\overline{W_{-1}(x)}=W_0(x)\,$ . 
It follows $\,\displaystyle
\frac{W_0(x)+W_{-1}(x)}{2}=-\ln ((1+\frac{1}{t})^{t+\frac{1}{2}})= 
-\frac{\alpha}{2}\cot\frac{\alpha}{2}\,$ which is real (for $\,x<0\,$). 
A: $\require{begingroup} \begingroup$
$\def\e{\mathrm{e}}\def\W{\operatorname{W}}\def\Wp{\operatorname{W_0}}\def\Wm{\operatorname{W_{-1}}}\def\i{\mathbf{i}}$
Using
the parametric representation,
of $\Wp(x),\,\Wm(x)$
we arrive at
\begin{align}
\Wp(\alpha)
&=-\tfrac\alpha2\,\cot\tfrac\alpha2-\tfrac\alpha2\cdot\i
\tag{1}\label{1}
,\\
\Wm(\alpha)
&=-\tfrac\alpha2\,\cot\tfrac\alpha2+\tfrac\alpha2\cdot\i
\tag{2}\label{2}
,\\
x(\alpha)&=
-\frac{\alpha}{2\sin\tfrac\alpha2}
\cdot\exp(-\tfrac\alpha2\,\cot\tfrac\alpha2)
\tag{3}\label{3}
\end{align}
for $x\le-\tfrac1\e$,
$\alpha\ge0$, since
\begin{align}
\lim_{\alpha\to0}x(\alpha)=-\tfrac1\e
\tag{4}\label{4}
.
\end{align}
Combining this with
the parametric representation
of $\Wp(x),\,\Wm(x)$
we have
\begin{align} 
\Wp(a)&=\frac{a\ln a}{1-a}
\tag{5}\label{5}
,\\
\Wm(a)&=\frac{\ln a}{1-a}
\tag{6}\label{6}
,\\
x(a)&=\frac{\ln a}{1-a}\cdot a^{\tfrac1{1-a}}
\tag{7}\label{7}
\end{align}
for $x\in[-\tfrac1\e,0]$,
$a\in[0,1]$.
The function of interest
\begin{align} 
f(x)&=\tfrac12\,(\Wp(x)+\Wm(x))
\tag{8}\label{8}
\end{align}
for $x<0$
is then represented as
two pieces
\begin{align} 
f_1(\alpha)&=-\tfrac\alpha2\,\cot\tfrac\alpha2
\tag{9}\label{9}
,\\
x(\alpha)&=
-\frac{\alpha}{2\sin\tfrac\alpha2}
\cdot\exp(-\tfrac\alpha2\,\cot\tfrac\alpha2)
,\quad \alpha>0
\tag{10}\label{10}
,\\
\text{and }\quad
f_2(a)&=\tfrac12\,\ln a\cdot\frac{1+a}{1-a}
\tag{11}\label{11}
,\\
x(a)&=\frac{\ln a}{1-a}\cdot a^{\tfrac1{1-a}}
,\quad a\in[0,1]
\tag{12}\label{12}
,\\
\lim_{\alpha=0}x(\alpha)&=
\lim_{a=1}x(a)=
-\tfrac1\e
\tag{13}\label{13}
,\\
\lim_{\alpha=0}f_1(\alpha)&=
\lim_{a=1}f_2(a)=-1
\tag{14}\label{14}
.
\end{align}
Omitting details of calculation,
this leads  to the desired result,
\begin{align} 
\left.
\frac{d f_1(\alpha)/d\alpha}{dx(\alpha)/d\alpha}
\right|_{\alpha=0}
&=
\left.
\frac{d f_2(a)/da}{dx(a)/da}
\right|_{a=1}
=-\tfrac23\e
\tag{15}\label{15}
,\\
\left.
\frac{d^2 f_1(\alpha)/d\alpha^2}{d^2x(\alpha)/d\alpha^2}
\right|_{\alpha=0}
&=
\left.
\frac{d^2 f_2(a)/da^2}{d^2x(a)/da^2}
\right|_{a=1}
=-\tfrac23\e
\tag{16}\label{16}
,
\end{align}
that means that $f(x)$ is indeed $C^2$-continuous at $x=-\tfrac1\e$.
Surprisingly, the value of first and second derivation
at this point is the same.
$\endgroup$
