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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:

enter image description here

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 $x<0$?

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    $\begingroup$ I am surprised that nobody answered your very interesting question. It is a nice observation. I suppose that you noticed the zero for $x=-\frac \pi 2$ $\endgroup$ Aug 3, 2017 at 14:59
  • $\begingroup$ @ClaudeLeibovici Thanks! I hope someone will provide an equally interesting answer. As for the zero---no, I had entirely missed that. In retrospect, it's not so strange: the equation $-\pi/2=we^w$ has the purely imaginary solutions $w=\pm i \pi/2$, and these evidently correspond to the two main branches. $\endgroup$ Aug 3, 2017 at 15:40
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    $\begingroup$ my question here seems to be related: math.stackexchange.com/questions/1652745/… $\endgroup$
    – tired
    Aug 3, 2017 at 21:25
  • $\begingroup$ @tired Indeed, I think that the answer to your question resolves the reality part of my question. (That it's smooth across $z=-1/e$, however, still seems mysterious.) $\endgroup$ Aug 3, 2017 at 21:28
  • $\begingroup$ For $\Im(z)\geq 0$ and near the branch point, you can express the average as a convergent expansion using the known series expansions of $W_0$ and $W_{-1}$ near the branch point. The terms involving the square roots of $e z+1$ cancel and you end up with a power series that is convergent around $z=1/e$. This should explain the analyticity at the branch point. $\endgroup$
    – Gary
    Nov 30, 2017 at 11:59

2 Answers 2

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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 $\,x<0\,$).

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  • $\begingroup$ +1. Do you have a reference where such a parameterization is introduced/mentioned? $\endgroup$
    – g.kov
    Sep 14, 2020 at 5:59
  • $\begingroup$ @g.kov : (Sorry for the delay, for a while I am not active here.) I don't know where such a parametrization is introduced, I only used the fact $\,{x_1}^{1/x_1}={x_2}^{1/x_2}\,$ where $\,x_1\neq x_2\,$ for $\,x_1:=(1+\frac{1}{t})^ t\,$ and $\,x_2:=(1+\frac{1}{t})^{1+t}\,$ . $\endgroup$
    – user90369
    Sep 28, 2020 at 14:57
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$\require{begingroup} \begingroup$ $\def\e{\mathrm{e}}\def\W{\operatorname{W}}\def\Wp{\operatorname{W_0}}\def\Wm{\operatorname{W_{-1}}}\def\i{\mathbf{i}}$

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.

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