2

Working on a problem in QFT, i was stumbeling about some expressions containing the Lambert-$W$ function. Playing around, i discovered experimentally

that the following statement seems to be true

$$ \Im (W_0(-x))=-\Im (W_{-1}(-x))\,\,\, \text{if} \,\,\, x>1/e $$

For large $x$ we can write

$W_0(-x)\approx\log(-x)$

and

$W_{-1}(-x)\approx\log(-x)-2 \pi i$

choosing now $\log(-x)=\pi i+\log(x)$

gives the desired result in this limit (but to be honest i don't see why this choice is the right one).

My question is now:

How can we prove the above statement? And (as a bonus) are there some additional relations like that?

tired
  • 12,688

1 Answers1

4

$W_0(-x)$ and $W_{-1}(-x)$ are both solution of the equation : $$we^w=-x$$ Let $w=a+ib$ then $$\begin{cases} a e^a \cos(b)-b e^a \sin(b) = -x \\ i\left(a e^a \sin(b)+b e^a \cos(b)\right) = 0 \\ \end{cases}$$ where $x\:,\:a\:,\:b$ are reals.

The system of equations is valid for the two considered solutions with a same couple $(a,b)$, but $i$ and $-i$ respectively.

Thus, two solutions are : $\quad \begin{cases} W_0(-x) = a+ib \\ W_{-1}(-x) = a-ib \\ \end{cases}$

$W_0(-x)$ and $W_{-1}(-x)$ are conjugates : $\begin{cases} \Re (W_0(-x))=\Re (W_{-1}(-x))\\ \Im (W_0(-x))=-\Im (W_{-1}(-x))\\ \end{cases}$ $\quad x>\frac{1}{e}$

JJacquelin
  • 69,089