The Lambert W function satisfies the identity


How do you prove that

$$ W(z) = \frac{z}{2\pi} \int_{-\pi}^{\pi} \frac{(1-\nu \cot(\nu))^2+\nu^2}{z+\nu \csc(\nu)e^{-\nu\cot(\nu)}} \, \mathrm{d}\nu $$

where $z$ is a real number and $z\geq-\frac{1}{e}$?

  • $\begingroup$ @MarkViola I found it on a Wikipedia article. The source the article gave was this $\endgroup$ – Ryan Parikh Sep 7 '19 at 16:26

A proof using complex analysis is provided on Theorem 3.1 on the article "Stieltjes, Poisson and other integral representations for functions of Lambert W" by German A. Kalugin, David J. Jeffrey, and Robert M. Corless. However, there are some preliminaries in the article not necessary for the proof for this theorem, hence I will try to summarize.

The proof uses the fact that the function $W(z)/z$ is holomorphic on the domain $D=\mathbb{C}\setminus (-\infty, -1/e]$. By Cauchy's integral formula, it follows that: $$\frac{W(z)}{z}=\frac{1}{2\pi i}\int_C \frac{W(t)}{t(t-z)}~dt$$ where $C$ is a positively oriented keyhole contour, as follows:

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Letting $r\to 0$ and $R\to \infty$, it can be shown that this reduces to (it can be shown that the contributions of each circle to the integral is zero): $$\frac{W(z)}{z}=\frac{1}{2\pi i} \left[\int_{-\infty}^{-1/e} \frac{W(t)}{t(t-z)}~dt+\int_{-1/e}^{-\infty} \frac{\overline{W(t)}}{t(t-z)}~dt\right]$$ Note that in the above we have used that for $x\in D$, we have that $W(\overline{x})=\overline{W(x)}$. Additionally, we have that for all $w\in \mathbb{C}$ that $\Im(w)=\frac{w-\overline{w}}{2i}$, hence the integral reduces to: $$\frac{W(z)}{z}=\frac{1}{\pi} \int_{-\infty}^{-1/e} \frac{\Im(W(t))}{t(t-z)}~dt=\frac{1}{\pi} \int_{-1/e}^{-\infty} \frac{\Im(W(t))}{t(z-t)}~dt$$ The article then proceeds to use the change of variable $\nu=\Im(W(t))$. The variables can be shown to be related by: $$t=-\nu \csc(\nu) e^{-\nu \cot(\nu)}$$ Differentiating gives: $$dt=-\csc(\nu)e^{-\nu \cot(\nu)}(\nu^2+(1-\nu\cot(\nu))^2)~d\nu$$ Hence, it follows that (since $t=-1/e$ gets mapped to $\nu=0$ and $t\to -\infty$ gets mapped to $\nu=\pi$): $$\frac{W(z)}{z}=\frac{1}{\pi} \int_0^{\pi} \frac{\nu^2+(1-\nu\cot(\nu))^2}{z+\nu \csc(\nu) e^{-\nu \cot(\nu)}}~d\nu$$ Since the integrand is even with respect to $\nu$, we obtain the desired result.

  • 2
    $\begingroup$ :p not so hard once you see the first integral, but knowing where to start without it is not so trivial +1 $\endgroup$ – Simply Beautiful Art Sep 7 '19 at 23:18

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