How to derive this representation of the Lambert W function?

Question

Lately I read this in some site about a closed-form representation of Lambert W function (all branch-cuts): $$\ln\bigg(\frac{W_k(z)-1}{ \ln(z)-1+2k\pi{i} }\bigg)=\frac{i}{2\pi}\int_0^\infty{\ln\bigg({\frac{t-\ln{t}+\ln{z}+(2k+1)\pi{i}}{t-\ln{t}+\ln{z}+(2k-1)\pi{i}}}\bigg)\frac{\mathrm{d}t}{t+1}}$$ where $\ln$ denotes the principal branch of natural logarithm.

This site doesn't contain any references.

Closed-form here refers to being able to be represented in a finite composition of integrals and all ''already-known and well-defined'' functions. Simply using of an iterative inverse denotion is not accepted.

I also read Closed-form representations of the Lambert W function by Alexander Kheyfits, in which he the author introduced a very similar representation solved by contour integration.

And btw is there some technique to derive a closed-form (integral is in desire) representation for almost any transcendental equations? I read some articles but they gave too many restrictions on the transcendental equation. (Riemann's method)

So the question is

1. Is this correct? If so, how to prove it?

2. Is there some technique to derive a closed-form representation for almost any transcendental equations? and What is it?

Any contribution is appreciated. :D

Answer 1

Let $$I= \frac{i}{2 \pi}\int_{0}^{\infty} \ln \left( \frac{t-\ln(t)+\ln(z)+(2k+1)\pi i}{t-\ln(t)+\ln(z)+(2k-1)\pi i} \right) \, \frac{\mathrm dt}{1+t},$$ where $z \in \mathbb{C}$, $k \in \mathbb{Z}$, and $\ln$ denotes the principal branch of the logarithm.

For reasons explained later, we need to exclude the case $k=-1$ and $z \in \left[- \frac{1}{e},0 \right)$.

Integration by parts shows that $$\small I = \frac{i}{2 \pi}\int_{0}^{\infty} \frac{(t-1) \ln(1+t)}{t} \left(\frac{1}{t- \ln(t)+\ln(z) +(2k-1)\pi i} - \frac{1}{t-\ln(t)+\ln(z)+(2k+1)\pi i} \right) \, \mathrm dt,$$ which is basically a generalization of the integral evaluated here by the user M.N.C.E.

As in the evaluation of that integral, let's add a parameter and differentiate under the integral sign.

Let $a$ be a positive parameter, and let $$ \small I(a) = \frac{i}{2 \pi}\int_{0}^{\infty} \frac{(t-1) \ln(1+at)}{t} \left(\frac{1}{t- \ln(t)+\ln(z) +(2k-1)\pi i} - \frac{1}{t-\ln(t)+\ln(z)+(2k+1)\pi i} \right) \, \mathrm dt.$$

Then $$ \small I^{\prime} (a) = \frac{i}{2 \pi} \int_{0}^{\infty} \frac{t-1 }{1+at} \left(\frac{1}{t- \ln(t)+\ln(z) +(2k-1)\pi i} - \frac{1}{t-\ln(t)+\ln(z)+(2k+1)\pi i} \right) \, \mathrm dt.$$

To evaluate $I^{\prime} (a)$, let's integrate the function $$f(w) = \frac{i}{2 \pi} \frac{-w-1}{1-aw} \frac{1}{-w-\ln(w)+\ln(z) + 2k \pi i } $$ around a keyhole contour.

Since we're using the principal branch of the logarithm, the branch cut for $\ln (w)$ should be along the negative real axis.

There are two simple poles inside the contour, one at $w= \frac{1}{a}$ and one at $w= W_{k}(z)$.

(See section 3 of the paper Unwinding the branches of the Lambert W function by Jeffrey, et al. for an explanation of why $ W_{k}(z)+ \ln (W_{k}(z)) = \ln(z) + 2 k \pi i $ except in the case $k = -1$ and $z \in [-\frac{1}{e},0)$.)

And for $w$ large in magnitude, $f(w)$ behaves like -$\frac{i}{2 \pi} \frac{1}{aw}$.

Letting the radius of the small circle about the origin go to zero, and then letting the radius of the big circle go to infinity, we get

$$ \begin{align} \small\oint f(w) \, \mathrm dw &= \small \frac{1}{a} + \color{red}{ \frac{i}{2 \pi} \int_{0}^{\infty} \frac{t-1}{1+at} \left(\frac{1}{t-\ln(t)+\ln(z)+(2k-1)\pi i}- \frac{1}{t-\ln(t)+\ln(z) +(2k+1) \pi i} \right) \, \mathrm dt} \\ &= 2 \pi i \, \frac{i}{2 \pi} \big( \operatorname{Res} \left[f(w), 1/a \right] + \operatorname{Res}\left[f(w), W_{k}(z)\right] \big) \\ &= -\left( \frac{\frac{1}{a^{2}}+ \frac{1}{a}}{-\frac{1}{a}+\ln(a)+\ln(z) +2k \pi i} + \frac{W_{k}(z)}{1-aW_{k}(z)} \right), \end{align}$$

from which it follows that $$I^{\prime} (a) =- \frac{\frac{1}{a^{2}}+ \frac{1}{a}}{-\frac{1}{a}+\ln(a)+\ln(z) +2k \pi i} + \frac{W_{k}(z)}{aW_{k}(z)-1} - \frac{1}{a} . $$

Integrating with respect to $a$, we get

$$\begin{align} I (a) &= -\ln \left(- \frac{1}{a} + \ln(a) + \ln(z) + 2 k \pi i \right) + \ln \left(a W_{k}(z)-1 \right) - \ln (a) +C \\ &= \ln \left(a W_{k}(z)-1 \right) -\ln \big(a \ln(z) -1 + 2 k \pi i a + a\ln(a) \big) +C. \end{align}$$

Since $\lim_{a \to 0^{+}} I(a)=0$, the constant of integration turns out to be zero.

Therefore, $$I = I(1) = \ln \left(W_{k}(z)-1 \right)- \ln \big(\ln(z)-1+ 2 k \pi i\big) .$$

Answer 2

And btw is there some technique to derive a closed-form (integral is in desire) representation for almost any transcendental equations? I read some articles but they gave too many restrictions on the transcendental equation. (Riemann's method)

$\DeclareMathOperator WW$Here is another representation that did not seem to appear online. A possible advantage is that is gives a direct integral representation for $\W_k(z)$instead of an exponent of an integral, like the question’s, or a ratio of integrals, like Kheyfit’s. One takes the series of $\W_k(z)$ function, which can be derived from Lagrange reversion and the Stirling number $S_n^{(m)}$of the first kind’s generating function:

$$\W_k(z)=w-r-\sum_{n=1}^\infty\sum_{m=1}^n\frac{(-w)^{-n}r^m}{m!} S_n^{(n-m+1)};w=\ln(z)+2\pi i k,r=\ln(w)$$

To convert it to an integral, use $S_n^{(m)}$’s contour integral representation, likely derived via inverse Z transform from its generating function:

$$S_n^{(m)}=\frac{n!}{2\pi i m!}\oint_{|z|=1}\ln^m(z+1) z^{-n-1} dz=\frac{n!}{2\pi m!}\int_{-\pi}^\pi\ln^m(e^{i t}+1) e^{-i nt} dt$$

where $z=e^{i t},-\pi\le t\le \pi$ was substituted. We then use the binomial theorem and the Maclaurin series for $\ln(1-z)$:

$$\W_k(z)=w-r-\frac1{2\pi}\int_{-\pi}^\pi\sum_{n=1}^\infty\sum_{m=1}^n\frac{(-w)^{-n}r^m e^{-i n t}\ln^{n-m+1}(e^{it}+1)n!}{(n-m+1)!m!}dt=w-r-\frac w{2\pi}\int_{-\pi}^\pi e^{it}\ln\left(\frac{e^{-it}(e^{it}w+\ln(e^{it}+1))(e^{it}w+r)}{w(e^{it}w+\ln(e^{it}+1)+r)}\right)dt$$

Using logarithm properties, expand out $w$ from $w(e^{it}w+\ln(e^{it}+1)+r)$. Now notice the integrand is graphically the same, except near given $z=0$ like for $-0.640084165<z<0$ if $k=-1$ and $z\in\Bbb R$, and $\frac w{2\pi}\ln(w)\int_{-\pi}^\pi e^{i t}dt=0$. This change makes the integral slightly less accurate numerically, but it still matches for many decimal places. Therefore:

$$\boxed{\W_k(z)= w-\ln(w)-\frac w{2\pi}\int_{-\pi}^\pi e^{it}\ln\left(\frac{(e^{it}w+\ln(e^{it}+1))(w+e^{-it}\ln(w))}{e^{it}w+\ln(e^{it}+1)+\ln(w)}\right)dt\mathop=^{w=\ln(z)+2\pi i k}\textstyle\ln(z)+2\pi i k-\ln(\ln(z)+2\pi i k)+\left(\frac{\ln(z)}{2\pi}+ik\right)\int_{-\pi}^\pi e^{it}\ln\left(\frac{(\ln(e^{it}+1)+e^{it}(\ln(z)+2\pi i k))(e^{-i t}\ln(\ln(z)+2\pi i k)+\ln(z)+2\pi i k)}{\ln(e^{it}+1)+e^{it}(\ln(z)+2\pi i k)+\ln(\ln(z)+2\pi i k)}\right)dt}$$

shown here:

where the following can be used for given $z,k$:

ReplaceAll[w->Log[z]+2 \[Pi] I*k][w-Log[w]]+NIntegrate[ReplaceAll[w->Log[z]+2 \[Pi] I*k][(E^(I t) w (Log[( (E^(I t) w+Log[1+E^(I t)]) ( w+E^(-I t)Log[w]))/( (E^(I t) w+Log[1+E^(I t)]+Log[w]))]))/(2 \[Pi])],{t,-\[Pi],\[Pi]}]