How to derive this representation of the Lambert W function? 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

*

*Is this correct? If so, how to prove it?


*Is there some technique to derive a closed-form representation for almost any transcendental equations? and What is it?
Any contribution is appreciated. :D
 A: 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,  $$I = I(1) = \ln \left(W_{k}(z)-1 \right)-  \ln \big(\ln(z)-1+ 2 k \pi i\big) .$$
