Definite integral with the Lambert W Function

Question

I am trying to find the integral $$ \int^{1}_{0} \frac{\mathrm{W} \left( -x\log x\right) }{-x\log x} \mathrm{d} x=\frac{\pi^{2} }{12} $$

By expanding the Lambert W-Function into a series, one obtains said result.

I am trying to find a solution using integration techniques, besides from a series expansion.

I attempted Feynman‘s trick but this is messy and I got stuck. I appreciate any help.

$$ I(a) = \int_0^1 \frac{\mathrm{W}(-a x \log x)}{-x \log x} \, dx $$ $$ I'(a) = \frac{d}{da} \int_0^1 \frac{\mathrm{W}(-a x \log x)}{-x \log x} \, dx = \int_0^1 \mathrm{W}'(-a x \log x) \, dx $$

$$ \mathrm{W}'(z) = \frac{1}{z (1 + \mathrm{W}(z))} \implies \mathrm{W}'(-a x \log x) = \frac{1}{-a x \log x (1 + \mathrm{W}(-a x \log x))} $$

$$ I'(a) =\int_0^1 \frac{1}{a}\frac{1}{-x \log x (1 + \mathrm{W}(-a x \log x))} dx $$

A solution with Feynman seems convoluted, to be honest.

Motivation: I was wondering about the infinite power tower - I really like the solution below but I would prefer one using no series expansion

https://math.stackexchange.com/q/3539090

Answer 1

Over $(0,1)$ the function $-x\log x$ is increasing over $I_1=(0,e^{-1}]$ and decreasing over $I_2=[e^{-1},1)$. The same applies to the function $W(-x \log x)$. The inverse function of $-x\log x$ over $I_1$ is $\frac{-z}{W_{-1}(-z)}$. The solution of $-x\log x=z$ with $x\in I_2$ is $\frac{-z}{W(-z)}$.

$$ \mathcal{J}_1=\int_{I_1}\frac{W(-x\log x)}{-x\log x}\,dx = \int_{0}^{e^{-1}}\frac{W(z)}{z}\cdot \frac{-1}{1+W_{-1}(-z)}\,dz $$

$$\mathcal{J}_2=\int_{I_2}\frac{W(-x\log x)}{-x\log x}\,dx = \int_{e^{-1}}^{0}\frac{W(z)}{z}\cdot \frac{-1}{1+W(-z)}\,dz.\tag{1}$$

so the original integral equals

$$ \mathcal{J}_1+\mathcal{J}_2=\int_{0}^{e^{-1}}\frac{W(z)}{z}\left(\frac{1}{1+W(-z)}-\frac{1}{1+W_{-1}(-z)}\right)\,dz.\tag{2}$$

By letting $z=-ue^u$ with $u$ ranging from $-1$ to $0$ we get

$$ \mathcal{J}_2 =\int_{-1}^{0}\frac{W(-ue^u)}{-u}\,du = \int_{0}^{1}\frac{W(ve^{-v})}{v}\,dv$$

and by letting $z=-ue^u$ with $u$ ranging from $-\infty$ to $-1$ we get

$$ \mathcal{J}_1 = \int_{-\infty}^{-1}\frac{W(-ue^u)}{-u}\,du = \int_{1}^{+\infty}\frac{W(ve^{-v})}{v}\,dv $$

hence

$$ \mathcal{J}_1+\mathcal{J}_2 = \int_{0}^{+\infty}W(v e^{-v})\frac{dv}{v} \tag{3}$$

where $ve^{-v}\in\left(0,e^{-1}\right]$ as $v$ ranges over $\mathbb{R}^+$. Now we may circumvent the usage of the exact coefficients of the power series of $W(x)$ by just saying that they exist and they are fixed by the Lagrange-Burmann inversion theorem. Assuming

$$ W(z) = \sum_{n\geq 0} c_n z^n $$

we have that $c_0=0$, and the radius of convergence of this power series equals $e^{-1}$, such that

$$ \int_{0}^{+\infty}W(v e^{-v})\frac{dv}{v} = \sum_{n\geq 1}c_n \int_{0}^{+\infty} v^{n-1} e^{-nv}\,dv = \sum_{n\geq 1}c_n \frac{n!}{n^{n+1}}.\tag{4}$$

By the residue theorem

$$ c_n = \frac{1}{2\pi i}\oint_{\|z\|=\varepsilon} \frac{W(z)}{z^{n+1}}\,dz$$

and the map sending $z$ to $ze^z$ deforms $\|z\|=\varepsilon$ into an homotopic curve, such that

$$ c_n = \frac{1}{2\pi i}\oint_{\|z\|=\varepsilon} \frac{W(z e^z)}{(ze^z)^{n+1}}(z+1)e^z\,dz = \frac{1}{2\pi i}\oint_{\|z\|=\varepsilon}\frac{(z+1)e^{-nz}}{z^n}\,dz,$$

$$ c_n = [z^{n-2}]e^{-nz}+ [z^{n-1}]e^{-nz} = n^{n-2} [z^{n-2}] e^{-z} + n^{n-1} [z^{n-1}] e^{-z}. \tag{5}$$

If we coordinate $(4)$ and $(5)$ we get

$$ \mathcal{J}_1+\mathcal{J}_2 = 1+\sum_{n\geq 2}\frac{n^{n-2}(-1)^{n-2}n!}{(n-2)!n^{n+1}}+\frac{n^{n-1}(-1)^{n-1}n!}{(n-1)!n^{n+1}}\tag{6}$$

and the claim follows by massive cancellation.