Compute $\int_{0}^{e}\frac{\ln(1-W(x))}{x}dx$

Question

So I made this integral for funzies, and I'm having trouble solving it.

$$\int_0^e\frac{\ln(1-W(x))}x dx$$

I may have made a mistake.

I let $u=W(x)$, then the integral becomes

$$\int_{0}^{1}\frac{(u+1)\ln(1-u)}{u}du$$

I used Taylor Series after, but then I got a weird sum. Anyone else wanna try?

Answer 1

$$\int_0^1\frac{(u+1)\ln(1-u)}udu=\int_0^1\ln(1-u)+\frac{\ln(1-u)}udu=I-\sum_{k=1}^\infty\frac1k\int_0^1u^{k-1}du=I-\sum_{k=1}^\infty\frac1{k^2}$$$I$ is easy to calculate, and the infinite sum is $\frac{\pi^2}6$.

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Answer 2

$$I = \int_0^e\frac{\ln(1-W(x))}x dx$$

$$u=W(x), x = ue^u, dx = (u+1)e^u du$$

$$\int_{0}^{1}\frac{\ln(1-u)}{ue^u}(u+1)e^u du = \int_{0}^{1}\frac{(u+1)\ln(1-u)}{u}du = \int_{0}^{1} \ln(1-u) du + \int_{0}^{1}\frac{\ln(1-u)}{u} du$$

$$\int_{0}^{1} \ln(1-u) du = (u-1)\ln(1-u) - u$$ $$\int_{0}^{1}\frac{\ln(1-u)}{u} du = - \operatorname{Li}_2{(u)}$$

$$\Rightarrow I = - \operatorname{Li}_2{(u)} + (u-1)\ln(1-u) - u + C \big\rvert_{x=0}^1$$

$$= - \operatorname{Li}_2{(1)} + (1-1)\ln(1-e) - 1 = -\frac{\pi^2}{6} - 1$$

Answer 3

For $|y|<1$, $$ \frac{1}{1-y}=\sum_{n=0}^{\infty} y^n \Leftrightarrow \ln (1-y)=-\sum_{n=0}^{\infty} \frac{y^{n+1}}{n+1}.$$ Plugging into the integrand, we have \begin{aligned} \int_0^1 \frac{(y+1) \ln (1-y)}{y} d y = & -\int_0^1 \frac{y+1}{y} \sum_{n=0}^{\infty} \frac{y^{n+1}}{n+1} d y \\ = & -\sum_{n=0}^{\infty} \frac{1}{n+1} \int_0^1(y+1) y^n d y \\ = & -\sum_{n=0}^{\infty} \frac{1}{n+1}\left(\frac{1}{n+2}+\frac{1}{n+1}\right) \\ = & -\sum_{n=0}^{\infty} \frac{1}{(n+1)(n+2)}-\sum_{n=0}^{\infty} \frac{1}{(n+1)^2} \\ = & -1-\frac{\pi^2}{6} \end{aligned}