Show that $\int_0^e \ln(1-\ln(x))dx=-e\gamma$

Question

Wolfram Alpha says $\int_0^e \ln(1-\ln(x))dx=-e\gamma$ which I thought was really interesting. I want to understand how to solve this integral. Here is my attempt:

$$\int_0^e \ln(1-\ln(x))dx$$ $$=\int_{-\infty}^0 \ln(1-u)xdu$$ $$=\int_{-\infty}^0 \ln(1-u)e^udu$$ $$=\ln(1-u)e^u\bigg|^0_{-\infty}+\int_{-\infty}^0 \frac{1}{1-u}e^udu$$ $$=\int_{-\infty}^0 \frac{e^u}{1-u}du$$

I feel like I am so close here but I can't figure out how to continue.

Answer 1

Substitute $t=1-\ln x$. Then, $ dx = e^{1-t}dt$ and \begin{align}\int_0^e \ln(1-\ln x)dx =e \int^\infty_0\ln t \ e^{-t}dt=-e\gamma \end{align}

Answer 2

Continuing from where I previously left off based off of Gary's suggestion.

$$\int_{-\infty}^0\frac{e^u}{1-u}du$$ $$=-\int_{\infty}^1\frac{e^{1-t}}{t}dt$$ $$=e\int^{\infty}_1\frac{e^{-t}}{t}dt$$ $$=eE_1(1)$$ $$=-e\gamma$$