Prove $\int_{0}^{1}\ln(\ln(x))\,{\rm d}x=-\gamma+\pi i$ 
How can we show that $$\int_{0}^{1}\ln(\ln(x))\,{\rm d}x=-\gamma+\pi i$$ is true or false?

I know that $$\text{Re}\left(\int_{0}^{1}\ln(\ln(x))\,{\rm d}x\right)=-\gamma$$ and $$\int_{0}^{1}\ln(\ln(x^{-1}))\,{\rm d}x=-\gamma$$but when I plugged this integral to Wolfram Alpha, it gave a decimal at the imaginary part that seems like $\pi$ as the imaginary part. Is there any ways to prove this? Using $\text{Li}(x)$ does not seem to work for any attempts. Do anyone have a suggestion that I could try to derive it?
 A: Just to complete the hint of Jyrki Lahtonen $$ \int_{0}^{1}\log\left(\log\left(x\right)\right)dx\stackrel{x=e^{-u}}{=}i\pi\int_{0}^{\infty}e^{-u}du+\int_{0}^{\infty}\log\left(u\right)e^{-u}du
 $$ and $$\int_{0}^{\infty}\log\left(u\right)e^{-u}du=\frac{d}{da}\left(\int_{0}^{\infty}u^{a}e^{-u}du\right)_{a=0}
 $$ $$=\psi\left(1\right)=-\gamma.$$
A: $\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
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$\ds{\int_{0}^{1}\ln\pars{\ln\pars{x}}\,\dd x =-\gamma + \pi\ic}$: True or
False ?.

$$\begin{array}{|c|}\hline\\
\mbox{I guess the $\ul{main\ point}$ of the OP question is:}
\\
\quad\mbox{What is the branch-cut W&A $\ul{choose}$ to perform the integration ?.}\quad 
\\ \\ \hline
\end{array}
$$

W&A chooses the branch-cut
$$
\begin{array}{|c|}\hline\\
\ds{\quad\ln\pars{z} = \ln\pars{\verts{z}} + \,\mathrm{arg}\pars{z}\ic\,,\quad
0 < \,\mathrm{arg}\pars{z} < 2\pi\,,\quad z \not= 0\quad}
\\ \\ \hline
\end{array}
$$
with two possible paths. Namely,
$$\,\mathcal{C}\pars{\pm\epsilon} \equiv
\braces{z = x \pm \epsilon\ic\,,\quad x \in \pars{0,1}\,,\quad \epsilon > 0}
$$
As $\ds{\ln\pars{x} < 0}$ when $\ds{x \in \pars{0,1}}$,
$\ds{\ul{\ \ln\pars{\ln\pars{x}} = \ln\pars{-\ln\pars{x}} + \pi\ic\ }}$ in the limit $\ds{\epsilon \to 0^{\pm}}$:
\begin{align}
\lim_{\epsilon\ \to\ 0^{+}}\,\,
\int_{\,\mathcal{C}\pars{\pm\epsilon}}\ln\pars{\ln\pars{z}}\,\dd z & =
\int_{0}^{1}\bracks{\ln\pars{-\ln\pars{x}} + \pi\ic}\,\dd x\ =\
\underbrace{\int_{0}^{1}\ln\pars{-\ln\pars{x}}\,\dd x}_{\ds{-\gamma}}\ +\ \pi\ic \\ & =
\fbox{$\ds{\ \color{#f00}{-\gamma + \pi\ic}\ }$}
\end{align}
