Evaluation of $\int_0^1 \frac{\log(1+x)}{1+x}\log\left(\log\left(\frac{1}{x}\right)\right) \ dx$

I need some hints, clues for getting the closed form of

$$\int_0^1 \frac{\log(1+x)}{1+x}\log\left(\log\left(\frac{1}{x}\right)\right) \ dx$$

Solutions Collecting From Web of "Evaluation of $\int_0^1 \frac{\log(1+x)}{1+x}\log\left(\log\left(\frac{1}{x}\right)\right) \ dx$"

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$\ds{\int_{0}^{1}{\ln\pars{1 + x} \over 1 + x}\ln\pars{\ln\pars{1 \over x}}
\,\dd x:\ {\large ?}}$

\begin{align}&\overbrace{\color{#c00000}{%
\int_{0}^{1}{\ln\pars{1 + x} \over 1 + x}\ln\pars{\ln\pars{1 \over x}}\,\dd x}}
^{\ds{\mbox{Set}\ x \equiv \expo{-t}\ \imp\ t = -\ln\pars{x}}}\ =\
\int_{\infty}^{0}{\ln\pars{1 + \expo{-t}} \over 1 + \expo{-t}}\,\ln\pars{t}\,
\pars{-\expo{-t}\,\dd t}
\\[3mm]&=\lim_{\mu \to -1}\partiald{}{\mu}\color{#00f}{\int_{0}^{\infty}\ln\pars{t}\expo{-t}
\pars{1 + \expo{-t}}^{\mu}\,\dd t}
\end{align}

\begin{align}&\color{#00f}{\int_{0}^{\infty}\ln\pars{t}\expo{-t}
\pars{1 + \expo{-t}}^{\mu}\,\dd t}
=\sum_{n = 0}^{\infty}{\mu \choose n}\int_{0}^{\infty}
\ln\pars{t}\expo{-\pars{n + 1}t}\,\dd t
\\[3mm]&=\sum_{n = 0}^{\infty}{-\mu + n – 1 \choose n}\pars{-1}^{n}\lim_{\epsilon \to 0}\partiald{}{\epsilon}\int_{0}^{\infty}t^{\epsilon}\expo{-\pars{n + 1}t}
\,\dd t
\\[3mm]&=\sum_{n = 0}^{\infty}{-\mu + n – 1 \choose n}\pars{-1}^{n}\lim_{\epsilon \to 0}\partiald{}{\epsilon}
\bracks{{1 \over \pars{n + 1}^{\epsilon + 1}}
\int_{0}^{\infty}t^{\epsilon}\expo{-t}\,\dd t}
\\[3mm]&=\sum_{n = 0}^{\infty}{-\mu + n – 1 \choose n}\pars{-1}^{n}\lim_{\epsilon \to 0}\partiald{}{\epsilon}
\bracks{{\Gamma\pars{\epsilon + 1} \over \pars{n + 1}^{\epsilon + 1}}}
\\[3mm]&=-\sum_{n = 0}^{\infty}{-\mu + n – 1 \choose n}\pars{-1}^{n}\,
{\ln\pars{n + 1} + \gamma \over n + 1}
\end{align}

Also,
$$
\lim_{\mu \to – 1}\partiald{}{\mu}{-\mu + n – 1 \choose n}
=-\Psi\pars{n + 1} – \gamma
$$

such that
\begin{align}&\color{#c00000}{%
\int_{0}^{1}{\ln\pars{1 + x} \over 1 + x}\ln\pars{\ln\pars{1 \over x}}\,\dd x}
=-\sum_{n = 1}^{\infty}\pars{-1}^{n}\,
{\bracks{\gamma + \Psi\pars{n}}\bracks{\gamma + \ln\pars{n}} \over n}
\\[3mm]&=-\gamma^{2}
\sum_{n = 1}^{\infty}{\pars{-1}^{n} \over n}
-\gamma\sum_{n = 1}^{\infty}\pars{-1}^{n}\,{\ln\pars{n} \over n}
-\gamma\sum_{n = 1}^{\infty}\pars{-1}^{n}\,{\Psi\pars{n} \over n}
-\sum_{n = 1}^{\infty}\pars{-1}^{n}\,{\Psi\pars{n}\ln\pars{n} \over n}
\end{align}

With
\begin{align}
\sum_{n = 1}^{\infty}{\pars{-1}^{n} \over n} & = -\ln\pars{2}
\\
\sum_{n = 1}^{\infty}\pars{-1}^{n}\,{\ln\pars{n} \over n}&=
\gamma\ln\pars{2} – \half\,\ln^{2}\pars{2}
\\[3mm]
\sum_{n = 1}^{\infty}\pars{-1}^{n}\,{\Psi\pars{n} \over n}&=
\gamma\ln\pars{2} + \half\,\ln^{2}\pars{2}
\end{align}

we’ll have
\begin{align}&\color{#66f}{\large%
\int_{0}^{1}{\ln\pars{1 + x} \over 1 + x}\ln\pars{\ln\pars{1 \over x}}\,\dd x}
\\[3mm]&=\color{#66f}{\large-\gamma^{2}\ln\pars{2}
-\sum_{n = 1}^{\infty}\pars{-1}^{n}\,{\Psi\pars{n}\ln\pars{n} \over n}}
\approx -0.2408
\end{align}

$\ds{\tt%
\mbox{So far, I was not able to evaluate the last sum. I’m still trying to}\ldots}$.

It seems related somehow to an
Euler Sum.

Substitute $x\mapsto e^{-x}$ and then $x\mapsto x/n$:
$$
\begin{align}\hspace{-1cm}
\int_0^1\frac{\log(1+x)}{1+x}\log(\log(1/x))\,\mathrm{d}x
&=\int_0^\infty\frac{\log(1+e^{-x})}{1+e^{-x}}\log(x)\,e^{-x}\,\mathrm{d}x\\
&=\sum_{n=2}^\infty\int_0^\infty(-1)^nH_{n-1}e^{-nx}\log(x)\,\mathrm{d}x\\
&=\sum_{n=2}^\infty\int_0^\infty(-1)^n\frac{H_{n-1}}ne^{-x}(\log(x)-\log(n))\,\mathrm{d}x\\
&=\sum_{n=2}^\infty(-1)^n\frac{H_{n-1}}n(-\gamma-\log(n))\tag{1}
\end{align}
$$
Using $(3)$ from this answer, we get
$$
\begin{align}
\sum_{n=2}^\infty(-1)^n\frac{H_{n-1}}n
&=\frac{\zeta(2)}{2}-\sum_{n=1}^\infty(-1)^{n-1}\frac{H_n}n\\
&=\frac12\log(2)^2\tag{2}
\end{align}
$$
Therefore, we get that $(1)$ is
$$
-\frac\gamma2\log(2)^2-\sum_{n=2}^\infty(-1)^n\frac{H_{n-1}}n\log(n)\tag{3}
$$


Comparison with Felix Marin’s Answer

Using $(8)$ from this answer, we get
$$
\begin{align}
\sum_{n=2}^\infty(-1)^n\frac{H_{n-1}}n\log(n)
&=\sum_{n=2}^\infty(-1)^n\frac{\psi(n)+\gamma}n\log(n)\\
&=\sum_{n=2}^\infty(-1)^n\frac{\psi(n)}n\log(n)+\gamma^2\log(2)-\gamma\frac{\log(2)^2}{2}\tag{4}
\end{align}
$$
Thus, plugging $(4)$ into $(3)$, we see that $(1)$ is
$$
-\gamma^2\log(2)-\sum_{n=2}^\infty(-1)^n\frac{\psi(n)}n\log(n)\tag{5}
$$
which is the same as Felix Marin gets.


Further Investigation

Perhaps the next step is to consider the derivative of
$$
\sum_{n=2}^\infty(-1)^n\frac{H_{n-1}}{n^x}\tag{6}
$$