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

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$
$\ds{\int_{0}^{1}{\ln\pars{1 + x} \over 1 + x}\ln\pars{\ln\pars{1 \over x}} \,\dd x:\ {\large ?}}$

\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
$$-\frac\gamma2\log(2)^2-\sum_{n=2}^\infty(-1)^n\frac{H_{n-1}}n\log(n)\tag{3}$$

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
$$-\gamma^2\log(2)-\sum_{n=2}^\infty(-1)^n\frac{\psi(n)}n\log(n)\tag{5}$$
$$\sum_{n=2}^\infty(-1)^n\frac{H_{n-1}}{n^x}\tag{6}$$