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

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

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  • $\begingroup$ have you tried u = 1+x? $\endgroup$ – Vishwa Iyer Jul 5 '14 at 17:57
  • $\begingroup$ Have you tried the sub. $\log(\frac{1}{x})=t$ $\endgroup$ – TheOscillator Jul 5 '14 at 18:36
  • $\begingroup$ I think the following might be of some use \begin{align*} I(a) & = \int_0^1 \frac{\log(1+ax)}{1+x}\log \log 1/x\mathrm{d}x \\ I'(a) & = \int_0^1 \frac{x \log \log x}{(ax+1)(x+1)}\mathrm{d}x \\ & = \frac{1}{a-1} \int_0^1 x \log \log \frac 1x\left( \frac{a}{ax+1} - \frac{x}{1+x} \right) \mathrm{d}x \\ & = \frac{1}{a-1} \int_0^1 \log \log \frac 1x\left( \frac{1}{1+x} - \frac{1}{1+ax}\right) \mathrm{d}x \\ & = \frac{1}{a-1}\frac{\log^2(2)}{2} - \frac{1}{a-1}\int_0^1 \frac{\log \log 1/x}{1+ax}\,\mathrm{d}x \end{align*} Then integrate back from 0 to 1, however convergence issuses =/ $\endgroup$ – N3buchadnezzar Jul 5 '14 at 18:55
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    $\begingroup$ There are lot of integrals related to the OP question. $\endgroup$ – Felix Marin Jul 5 '14 at 20:45
  • $\begingroup$ $$\int_0^1 \frac{\log(1+x)}{x}\log\left(\log\left(\frac{1}{x}\right)\right) \ dx = \frac{\pi^2}{12} \left(2 \ln 2 - 12 \ln A + \ln \pi \right),$$ where $A$ is the Glaisher–Kinkelin constant $\endgroup$ – user153012 Nov 4 '14 at 0:56
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$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{\int_{0}^{1}{\ln\pars{1 + x} \over 1 + x}\ln\pars{\ln\pars{1 \over x}} \,\dd x:\ {\large ?}}$

\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.

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  • $\begingroup$ Could you kindly point me towards a definition of the function $\Psi(n)$? I'm drawing a blank. $\endgroup$ – David H Jul 5 '14 at 22:18
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    $\begingroup$ @DavidH $\displaystyle{\large \Psi\pars{z} \equiv \ln\,'\left(\,\Gamma\left(z\right)\,\right)}$ is the digamma function. You'll find a good starting point in this table. $\endgroup$ – Felix Marin Jul 5 '14 at 22:45
  • $\begingroup$ Got it. As an historical aside, I'm somewhat curious as to the etymology behind labeling the digamma function by the capital Greek letter 'psi' as opposed to, oh I dunno, the Greek letter 'digamma'? :) Also, thanks in particular for the reference. $\endgroup$ – David H Jul 5 '14 at 23:07
  • $\begingroup$ @DavidH There is some historic report over here. $\endgroup$ – Felix Marin Jul 5 '14 at 23:15
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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 $$ \sum_{n=2}^\infty(-1)^n\frac{H_{n-1}}{n^x}\tag{6} $$

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  • $\begingroup$ @Kirill: that is what I used. $\endgroup$ – robjohn Oct 29 '14 at 18:04

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