A closed form of $\int_0^1\frac{\ln\ln\left({1}/{x}\right)}{x^2-x+1}\mathrm dx$

Question

This integral has been bugging me since yesterday:

$$\int_0^1\frac{\ln\ln\left({1}/{x}\right)}{x^2-x+1}\mathrm dx$$

I've tried substitution $y={1}/{x}$ and $e^y={1}/{x}$, but those didn't help much. Wolfram Alpha gives me result: $-0.67172$. Could anyone here please help me to obtain the closed form of the integral preferably (if possible) with elementary ways (high school methods)? Any help would be greatly appreciated. Thank you.

Answer 1

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ With $\ds{r \equiv {1 + \root{3}\ic \over 2} = \expo{\pi\ic/3}}$ \begin{align}&\color{#c00000}{ \int_{0}^{1}{\ln\pars{\ln\pars{1/x}} \over x^{2} - x + 1}\,\dd x} =\int_{0}^{1}{\ln\pars{\ln\pars{1/x}} \over \pars{x - r}\pars{x - r^{*}}}\,\dd x \\[3mm]&=\int_{0}^{1}\ln\pars{\ln\pars{1/x}} \pars{{1 \over x - r} - {1 \over x - r^{*}}}\,{1 \over r - r^{*}}\,\dd x \\[3mm] & = {1 \over \Im\pars{r}}\,\Im\int_{0}^{1} {\ln\pars{\ln\pars{1/x}} \over x - r}\,\dd x \end{align}

With $\ds{x \equiv \expo{-t}}$: \begin{align}&\color{#c00000}{ \int_{0}^{1}{\ln\pars{\ln\pars{1/x}} \over x^{2} - x + 1}\,\dd x} ={2\root{3} \over 3}\Im\int_{\infty}^{0}{\ln\pars{t} \over \expo{-t} - r} \,\pars{-\expo{-t}\,\dd t} \\[3mm]&=-\,{2\root{3} \over 3}\Im\bracks{{1 \over r}\int_{0}^{\infty} {\ln\pars{t}\expo{-t} \over 1 - \expo{-t}/r}\,\dd t} \\[3mm]&=-\,{2\root{3} \over 3}\Im\bracks{{1 \over r} \sum_{n = 1}^{\infty}{1 \over r^{n - 1}}\int_{0}^{\infty} \ln\pars{t}\expo{-nt}\,\dd t} \\[3mm]&=-\,{2\root{3} \over 3}\Im\lim_{\mu\ \to\ 0}\partiald{}{\mu}\bracks{ \sum_{n = 1}^{\infty}r^{-n}\int_{0}^{\infty}t^{\mu}\expo{-nt}\,\dd t} \\[3mm]&=-\,{2\root{3} \over 3}\Im\lim_{\mu\ \to\ 0}\partiald{}{\mu}\bracks{ \sum_{n = 1}^{\infty}{r^{-n} \over n^{\mu + 1}}\Gamma\pars{\mu + 1}} \\[3mm]&=-\,{2\root{3} \over 3}\Im\lim_{\mu\ \to\ 0}\partiald{}{\mu}\bracks{ \Gamma\pars{\mu + 1}{\rm Li}_{\mu + 1}\pars{r^{*}}} \\[3mm]&=-\,{2\root{3} \over 3}\Im\lim_{\mu\ \to\ 0}\partiald{}{\mu}\bracks{ \Gamma\pars{\mu + 1}{\rm Li}_{\mu + 1}\pars{\expo{-\pi\ic/3}}} \end{align}

$\ds{{\rm Li}_{1}\pars{z} = -\ln\pars{1 - z}}$. Derivatives of the PolyLogarithm, respect of the order, can be evaluated from its integral representation.

Also, see Hurwitz Zeta Function.

Answer 2

Let $x=e^{-u}$. Then: \begin{eqnarray} \int_0^1\frac{\ln\ln\left(\frac{1}{x}\right)}{x^2-x+1}dx&=&\int_0^\infty\frac{e^{-u}\ln u}{e^{-2u}-e^{-u}+1}du \end{eqnarray} Let: \begin{eqnarray} I(a)&=&\int_0^\infty\frac{e^{-u}u^a}{e^{-2u}-e^{-u}+1}du=\int_0^\infty\frac{e^{-u}(1+e^{-u})u^a}{1+e^{-3u}}du\\ &=&\int_0^\infty\sum_{n=0}^\infty(-1)^ne^{-(3n+1)u}(1+e^{-u})u^a\ du\\ &=&\int_0^\infty\sum_{n=0}^\infty(-1)^n(e^{-(3n+1)u}+e^{-(3n+2)u})u^a\ du\\ &=&\Gamma(a+1)\sum_{n=0}^\infty(-1)^{n}\left(\frac{1}{(3n+1)^{a+1}}+\frac{1}{(3n+2)^{a+1}}\right)\\ &=&6^{-a-1}\Gamma(a+1)\left(\zeta(a+1,\frac{1}{6})+\zeta(a+1,\frac{1}{3})-\zeta(a+1,\frac{2}{3})-\zeta(a+1,\frac{5}{6})\right) \end{eqnarray} Hence: \begin{eqnarray} I'(0)&=&\frac{1}{6}\left[(\gamma-\ln 6)\left(\psi_0(\frac16)-\psi_0(\frac56)+\psi_0(\frac13)-\psi_0(\frac23)\right)\right.\\ &&\left.-\gamma_1(\frac16)-\gamma_1(\frac13)+\gamma_1(\frac23)+\gamma_1(\frac56)\right]\\ &=&\frac{1}{6}\left[-\frac{4\pi(\gamma-\ln 6)}{\sqrt 3}-\gamma_1(\frac16)-\gamma_1(\frac13)+\gamma_1(\frac23)+\gamma_1(\frac56)\right]\\ \end{eqnarray} where $\gamma_k(x)$ is the $k$-th Stieltjes $\Gamma$ constant. Using: $$ \psi_0(1-z)-\psi_0(z)=\pi\cot(\pi z) $$ it is easy to get: $$ \psi_0(\frac16)-\psi_0(\frac56)+\psi_0(\frac13)-\psi_0(\frac23)=-\frac{4\pi(\gamma-\ln 6)}{\sqrt 3}. $$ For $\gamma_1(\frac{p}{q})$, we have to use the following formula: \begin{eqnarray} \gamma_1(1,\frac{p}{q})-\gamma_1(1-\frac{p}{q})=-\pi(\log(2\pi q)+\gamma)\cot\frac{\pi p}{q}-2\pi\sum_{j=1}^{q-1}\ln\Gamma(\frac jq)\sin\left(\frac{2\pi jp}{q}\right) \end{eqnarray} from this. First, I have: \begin{eqnarray} \gamma_1(\frac{1}{3})-\gamma_1(\frac{2}{3})&=&-\frac{\pi}{2\sqrt 3}[2\gamma-\ln 3+8\ln(2\pi)-12\ln(\ln\Gamma(\frac13))],\\ \gamma_1(\frac{1}{6})-\gamma_1(\frac{5}{6})&=&-\pi\sqrt 3\left[\gamma+\ln\left(\frac{12\pi\Gamma(\frac23)\Gamma(\frac56)}{\Gamma(\frac16)\Gamma(\frac13)}\right)\right]=-\pi\sqrt 3\left[\gamma+\ln\left(\frac{4\cdot 2^{2/3}\pi^3}{3\sqrt3\Gamma^5(\frac13)}\right)\right]. \end{eqnarray} Putting all the results together, finally we have: \begin{eqnarray} I'(0)&=&\frac{\pi}{12\sqrt 3}\left[\ln\frac{268435456}{531441}+32\ln\pi-48\ln\left(\Gamma(\frac13)\right)\right]. \end{eqnarray} The numerical value is $-0.671719601885875$ which is the same as that from Mathematica command:

NIntegrate[Log[Log[1/x]]/(x^2 - x + 1)}, \{x, 0, 1\}]