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

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


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\}]

I'm aware that this is a really old question, but searching through most of the questions similar to the nested logarithmic integral, I haven't found this approach yet. And I feel that this approach is a lot more straightforward and doesn't require the use of any complex constants.

Let$$\mathfrak{I}=\int\limits_0^1\mathrm dx\,\frac {\log\log\left(\frac 1x\right)}{1+2x\cos a+x^2}$$denote the general form of the integral. Since$$\cos a=\frac {e^{ia}+e^{-ia}}2$$Therefore$$\begin{align*}\mathfrak{I} & =\int\limits_0^1\mathrm dx\,\frac {\log\log\left(\frac 1x\right)}{(1+xe^{ia})(1+xe^{-ia})}\\ & =\int\limits_0^1\mathrm dx\,\log\log\left(\frac 1x\right)\sum\limits_{n\geq0}\sum\limits_{m\geq0}(-1)^{n+m}x^{n+m}e^{(n-m)ia}\end{align*}$$

The double sum can be simplified by writing the sum out and expanding$$1-x(e^{ia}+e^{-ia})+x^2(e^{2ia}+1+e^{-2ia})-x^3(e^{3ia}+e^{ia}+e^{-ia}+e^{-3ia})+\cdots$$If we denote $a_n$ as the coefficient of $x^n$, then we can rewrite the coefficient as$$a_n=(-1)^n\sum\limits_{k=0}^ne^{(n-2k)ia}=(-1)^n\frac {\sin a(n+1)}{\sin a}$$ Therefore the integral becomes$$\mathfrak{I}=\frac 1{\sin a}\sum\limits_{n\geq0}(-1)^n\sin a(n+1)\int\limits_0^1\mathrm dx\, x^n\log\log\left(\frac 1x\right)$$Thie intermediate integral is much easier to consider. First set $x\mapsto-\log x$ and then make the transformation $x\mapsto x(n+1)$ to get$$\begin{align*}\int\limits_0^1\mathrm dx\,x^n\log\log\left(\frac 1x\right) & =\frac 1{n+1}\int\limits_0^{\infty}\mathrm dx\, e^{-x}\log x-\frac {\log(n+1)}{n+1}\\ & =-\frac {\gamma+\log(n+1)}{n+1}\end{align*}$$where $\gamma$ is the Euler Mascheroni constant. Splitting up the infinite sum, then$$\begin{align*}\mathfrak{I}=-\frac {\gamma a}{2\sin a}-\frac 1{\sin a}\sum\limits_{n\geq1}(-1)^{n-1}\frac {\sin an\log n}n\end{align*}$$where we have used$$\sum\limits_{n\geq1}(-1)^{n-1}\frac {\sin an}n=\frac a2$$

The remaining log - sine sum can be evaluated. There is a closed form in terms of the gamma function. Essentially$$\log\Gamma(x)=\left(\frac 12-x\right)(\gamma+\log 2)+(1-x)\log\pi-\frac 12\log\sin\pi x+\frac 1{\pi}\sum\limits_{n\geq1}\frac {\sin 2\pi n x\log n}n$$

The only difference between the above identity and our sum is the alternating $(-1)^n$ term. This, however, can easily be remedied by considering the expansion$$\begin{align*}\sin n(\pi-a) & =\sin\pi n\cos an-\sin an\cos\pi a\\ & =-\sin a n\cos\pi n\\ & =(-1)^{n-1}\sin an\end{align*}$$

Therefore, evaluating $x$ at $\tfrac 12-\tfrac a{2\pi}$ gives the sum to be$$\sum\limits_{n\geq1}(-1)^{n-1}\frac {\log n\sin an}n=\pi\log\Gamma\left(\frac 12-\frac a{2\pi}\right)-\frac a2(\gamma+\log 2)-\frac {\pi}2\log2\pi+\frac {\pi}2\log\cos\left(\frac a2\right)$$Hence, by properties of the natural log,$$\int\limits_0^1\mathrm dx\,\frac {\log\log\left(\frac 1x\right)}{1+2x\cos a+x^2}\color{blue}{=\frac {\pi}{\sin a}\log\left[\frac {\sqrt\pi(2\pi)^{\frac a{2\pi}}}{\Gamma\left(\frac 12-\frac a{2\pi}\right)\sqrt{\cos\left(\frac a2\right)}}\right]}$$

Now set $a=\tfrac {2\pi}3$ so that $\cos a=-\frac 12$ to get the integral in question.

Tags:

Integration

Calculus

Closed Form

Definite Integrals

Improper Integrals

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