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

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.

• Closed form via mathematica $$\frac{\pi \left(\log \left(\frac{128}{27}\right)+8 \log (\pi )-12 \log \left(\Gamma \left(\frac{1}{3}\right)\right)\right)}{3 \sqrt{3}}$$ – Brad Jul 22 '14 at 4:59
• @Brad Thanks, but how to get it? Anyway, you didn't complete this answer. – Anastasiya-Romanova 秀 Jul 22 '14 at 5:01
• See the paper of Victor Adamchik (freely avalable):repository.cmu.edu/cgi/…, More precisely Example 3.4. – Omran Kouba Jul 22 '14 at 6:52
• Thank you so much Prof. @OmranKouba (* ˘⌣˘)◞[]♥[]ヽ(•‿• ) – Anastasiya-Romanova 秀 Jul 22 '14 at 7:18
• @Anastasiya-Romanova This integral is one of Malmsten's integrals. You find several complete solutions for this and similar integrals in this paper. Note that the unique solution which does not require the knowledge of higher analysis is perhaps that given by Malmsten in 1842, other solutions being certainly more complicated. – Iaroslav Blagouchine Dec 2 '14 at 3:19


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\}]

• In fact, you use the same ingredients as Adamchik. The missing fact is that $$\frac{\Gamma(\frac{2}{3})\Gamma(\frac{5}{6})}{\Gamma(\frac{1}{6})}=\frac{4\ 2^{2/3} \pi ^3}{3 \sqrt{3} \Gamma \left(\frac{1}{3}\right)^5}$$ which yields the same simplified answer. This follows from the duplication and complement formulas for $\Gamma$. – Omran Kouba Aug 22 '14 at 5:17
• @OmranKouba, thanks. – xpaul Aug 22 '14 at 15:30

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.