# Integral ${\large\int}_0^1\frac{\ln^2\ln\left(\frac1x\right)}{1+x+x^2}dx$

Gradshteyn & Ryzhik, 7th ed., p. 570, formula 4.325(5) give the following definite integral: \begin{align*}{\large\int}_0^1\frac{\ln\ln\left(\frac1x\right)}{1+x+x^2}dx&=\frac\pi{\sqrt3}\ln\frac{\sqrt{2\pi}\,\Gamma\left(\frac23\right)}{\Gamma\left(\frac13\right)}\\&=\frac\pi{\sqrt3}\left(\frac{4\ln2\pi}3-\frac{\ln3}2-2\ln\Gamma\left(\tfrac13\right)\right)\end{align*} This and other similar integrals are discussed in several papers:

• Vardi, Integrals, an introduction to analytic number theory. Am. Math. Mon. 95, 308–315 (1988)
• Adamchik, A class of logarithmic integrals. Proceedings ISSAC, 1–8, 1997
• Medina, Moll, A class of logarithmic integrals, Ramanujan J. 20 (2009), no. 1, 91–126
• Blagouchine, Rediscovery of Malmsten's integrals, their evaluation by contour integration methods and some related results, Ramanujan J. 2014; 35: 21

Is it possible to find a closed form for a similar integral having the square of the logarithm in the numerator? $${\large\int}_0^1\frac{\ln^2\ln\left(\frac1x\right)}{1+x+x^2}dx$$

• Maybe, but what for? – Start wearing purple Oct 18 '14 at 22:59
• @O.L. Why would you want to know that? – X.C. Oct 18 '14 at 23:04
• I was always curious about the motivation of people asking whether it is possible to find a closed form of integrals of type $\displaystyle\int_0^1\frac{\ln^2\ln\frac1x}{1+x+x^2}dx$. – Start wearing purple Oct 18 '14 at 23:07
• @O.L. Well, different people may have different motivations, so I cannot say for everybody. I just hope the answer will be aesthetically pleasing and I could spend some time looking at it in delight. – X.C. Oct 18 '14 at 23:22
• There is something similar over here. – Felix Marin Oct 29 '14 at 17:47

You may write \begin{align*} {\large\int}_0^1\frac{\ln^2\ln\left(\frac1x\right)}{1+x+x^2}dx&={\large\int}_0^\infty \left(1-e^{-t}\right)\frac{\ln^2t}{1-e^{-3t}}e^{-t}dt\\ &=\sum_{n=0}^\infty{\large\int}_0^\infty \left(e^{-t}-e^{-2t}\right)e^{-3nt}\ln^2t\:dt\\ &=\sum_{n=0}^\infty \left({\large\int}_0^\infty e^{-(3n+1)t}\ln^2t\:dt-{\large\int}_0^\infty e^{-(3n+2)t}\ln^2t\:dt\right)\\ &=\sum_{n=0}^\infty \left. \partial_{s}^2 \left({\large\int}_0^\infty t^se^{-(3n+1)t}\:dt-{\large\int}_0^\infty t^s e^{-(3n+2)t}\:dt\right)\right|_{s=0}\\ &=\left. \partial_{s}^2 \left(\Gamma(s+1) \left(\sum_{n=0}^\infty \frac{1}{(3n+1)^{s+1}}-\sum_{n=0}^\infty \frac{1}{(3n+2)^{s+1}}\right)\right)\right|_{s=0}\\ &=\left. \partial_{s}^2 \left(\frac{\Gamma(s+1)}{3^{s+1}} \left(\zeta\left(s+1,\frac13\right)-\zeta\left(s+1,\frac23\right)\right)\right)\right|_{s=0}\\ \end{align*} Then, using the Laurent series expansion of the Hurwitz zeta function near $1$, $$\zeta(s+1,a) = \frac1s+\sum_{k=0}^{\infty}\frac{(-1)^{k}}{k!} \gamma_{k}(a) s^{k}, \quad a>0, \, s\to 0,$$ with $\displaystyle \gamma_{0}(a)= - \psi(a)=-\Gamma'(a)/\Gamma(a)$, we get
$$\color{#0052A3}{\frac{\pi^3\sqrt{3}}{54}+\frac{\sqrt{3}\pi}{9} (\gamma+\ln 3)^2+\frac{2}{3} (\gamma+\ln 3)\! \left(\gamma_1\!\!\left(\frac{1}{3}\right)-\gamma_1\!\!\left(\frac{2}{3}\right)\right)+\frac{1}{3}\!\!\left(\gamma_2\!\!\left(\frac{1}{3}\right)-\gamma_2\!\!\left(\frac{2}{3}\right)\right)}.$$
$$\color{#3366FF}{\gamma_1\left(\tfrac13\right)-\gamma_1\left(\tfrac23\right)=\frac\pi{\sqrt 3}\left(6\ln\Gamma\left(\tfrac13\right)-\gamma+\frac{\ln 3}2-4\ln(2\pi)\right)}.$$
• It's known that $\gamma_1\left(\tfrac13\right)=\gamma_1-\left(\frac{3\gamma}2+\frac{3\ln 3}4\right)\cdot\ln 3+\frac\pi{\sqrt 3}\left(3\ln\Gamma\left(\tfrac13\right)-\frac\gamma2+\frac{\ln 3}4-2\ln(2\pi)\right)$ and $\gamma_1\left(\tfrac23\right)=\gamma_1-\left(\frac{3\gamma}2+\frac{3\ln 3}4\right)\cdot\ln 3-\frac\pi{\sqrt3}\left(3\ln\Gamma\left(\tfrac13\right)-\frac\gamma2+\frac{\ln 3}4-2\ln(2 \pi)\right)$. So, $\gamma_1\left(\tfrac13\right)-\gamma_1\left(\tfrac23\right)=\frac\pi{\sqrt 3}\left(6\ln\Gamma\left(\tfrac13\right)-\gamma+\frac{\ln 3}2-4\ln(2\pi)\right)$. – Vladimir Reshetnikov Oct 19 '14 at 4:15