I did some numeric experiments with integrals involving double logarithms (because they received much interest both on this site and in published papers, sometimes under names of Malmsten—Vardi—Adamchik integrals).

It appears that $${\large\int}_0^1\ln\ln\left(\frac{1+x}{1-x}\right)\cdot\frac{\ln x}{1-x^2}\,dx\stackrel{\color{gray}?}=\frac{\pi^2}{24}\,\ln\left(\frac{A^{36}}{16\,\pi^3}\right),$$ where $A=\exp\left(\frac1{12}-\zeta'(-1)\right)$ is the Glaisher—Kinkelin constant (I have more than $1000$ decimal digits confirming this conjecture). How can we prove it?

Vladimir Reshetnikov
  • 45,303
  • 7
  • 151
  • 282
  • 3
    It is probably irrelevant (and you might know it already), but by the change of variable $u=\log\frac{1+x}{1-x}$ we get $\frac{1}{2}\int_0^\infty \log[u]\log[\tanh[u/2]]\mathrm du$. I tried expanding the tangent through an infinite product $\tanh x=x\prod_k \left(\frac{1-2k}{k}\right)^2\frac{\pi^2 k^2+x^2}{\pi^2(1-2k)^2+4x^2}$, but it was pointless. Ill keep on trying and if I get something sensible Ill comment. – AccidentalFourierTransform Dec 20 '15 at 22:48
  • 2
    I have to leave now. But I note: $\ln \frac{1+x}{1-x} = 2 \,\mathrm{arctanh}\, x$ and $\frac{\mathrm{d}}{\mathrm{d}x} \,\mathrm{arctanh}\, x = \frac{1}{1-x^2}$. – Eric Towers Dec 20 '15 at 23:16

2 Answers2


Using the substitution suggested by @AccidentalFourierTransform , and the integral $\displaystyle \,\,\, \int \frac{1}{\sinh x} dx=\ln\tanh\left(\frac{x}{2}\right)+C$, we have

$$\begin{align} -2\int_0^1 \ln \ln \left(\frac{1+x}{1-x}\right) \frac{\ln x}{1-x^2}dx \\&= -\int_0^{\infty}\ln x \,\,\ln \tanh\left(\frac{x}{2}\right)dx \\&=\int_0^{\infty}\ln x \int_x^{\infty}\frac1{\sinh t} dt dx \\&=\int_0^{\infty} \int_0^{t}\ln x \frac1{\sinh t} dx dt \\&=\int_0^{\infty} \frac{x\ln x-x}{\sinh x} dx \\&=\int_0^{\infty} (x\ln x -x) \frac{2 e^{-x}}{1-e^{-2x}}\,dx \\&=\int_0^{\infty} (x\ln x-x)2\sum_{n=0}^{\infty} e^{-x(2n+1)}dx \\&=2\sum_{n=0}^{\infty} \left(\frac{d}{ds} \frac{\Gamma(s)}{(2n+1)^s} \Bigg{|}_{s=2}-\frac1{(2n+1)^2}\right) \\&=2\sum_{n=0}^{\infty} \left(\frac{\Gamma'(2)-1}{(2n+1)^2}-\frac{\ln(2n+1)}{(2n+1)^2}\right) \\&=2(\Gamma'(2)-1)\frac{\pi^2}{8}-2\left(\sum_{n=1}^{\infty}\frac{\ln(n)}{n^2}-\sum_{n=1}^{\infty} \frac{\ln(2n)}{(2n)^2} \right) \\&=2(\Gamma'(2)-1)\frac{\pi^2}{8}+2\zeta'(2)+\frac12\left(\ln2 \frac{\pi^2}{6}-\zeta'(2)\right) \\&=-\frac{\pi^2}{12}\ln\left(\frac{A^{36}}{16\pi^3}\right) \end{align}$$

Noam Shalev - nospoon
  • 6,140
  • 25
  • 39

Hint. One may observe that, by the change of variable $u=\dfrac{1-x}{1+x}$ one gets $$ I={\int}_0^1\log\log\left(\frac{1+x}{1-x}\right)\cdot\frac{\log x}{1-x^2}\,dx=\int_0^1\log\left(-\log u\right)\cdot\frac{\log (1-u)-\log (1+u)}{2u}\,du. \tag1 $$ Then by a standard Taylor expansion, we have $$ \frac{\log (1-u)-\log (1+u)}{2u}= -\sum_{n=0}^{\infty} \frac{u^{2n}}{2n+1}, \qquad |u|<1,\tag2 $$ giving $$ I=-\sum_{n=0}^{\infty} \frac{1}{2n+1}\int_0^1u^{2n}\log\left(-\log u\right)\:du.\tag3 $$ The latter integral is easily obtained using the well-known integral representation of the Euler gamma function $$ \frac{\Gamma(s)}{(a+1)^s}=\int_0^\infty t^{s-1} e^{-(a+1)t}\:dt. \tag4 $$ By differentiating $(4)$ with respect to $s$ and putting $s=1$ we produce $$ \int_0^1u^a\log\left(-\log u\right)\:du=-\frac{\gamma+\log(a+1)}{a+1} \tag5 $$ leading to $$ I=\sum_{n=0}^{\infty} \frac{\gamma+\log(2n+1)}{(2n+1)^2}=\left.\left(\gamma-\frac{d}{ds} \right)\left(\left(1-2^{-s}\right)\zeta(s)\right)\right|_{s=2}=\frac{\pi^2}{24}\,\ln\left(\frac{A^{36}}{16\,\pi^3}\right)\tag6 $$ as announced.

These integrals have been studied by Adamchik, Vardi, Moll and many others. One may have a look at this interesting paper.

Olivier Oloa
  • 118,697
  • 18
  • 195
  • 315