There are well-known closed-form evaluations for integrals of the form $\int_0^1 a(x) \ln \left(\ln \left(\frac{1}{x}\right)\right) \, dx $ for certain algebraic functions $a(x)$. For example, an evaluation of this form is given in the following link: Evaluating $\int_0^1 \log \log \left(\frac{1}{x}\right) \frac{dx}{1+x^2}$. Closed-form evaluations for integrals of the form $\int_0^1 a(x) \ln \left(\ln \left(\frac{1}{x}\right)\right) \, dx$ may be used to evaluate Glaisher-type infinite products. For example, the evaluation of $ \int_{0}^{1} \frac{\ln \left(\ln \left(\frac{1}{x}\right)\right)}{x^{2}+1} \, dx $ may be used to prove that:

\begin{equation*} \prod _{i=0}^{\infty } \frac{ (4 i+3)^{\frac{1}{4 i+3}} }{ (4 i+1)^{\frac{1}{4 i+1}}} = \frac{ 2^{ \frac{ \pi}{2} } e^{\frac{\gamma \pi }{4}} \pi ^{ \frac{3 \pi}{4}}}{ \Gamma^{\pi } \left(\frac{1}{4}\right)}. \end{equation*}

Mathematica does not seem to be able to evaluate integrals of the form $\int_0^1 T(x) \ln \left(\ln \left(\frac{1}{x}\right)\right) \, dx $ for certain elementary transcendental functions $T\left(x\right)$. For example, Mathematica is unable to evaluate the following integrals, and the Inverse Symbolic Calculator is not currently able to identify the following numbers:

\begin{align*} \int_0^1 \ln (1-x) \ln \left(\ln \left(\frac{1}{x}\right)\right) \, dx & = 1.834962542065861... \\ \int_0^1 \ln (x+1) \ln \left(\ln \left(\frac{1}{x}\right)\right) \, dx & = -0.4553656688368576... \\ \int_0^1 \tan^{-1}(x) \ln \left(\ln \left(\frac{1}{x}\right)\right) \, dx & = -0.521812852476476... \end{align*}

I am interested in evaluating $\int_0^1 \ln (1-x) \ln \left(\ln \left(\frac{1}{x}\right)\right) \, dx$, but it does not seem to be feasible to use integration by parts to evaluate this integral, and it is not obvious to me how to evaluate this integral using substitution. I have also considered expanding the integrand of this integral using power series to try to evaluate this integral.

A new closed-form evaluation of this integral would be interesting, partly because such an evaluation may be used to construct new Glaisher-type infinite products.

A similar problem is given in the following link: Integral ${\large\int}_0^1\frac{\ln^2\ln\left(\frac1x\right)}{1+x+x^2}dx$. I've tried to mimic the strategy used in the answer given in this link, but using the substitution $t=\ln\left(\frac{1}{x}\right)$ in this case yields the following integral, which Mathematica is unable to evaluate:

\begin{equation*} \int_0^{\infty } e^{-t} \ln \left(1-e^{-t}\right) \ln (t) \, dt. \end{equation*}

It is also natural to ask: What are some non-trivial examples of elementary transcendental functions $T\left(x\right)$ such that there is a closed-form evaluation of $\int_0^1 T\left(x\right) \ln \left(\ln \left(\frac{1}{x}\right)\right) \, dx$? What is $\int_0^1 \sin^{-1} x \ln \left(\ln \left(\frac{1}{x}\right)\right) \, dx$? What is $\int_0^1 \ln(x^{2} +1) \ln \left(\ln \left(\frac{1}{x}\right)\right) \, dx$?

  • 2
    $\begingroup$ Using taylor series for $ln(1-x)$ maybe would help $\endgroup$ Apr 8, 2016 at 1:32
  • $\begingroup$ @ArchisWelankar Thanks for your comment! It is easily seen that $\int_{0}^{1} x^n \ln \left(\ln \left(\frac{1}{x}\right)\right) \, dx$ is equal to $\frac{1}{n+1} (-\gamma -\ln (n+1))$ for all $n \in \mathbb{N}_{0}$. Given the Maclaurin series $\ln (1-x) = \sum _{n=0}^{\infty } -\frac{x^{n+1}}{n+1}$, it is clear that the problem of evaluating $\int_{0}^{1} \ln(1-x) \ln \left( \ln \left( \frac{1}{x} \right) \right) dx$ reduces to the difficult problem of evaluating the sum $\sum _{n=0}^{\infty } \frac{\ln (n+2)+\gamma }{(n+2) (n+1)} = 1.83496254206...$. $\endgroup$ Apr 8, 2016 at 2:09

2 Answers 2


We have the following closed form.

Proposition. $$ \int_0^1\log \left(1-x\right)\log \left(-\log x\right)\:dx=\gamma-\gamma_1+\gamma_1(1,0)\tag1 $$

where $\displaystyle \gamma$ is the Euler-Mascheroni constant, where $\gamma_1$ is the Stieltjes constant, $$\gamma_1 = \lim_{N\to+\infty}\left(\sum_{n=1}^N \frac{\log n}n-\int_1^N\frac{\log t}t\:dt\right)$$ and where $\gamma_1(a,b)$ is the poly-Stieltjes constant (see here), $$\gamma_1(a,b) = \lim_{N\to+\infty}\left(\sum_{n=1}^N \frac{\log (n+a)}{n+b}-\int_1^N\frac{\log t}t\:dt\right)\!.$$

Proof. One may recall the classic integral representation of the Euler gamma function $$ \frac{\Gamma(s)}{(a+1)^s}=\int_0^\infty t^{s-1} e^{-(a+1)t}\:dt, \qquad s>0,\, a>-1. \tag2 $$ By differentiating $(2)$ with respect to $s$, putting $s=1$ and making the change of variable $x=e^{-t}$, we get $$ \int_0^1x^a\log\left(-\log x\right)\:dx=-\frac{\gamma+\log(a+1)}{a+1},\qquad a>-1, \tag3 $$

where $\displaystyle \gamma$ is the Euler-Mascheroni constant. We are allowed to insert the standard Taylor series expansion, $$ \log (1-x)= -\sum_{n=1}^{\infty} \frac{x^n}n, \qquad |x|<1,\tag4 $$ into the given integral, then using $(3)$ we obtain $$ \begin{align} \int_0^1\log \left(1-x\right)\log \left(-\log x\right)\:dx&=-\int_0^1\sum_{n=1}^{\infty}\frac{x^n}n \:\log (-\log x)\:dx\\ &=-\sum_{n=1}^{\infty} \frac1n\int_0^1 x^n\log (-\log x)\:dx\\ &=\sum_{n=1}^{\infty} \frac{\gamma+\log(n+1)}{n(n+1)}\\ &=\gamma \sum_{n=1}^{\infty}\frac1{n(n+1)}+\sum_{n=1}^{\infty} \frac{\log(n+1)}{n(n+1)}\\ &=\gamma +\sum_{n=1}^{\infty} \frac{\log(n+1)}{n(n+1)},\tag5 \end{align} $$ and we may conclude with Theorem $2$ here to get

$$ \begin{align} \sum_{n=1}^{\infty} \frac{\log (n+1)}{n(n+1)}=\gamma_1({1,0})-\gamma_1,\tag6 \end{align} $$

since $\gamma_1(1,1)=\gamma_1$.

Remark. I am inclined to believe that the poly-Stieltjes constants will turn out to be a tool for many of the considered integrals.


This is not really an answer (OlivierOloa already gave a great one), but just to provide the full picture for the first integral:

$$\int_0^1 \ln (1-x) \ln \left(\ln \left(\frac{1}{x}\right)\right) \, dx-\gamma=$$

$$=\sum_{n = 2}^{\infty} \frac{(-1)^n \zeta(n)}{n-1}=\sum_{k=1}^{\infty} \frac{\ln (1+\frac{1}{k})}{k}=\int_{1}^{\infty} \frac{\ln ([x]+1)}{x^2}dx=$$

$$=\frac{\pi^2}{4}-1-4\int_{0}^{\infty} \frac{\arctan x}{1+x^{2}} \frac{dx}{e^{\pi x}+1}=\int^{\infty}_0 \frac{\log x+\Gamma (0,x) + \gamma}{e^x-1}~dx$$

Here $[]$ is the floor function, $\Gamma(0,x)$ is the incomplete Gamma function.

This value starts to become quite popular here. See this answer and the links therein.


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