What is a closed form of $\int_0^1\ln(-\ln x)\ \text{li}\ x\ dx$

Question

Let $\operatorname{li} x$ denote the logarithmic integral: $$\operatorname{li} x=\int_0^x\frac{dt}{\ln t}.$$ Is it possible to find a closed form of the following integral? $$\int_0^1\ln(-\ln x) \operatorname{li} x\ dx$$ I tried integration by parts, but only ended up with more complicated expressions.

Answer 1

This integral can be expressed in a closed form: $$\int_0^1\ln(-\ln x)\operatorname{li}x\ dx=\frac{\pi^2}{12}+\frac{(\ln2)^2}2+\gamma\,\ln2,\tag1$$ where $\gamma$ is the Euler-Mascheroni constant.

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Proof:

Let $$\mathcal{I}(a)=\int_0^1(-\ln x)^a\operatorname{li}x\ dx,\tag2$$ then, taking a derivative with respect to $a$, we get $$\mathcal{I}'(0)=\int_0^1\ln(-\ln x)\operatorname{li}x\ dx.\tag3$$ Mathematica can express the integral $(2)$ in a closed form for explicitly given non-negative integer values of $a$, but not in a general form with a parameter. We could use those values to get a conjectured general formula, and indeed one could get the result $(1)$ this way. Numerical integration shows that $$\Bigg|\frac{\pi^2}{12}+\frac{(\ln2)^2}2+\gamma\,\ln2-\int_0^1\ln(-\ln x)\operatorname{li}x\ dx\Bigg|<10^{-1000}\tag4,$$ so the result looks plausible.

But here we will follow a different approach to get a rigorous proof (modulo possible bugs in Mathematica tuned exactly to hide that tiny difference). Take the formula $(12)$ from the MathWorld article: $$\operatorname{li}x=\operatorname{Ei}(\ln x),\tag5$$ where $\operatorname{Ei}z$ is the exponential integral, and plug it into $(2)$. In this form the integral can be evaluated in Mathematica, that gives us $$\mathcal{I}(a)=-\frac{_2F_1(a+1,a+1;a+2;-1)\ \Gamma(a+1)}{a+1},\tag6$$ and, taking a derivative with respect to $a$, $$\mathcal{I}'(0)=(1+\gamma)\ln2-\left[\frac{d}{da}{_2F_1}(a+1,a+1;a+2;-1)\right]_{a=0}.\tag7$$ To find the derivative of the hypergeometric function we need to replace it with its definition (formula $(8)$ in this MathWorld article): $$_2F_1(a+1,a+1;a+2;-1)=\sum_{n=0}^\infty\frac{(a+1)\ \Gamma(a+n+1)}{(a+n+1)\ \Gamma(a+1)\ n!}(-1)^n.\tag8$$ So, $$\left[\frac{d}{da}{_2F_1}(a+1,a+1;a+2;-1)\right]_{a=0}=\sum_{n=0}^\infty\frac{(n+1)(\gamma+\psi(n+1))+n}{(n+1)^2}(-1)^n,\tag9$$ where $\psi(n+1)=\frac{\Gamma'(n+1)}{\Gamma(n+1)}=-\gamma+\sum_{m=1}^n\frac1m\,$ is the digamma function. Evaluating this sum with Mathematica we get $$\left[\frac{d}{da}{_2F_1}(a+1,a+1;a+2;-1)\right]_{a=0}=\ln2-\frac{(\ln2)^2}2-\frac{\pi^2}{12}\tag{10}.$$ Plugging this back in $(7)$ and $(3)$ we get the final result $(1)$.