TITLE: What is a closed form of $\int_0^1\ln(-\ln x)\ \text{li}\ x\ dx$
QUESTION [27 upvotes]: 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.

REPLY [2 votes]: Solution by Khalef Ruhemi
From the definition 
$$\operatorname{li}(x)=\int_0^x\frac{\ dt}{\ln t}=-\int_0^x\int_0^\infty t^y\ dt\ dy=-\int_0^\infty\frac{x^{y+1}}{1+y}\ dy=-\int_0^\infty\frac{e^{-(1+y)\ln(1/x)}}{1+y}\ dy$$
where we used $$\frac{1}{\ln t}=-\int_0^\infty t^y\ dy, \quad t\in(0,1)$$
It follows that 
\begin{align}
I&=-\int_0^1\left(\ln\left(\ln(1/x)\right)\int_0^\infty\frac{e^{-(1+y)\ln(1/x)}}{1+y}\ dy\right)\ dx,\quad \ln(1/x)=t\\
&=-\int_0^\infty\left(\frac1{1+y}\int_0^\infty e^{-(2+y)t}\ln t \ dt\right)\ dy\\
&=\int_0^\infty\frac1{1+y}\left(\frac{\gamma}{2+y}+\frac{\ln(2+y)}{2+y}\right)\ dy,\quad \frac1{1+y}=x\tag{*}\\
&=\gamma\int_0^1\frac{\ dx}{1+x}+\int_0^1\frac{\ln(1+x)-\ln x}{1+x}\ dx\\
&=\gamma\ln2+\frac12\ln^22+\frac12\zeta(2)
\end{align}

Regarding step $(*)$, we know that $$\int_0^\infty t^p e^{-(y+2)t}\ dt=\Gamma(p+1)(2+y)^{-p-1}$$
Differentiate both sides with respect to $p$ to get 
$$\int_0^\infty t^p e^{-(y+2)t}\ln t\ dt=\Gamma'(p+1)(2+y)^{-p-1}-\Gamma(p+1)(2+y)^{-p-1}\ln(2+y)$$
Set $p=0$ we get
$$\int_0^\infty e^{-(2+y)t}\ln t\ dt=-\frac{\gamma}{2+y}-\frac{\ln(2+y)}{2+y}$$