TITLE: A difficult integral about function $\ln x$ and $\ln\ln x$
QUESTION [6 upvotes]: Some days ago, I met a difficult integral $$\int_0^1 {\left( {1 + \ln x} \right)\ln \left( {1 + x} \right)\ln \ln \frac{1}{x} \,{\rm{d}} x} .$$
I considered
$$\int_0^1 {{x^n}\ln \ln \frac{1}{x}\,{\rm{d}} x}  = \int_0^\infty  {{e^{ - \left( {n + 1} \right)x}}\ln x\,{\rm{d}} x}  =  - \frac{{\gamma  + \ln \left( {n + 1} \right)}}{{n + 1}}.$$
But I have no idea to continue it! Could you show me how to compute it?

REPLY [6 votes]: To replace $x$ with $e^{-t}$ is a good idea even in the first place. That leads to:
$$ I = \int_{0}^{+\infty}e^{-t}(1-t)\log(1+e^{-t})\log t\,dt\tag{1}$$
but since:
$$ \int_{0}^{+\infty}e^{-(n+1)t}(1-t)\log t\,dt = -\frac{1}{(n+1)^2}\left(1+n\gamma+n\log (n+1)\right)\tag{2}$$
we have:
$$ I = \sum_{n\geq 1}\frac{(-1)^n}{n(n+1)^2}\left(1+n\gamma+n\log (n+1)\right)\tag{3}$$
so:
$$ I = 2-2\log 2-\gamma-\zeta(2)\left(\frac{1}{2}+\log\sqrt{4\pi}\right)+\pi^2\log A\tag{4}$$
where $A$ is the Glaisher-Kinkelin constant.