<s>[INST] Log Log Integrals II The integral
\begin{align}
I_{4} = \int_{0}^{1} \ln(1-x) \ \ln^{2}\left( \ln\left(\frac{1}{x}\right) \right) \ \frac{dx}{x}
\end{align}
can be expressed as
\begin{align}
I_{4} = \zeta^{''}(2) - \frac{\gamma^{2} \pi^{2}}{6} - \frac{\gamma \pi^{2}}{3} \ln\left( \frac{2 \pi}
{A^{12}} \right) + \frac{\pi^{4}}{36}
\end{align}
where $A$ is the Glaisher-Kinkelin constant.
The integrals 
\begin{align}
I_{5} = \int_{0}^{1} \frac{\ln(1-x)}{\sqrt{\ln\left(\frac{1}{x}\right)}} \ \ln^{2}\left( 
\ln \left(\frac{1}{x}\right) \right) \ \frac{dx}{x}
\end{align}
and
\begin{align}
I_{6} = \int_{0}^{1} \frac{\ln(1-x)}{\ln^{3/2}\left(\frac{1}{x}\right)} \ \ln^{2}\left( 
\ln \left(\frac{1}{x}\right) \right) \ \frac{dx}{x}
\end{align}
can be expressed in terms of $\partial_{s}^{2}\zeta(s)|_{s=3/2}$ and $\partial_{s}^{2}\zeta(s)|_{s=1/2}$, 
respectively. Can these integrals be evaluated in closed form expression without 
the use of derivatives of the Zeta function? 
If the integrals can be evaluated in such a way what is the resulting value?
 [/INST] For the first integral, \begin{align}
I_{5} &= \int_{0}^{1} \frac{\ln(1-x)}{\sqrt{\ln\left(\frac{1}{x}\right)}} \ \ln^{2}\left( 
\ln \left(\frac{1}{x}\right) \right) \ \frac{dx}{x}\\&=-\frac{\sqrt{\pi}}{2}(\pi^2+2\gamma+4\log2)^2\zeta\left(\frac32\right)+2\sqrt{\pi}(\gamma+2\log2)\zeta'\left(\frac32\right)-\sqrt{\pi}\zeta''\left(\frac32\right)
\end{align} there is only one $\zeta''$ term so it doesn't seem likely to be able to remove it.
The second integral does not converge. But you might be interested in $$\zeta'\left(\frac12\right)=\frac{1}{4}(\pi+2\gamma+2\log8\pi)\zeta\left(\frac12\right)\\\zeta''\left(\frac12\right)=\left(2G+\frac{5\pi^2+4\gamma\pi+4\gamma^2}{16}+\frac{\log8\pi}{4}(\log8\pi+\pi+2\gamma)\right)\zeta\left(\frac12\right)$$
</s>