Log Log Integrals Evaluate the integrals
\begin{align}
I_{1} &= \int_{0}^{1} \ln^{2n}(x) \ \ln\left(\ln\left(\frac{1}{x}\right)\right) dx
\end{align}
and
\begin{align}
I_{2} = \int_{0}^{1} \ln^{2n}(x) \ \ln^{2}\left(\ln\left(\frac{1}{x}\right)\right) dx.
\end{align}
From the resulting values of the integrals above is it possible to evaluate the integral
\begin{align}
I_{3} = \int_{0}^{1} \left( x^{a} + \frac{1}{x^{a}} \right) \ \ln^{p}\left(\ln\left(\frac{1}{x}\right)\right) dx
\end{align}
where $p=1,2$ in a compact form? Please show all work in in the solutions. 

$I_j$ with $j = 1,2$ can be calculated as follows:
Substitute $u = -\ln x$, $\,dx = e^{-u} \,du$ to obtain 
\begin{align}
I_1 &=
\int_0^{\infty} x^{2n}e^{-x} \ln x \,dx = \partial_{\mu}|_0\int_0^{\infty} x^{2n+\mu}e^{-x}\,dx \\
&=\partial_{\mu}|_0 \,\Gamma(2n+\mu+1) \\
&= \Gamma(2n+1) \ \psi(2n+1)
\end{align}
and similarly, by differentiating twice before setting $\mu=0$, we obtain 
\begin{align}
I_{2} = \Gamma(2n+1) \ \left[\psi(2n+1)^2+\psi_1(2n+1)\right].
\end{align}
Now we will use these results (with $n = 0$) to prove some lemmas.
First notice that for $b>0$, we have
\begin{align}
\int_0^{\infty} e^{-bx}  \ln x \,dx &= \frac{1}{b}\int_0^{\infty} e^{-x}  (\ln x - \ln b)\,dx \\
&= - \frac{\ln b}{b}+ \frac{1}{b}\int_0^{\infty} x^0e^{-x}  \ln x \,dx \\ 
&= - \frac{\ln b}{b}+ \frac{1}{b}\psi(1)\Gamma(1) \\
&= - \frac{\ln b+\gamma}{b}
\end{align}
where I used the known value $\psi(1) = -\gamma$.
Also note that 
\begin{align}
\int_0^{\infty} e^{-bx}  \ln^2 x \,dx &= \frac{1}{b}\int_0^{\infty} e^{-x}  \left(\ln^2 x - 2 \ln b \ln x + \ln^2 b\right)\,dx \\
&= \frac{1}{b}\int_0^{\infty} e^{-x}  (\ln^2 x - 2 \ln b \ln x + \ln^2 b)\,dx \\
&= \frac{1}{b} \left(\pi^2/6+\gamma^2 + 2 \gamma \ln b + \ln^2 b \right) \\ 
&= \frac{\pi^2/6+(\gamma+\ln b)^2}{b}
\end{align}
where I used that $\psi_1(1) = \zeta(2) = \pi^2/6$.
Now substitute $u = -\ln x$ to obtain for $p = 1$
\begin{align}
\int_0^{1} \left(x^a + x^{-a} \right) \ln\left( \ln \frac{1}{x} \right) \,dx
&= \int_0^{\infty} \left(e^{-(1+a)x} + e^{-(1-a)x}  \right)  \ln x \,dx \\
&= \frac{2 \gamma}{1-a^{2}} - \frac{\ln (1-a)}{1-a}- \frac{\ln (1+a)}{1+a},
\end{align}
where $a \neq 1$, and for $p = 2$
\begin{align}
\int_0^{1} \left(x^a + x^{-a} \right) \ln^2\left( \ln \frac{1}{x} \right) \,dx &= \int_0^{\infty} \left(e^{-(1+a)x} + e^{-(1-a)x}  \right)  \ln^2 x \,dx \\
&= \frac{\pi^{2}}{3(1-a^{2})} + \frac{(\gamma+\ln (1-a))^2}{1-a} + \frac{(\gamma+\ln (1+a))^2}{1+a},
\end{align}
where $a \neq 1$. 
