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. 
 A: $I_j$ 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 $a \neq 1$. 
A: $\newcommand{\+}{^{\dagger}}
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$\ds{{\cal I}_{n}\pars{\mu}
    \equiv\int_{0}^{1}\ln^{2n}\pars{x}\ln^{\mu}\pars{\ln\pars{1 \over x}}\,\dd x:\
    {\large ?}.\qquad n,\mu \in {\mathbb N}}$.

With $\ds{x = \expo{-t}\quad\imp\quad t = -\ln\pars{x}}$:
  \begin{align}
{\cal I}_{n}\pars{\mu}&
=\int_{\infty}^{0}\ln^{2n}\pars{\expo{-t}}\ln^{\mu}\pars{\ln\pars{\expo{t}}}
\,\bracks{\expo{-t}\pars{-\dd t}}
=\int_{0}^{\infty}t^{2n}\ln^{\mu}\pars{t}\expo{-t}\,\dd t
\\[3mm]&=\lim_{\alpha \to 2n}\partiald[\mu]{}{\alpha}
\int_{0}^{\infty}t^{\alpha}\expo{-t}\,\dd t
=\lim_{\alpha \to 2n}\partiald[\mu]{\Gamma\pars{\alpha + 1}}{\alpha}
\end{align}
  where $\ds{\Gamma\pars{z}}$ is the Gamma Function.

\begin{align}
{\cal I}_{n}\pars{\mu}&
=\lim_{\alpha \to 2n}\left\lbrace%
\begin{array}{ll}
\Gamma\pars{\alpha + 1}\Psi\pars{\alpha + 1}\,, & \quad\mu = 1 
\\[2mm]
\Gamma\pars{\alpha + 1}
\bracks{\Psi^{2}\pars{\alpha + 1} + \Psi'\pars{\alpha + 1}}\,, & \quad\mu = 2 
\end{array}\right.
\end{align}
$\ds{\Psi\pars{z} \equiv \totald{\ln\pars{\Gamma\pars{z}}}{z}}$ is the Digamma Function.

\begin{align}
\color{#00f}{\large{\cal I}_{n}\pars{\mu}}&
=\color{#00f}{\large\left\lbrace%
\begin{array}{ll}
\pars{2n}!\,\Psi\pars{2n + 1}\,, & \quad\mu = 1 
\\[2mm]
\pars{2n}!\,\bracks{\Psi^{2}\pars{2n + 1} + \Psi'\pars{2n + 1}}\,, & \quad\mu = 2 
\end{array}\right.}
\end{align}
  where we used
  Recurrence Formula ${\bf\mbox{6.1.15}}$:
  $\ds{\Gamma\pars{m + 1} = m!\,,\quad m \in {\mathbb N}}$. 

$\color{#f00}{\ds{I_{3}\equiv\int_{0}^{1}\pars{x^{a} + {1 \over x^{a}}}
\ln^{p}\pars{\ln\pars{1 \over x}}\,\dd x:\ {\large ?}}}$
\begin{align}
I_{3}&=\int_{\infty}^{0}\pars{\expo{-at} + \expo{at}}
\ln^{p}\pars{\ln\pars{\expo{t}}}\,\pars{-\expo{-t}\,\dd t}
\\[3mm]&=\int_{0}^{\infty}\ln^{p}\pars{t}\expo{-\pars{1 + a}t}\,\dd t
\int_{0}^{\infty}\ln^{p}\pars{t}\expo{-\pars{1 - a}t}\,\dd t
\\[3mm]&=\lim_{\alpha \to 0}\partiald[p]{}{\alpha}\bracks{%
\int_{0}^{\infty}t^{\alpha}\expo{-\pars{1 + a}t}\,\dd t
\int_{0}^{\infty}t^{\alpha}\expo{-\pars{1 - a}t}\,\dd t}
\end{align}
Both integrals converge simultaneously when $\ds{\Re\pars{\alpha} > - 1}$ and
$\ds{\verts{\Re\pars{a}} < 1}$. In that case
\begin{align}
I_{3}&=\lim_{\alpha \to 0}\partiald[p]{}{\alpha}\bracks{%
\pars{1 + a}^{-\alpha - 1} + \pars{1 - a}^{-\alpha - 1}}\Gamma\pars{\alpha + 1}
\end{align}

$$\color{#00f}{\large%
\left.I_{3}\right\vert_{p\ =\ 1}
=-\,{\pars{1 - a}\ln\pars{1 + a} + \pars{1 + a}\ln\pars{1 - a} + 2\gamma
\over 1 - a^{2}}}\,,\quad \verts{\Re\pars{a}} < 1
$$

$$\color{#00f}{\large%
\left.I_{3}\right\vert_{p\ =\ 2} = {Num \over 3\pars{1 - a^{2}}}}
$$

\begin{align}
\color{#00f}{Num}&=\color{#00f}{3 (a+1) \ln^2(1-a)-3 (a-1) \ln^2(a+1)+6 \gamma  (a+1) \ln(1-a)}
\\[3mm]&\color{#00f}{-6 \gamma(a-1)\ln(a+1)+6 \gamma ^2+\pi^{2}}
\,,\quad \verts{\Re\pars{a}} < 1
\end{align}
  $\ds{\gamma}$ is the Euler-Mascheroni Constant ${\bf\mbox{6.1.3}}$.

