TITLE: Integral $ \int_0^1 \frac{\ln \ln (1/x)}{1+x^{2p}} dx$...Definite Integral
QUESTION [6 upvotes]: Calculate
$$
I_1:=\int_0^1 \frac{\ln \ln (1/x)}{1+x^{2p}} dx, \ p \geq 1.
$$
I am trying to solve this integral $I_1$.  I know how to solve a related integral $I_2$
$$
I_2:=\int_0^1 \frac{\ln \ln (1/x)}{1+x^2} dx=\frac{\pi}{4}\bigg(2\ln 2 +3\ln \pi-4\ln\Gamma\big(\frac{1}{4}\big) \bigg)
$$
but I am not sure how to use that result here. In this case I just use the substitution $x=e^{-\xi}$ and than use a series expansion.  The result is
$$
I_2=\int_0^\infty \frac{\xi^s e^{-\xi}}{1+e^{-2\xi}} d\xi=\sum_{n=0}^\infty (-1)^n \frac{\Gamma(s+1)}{(2n+1)^{s+1}}=\Gamma(s+1)L(s+1,\chi_4)
$$
where L is the Dirichlet L-Function where $\chi_4$ is the unique non-principal character.  This result is further simplified but takes some work.  I am interested in the general case above, $I_1$ Thanks

REPLY [11 votes]: $\newcommand{\+}{^{\dagger}}
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$\ds{\int_{0}^{1}{\ln\pars{\ln\pars{1/x}} \over 1 + x^{2p}}\,\dd x\,,\qquad
     p \geq 1:\ {\large ?}}$

\begin{align}
&\int_{0}^{1}{\ln\pars{\ln\pars{1/x}} \over 1 + x^{2p}}\,\dd x
=\overbrace{\int_{\infty}^{1}{\ln\pars{\ln\pars{x}} \over 1 + x^{-2p}}
\,\pars{-\,{\dd x \over x^{2}}}}^{\ds{x\ \to\ {1 \over x}}}
=\int_{1}^{\infty}{\ln\pars{\ln\pars{x}}x^{-2} \over 1 + x^{-2p}}\,\dd x
\\[3mm]&=\underbrace{\int_{0}^{\infty}
{\ln\pars{t}\expo{-2t} \over 1 + \expo{-2pt}}\,\expo{t}\,\dd t}
_{\ds{x\ \equiv \expo{t}}}
=\lim_{\mu \to 0}\partiald{}{\mu}\int_{0}^{\infty}t^{\mu}\expo{-t}\,
{1 \over 1 + \expo{-2pt}}\,\dd t
\\[3mm]&=\lim_{\mu \to 0}\partiald{}{\mu}\int_{0}^{\infty}t^{\mu}\expo{-t}\,
\sum_{\ell = 0}^{\infty}\pars{-1}^{\ell}\expo{-2\ell pt}\,\dd t
=\lim_{\mu \to 0}\partiald{}{\mu}\sum_{\ell = 0}^{\infty}
\pars{-1}^{\ell}\int_{0}^{\infty}t^{\mu}\expo{-\pars{2\ell p + 1}t}\,\dd t
\\[3mm]&=\lim_{\mu \to 0}\partiald{}{\mu}\sum_{\ell = 0}^{\infty}
{\pars{-1}^{\ell} \over \pars{2\ell p + 1}^{\mu + 1}}
\underbrace{\int_{0}^{\infty}t^{\mu}\expo{-t}\,\dd t}_{\ds{=\ \Gamma\pars{\mu + 1}}}
\end{align}
  where $\ds{\Gamma\pars{z}}$ is the
  Gamma Function.

$$
\int_{0}^{1}{\ln\pars{\ln\pars{1/x}} \over 1 + x^{2p}}\,\dd x
=\lim_{\mu \to 0}\partiald{}{\mu}\bracks{\Gamma\pars{\mu + 1}%
\sum_{\ell = 0}^{\infty}{\pars{-1}^{\ell} \over \pars{2\ell p + 1}^{\mu + 1}}}
\tag{1}
$$

Let's reduce the $\ds{\ell}$-sum in the right hand side:
  \begin{align}
&\sum_{\ell = 0}^{\infty}{\pars{-1}^{\ell} \over \pars{2\ell p + 1}^{\mu + 1}}
=\sum_{\ell = 0}^{\infty}\braces{%
{1 \over \bracks{2\pars{2\ell}p + 1}^{\mu + 1}}-
{1 \over \bracks{2\pars{2\ell + 1}p + 1}^{\mu + 1}}}
\\[3mm]&={1 \over \pars{4p}^{\mu + 1}}\sum_{\ell = 0}^{\infty}\braces{%
{1 \over \bracks{\ell + 1/\pars{4p}}^{\mu + 1}}-
{1 \over \bracks{\ell + 1/2 + 1/\pars{4p}}^{\mu + 1}}}
\\[3mm]&={1 \over \pars{4p}^{\mu + 1}}\bracks{%
\zeta\pars{\mu + 1,{1 \over 4p}} - \zeta\pars{\mu + 1,\half + {1 \over 4p}}}
\end{align}
  where $\ds{\zeta\pars{s,q}}$ is the
  Generalizated Zeta Function or/and Hurwitz Zeta Function.

$\pars{1}$ is reduced to:
\begin{align}
&\!\!\!\!\!\int_{0}^{1}{\ln\pars{\ln\pars{1/x}} \over 1 + x^{2p}}\,\dd x
=\lim_{\mu \to 0}\partiald{}{\mu}\braces{%
{\Gamma\pars{\mu + 1} \over \pars{4p}^{\mu + 1}}\bracks{%
\zeta\pars{\mu + 1,{1 \over 4p}} - \zeta\pars{\mu + 1,\half + {1 \over 4p}}}}\tag{2}
\end{align}

Also ( see this page ):
  \begin{align}
&\!\!\!\!\!\!\!\!
\zeta\pars{\nu + 1,{1 \over 4p}} - \zeta\pars{\nu + 1,\half + {1 \over 4p}}
=
\sum_{n = 0}^{\infty}{\pars{-1}^{n} \over n!}\,
\bracks{\gamma_{n}\pars{1 \over 4p} - \gamma_{n}\pars{\half + {1 \over 4p}}}\nu^{n}
\tag{3}
\end{align}
  where $\ds{\gamma_{n}\pars{a}}$ is a
  Generalizated Stieltjes Constant.

With the expression $\pars{3}$, $\pars{2}$ is reduced to:
\begin{align}
&\int_{0}^{1}{\ln\pars{\ln\pars{1/x}} \over 1 + x^{2p}}\,\dd x
\\[3mm]&=\sum_{n = 0}^{\infty}{\pars{-1}^{n} \over n!}\,
\bracks{\gamma_{n}\pars{1 \over 4p} - \gamma_{n}\pars{\half + {1 \over 4p}}}\
\overbrace{\braces{\lim_{\mu \to 0}\partiald{}{\mu}\bracks{%
\Gamma\pars{\mu + 1}\mu^{n} \over \pars{4p}^{\mu + 1}}}}
^{\ds{-\bracks{\gamma + \ln\pars{4p}}\delta_{n,0} + \delta_{n,1} \over 4p}}
\\[3mm]&={\gamma + \ln\pars{4p} \over 4p}\,\bracks{%
\gamma_{0}\pars{\half + {1 \over 4p}} - \gamma_{0}\pars{1 \over 4p}}
+
{1 \over 4p}\,
\bracks{\gamma_{1}\pars{\half + {1 \over 4p}} - \gamma_{1}\pars{1 \over 4p}}
\end{align}
where $\ds{\gamma}$ is the
Euler-Mascheroni Constant.
According to the Blagouchine paper:
$\ds{\gamma_{0}\pars{v} = -\Psi\pars{v}}$ where $\ds{\Psi\pars{v}}$ is the
Digamma Function.

Finally, we arrive to this answer main result:
  \begin{align}
&\color{#00f}{\large\int_{0}^{1}{\ln\pars{\ln\pars{1/x}} \over 1 + x^{2p}}\,\dd x}
\\[3mm]&=\color{#00f}{{\gamma + \ln\pars{4p} \over 4p}\,\bracks{%
\Psi\pars{1 \over 4p} - \Psi\pars{\half + {1 \over 4p}}}
+
{1 \over 4p}\,
\bracks{\gamma_{1}\pars{\half + {1 \over 4p}} - \gamma_{1}\pars{1 \over 4p}}}
\end{align}
  The constants $\ds{\braces{\gamma_{1}\pars{a}}}$ can be calculated for rational values of $a$ by means of a rather cumbersome expression (see formula $\pars{26}$ in
  Blagouchine paper ). When $\ds{p = 1}$, the results is somehow simple since we can use formula $\pars{11}$ of
  Blagouchine paper which is valid when
  $\ds{{1 \over 4p} + \pars{\half + {1 \over 4p}} = 1}$.

ADDENDUM
Recently, the paper by Professor Blagouchine was published in Journal of Number Theory as he
told me via a comment. See the following link: 
A theorem for the closed-form evaluation of the first generalized Stieltjes constant at rational arguments and some related summations .