Show that $\displaystyle{\int_0^1 \log \log \left(\frac{1}{x}\right) \frac{dx}{1+x^2} = \frac{\pi}{2}\log \left(\sqrt{2\pi} \Gamma\left(\frac{3}{4}\right) / \Gamma\left(\frac{1}{4}\right)\right)}$
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Solve the integral $S_k = (-1)^k \int_0^1 (\log(\sin \pi x))^k dx$
I cannot think of a change of variable nor other integrating methods. Maybe there is a known method that I am missing.
By the substitution $x = e^{-t}$, we find that
$$\begin{align*} \int_{0}^{1} \frac{(\log (1/x))^s}{1+x^2} \; dx &= \int_{0}^{\infty} \frac{t^s e^{-t}}{1 + e^{-2t}} \; dt \\ &= \int_{0}^{\infty} \sum_{n=0}^{\infty} (-1)^n t^s e^{-(2n+1)t} \; dt \\ &= \sum_{n=0}^{\infty} (-1)^n \, \frac{\Gamma(s+1)}{(2n+1)^{s+1}} \\ &= \Gamma(s+1)L(s+1, \chi_4), \end{align*}$$
where $L(s, \chi_4)$ is the Dirichlet L-function of the unique non-principal character $\chi_4$ to the modulus 4. Often it is denoted as $\beta(s)$ and called the Dirichlet beta function. Thus differentiating both sides with respect to $s$ and plugging $s = 0$, we obtain a representation formula
$$\int_{0}^{1} \frac{\log \log (1/x)}{1+x^2} \; dx = \psi_0(1) \beta(1) + \beta'(1),$$
and the problem reduces to find the value of $\beta(1)$ and $\beta'(1)$. Note that $\beta(1) = \frac{\pi}{4}$ is just the Gregory series. For $\beta'(1)$, we first notice that the following functional equation holds.
$$ \beta(s)=\left(\frac{\pi}{2}\right)^{s-1} \Gamma(1-s) \cos \left( \frac{\pi s}{2} \right)\,\beta(1-s). $$
This follows from the general theory of Dirichlet L-functions, and one can consult with any analytic number theory textbook to find its proof. Therefore it is sufficient to calculate $\beta'(0)$. For $0 < s$, we have
$$\begin{align*} -\beta'(s) &= \sum_{n=1}^{\infty} \left[ \frac{\log(4n+1)}{(4n+1)^s} - \frac{\log(4n-1)}{(4n-1)^s} \right] \\ &= \sum_{n=1}^{\infty} \frac{1}{(4n)^s} \left[ \log \left( \frac{4n+1}{4n-1} \right) - \frac{1}{2n} \right] + 2^{-2s-1}\zeta(s+1) \\ & \qquad + \sum_{n=1}^{\infty} \left( \frac{1}{(4n+1)^s} - \frac{1}{(4n)^s} \right) \log (4n+1) \\ & \qquad + \sum_{n=1}^{\infty} \left( \frac{1}{(4n)^s} - \frac{1}{(4n-1)^s} \right) \log (4n-1) \\ & =: A(s) + 2^{-2s-1}\zeta(s+1) + B(s) + C(s). \end{align*}$$
We first estimate $B(s)$. As $n \to \infty$, we have
$$ \log \left( \frac{4n}{4n+1} \right) = -\frac{1}{4n} + O\left( \frac{1}{n^2} \right), \quad \log \left( \frac{4n}{4n-1} \right) = \frac{1}{4n} + O\left( \frac{1}{n^2} \right). $$
Thus when $s \to 0$,
$$\begin{align*} B(s) &= \sum_{n=1}^{\infty} \frac{1}{(4n)^s} \left[ \exp\left( s \log \left( \frac{4n}{4n+1} \right) \right) - 1 \right] \left[ \log (4n) - \log \left(\frac{4n}{4n+1} \right) \right] \\ &= \sum_{n=1}^{\infty} \frac{1}{(4n)^s} \left[ - \frac{s}{4n} + O \left(\frac{s^2}{n^2} \right) \right] \left[ \log (4n) + O \left(\frac{1}{n} \right) \right] \\ &= -s 2^{-2s-2} \sum_{n=1}^{\infty} \frac{1}{n^{s+1}} \log (4n) + O(s) \\ &= s 2^{-2s-2} \left[ \zeta'(s+1) - \zeta(s+1) \log 4 \right] + O(s). \end{align*}$$
Similar consideration also shows that
$$ C(s) = s 2^{-2s-2} \left[ \zeta'(s+1) - \zeta(s+1) \log 4 \right] + O(s).$$
Thus we have
$$ 2^{-2s-1}\zeta(s+1) + B(s) + C(s) = 2^{-2s-1} \left[ \zeta(s+1) + s \zeta'(s+1) - s \zeta(s+1) \log 4 \right] + O(s). $$
But since
$$\zeta(1+s) = \frac{1}{s} + \gamma + O(s),$$
we have
$$ \lim_{s\downarrow 0} \left( 2^{-2s-1}\zeta(s+1) + B(s) + C(s) \right) = \frac{\gamma}{2} - \log 2.$$
For $A(s)$, the summands are positive with possible finite exceptional terms. Thus the Monotone Convergence Theorem guarantees that
$$ \lim_{s\downarrow 0} A(s) = \sum_{n=1}^{\infty} \left[ \log \left( \frac{4n+1}{4n-1} \right) - \frac{1}{2n} \right]. $$
Let $L$ denote this limit. Then by Stirling's formula,
$$\begin{align*} e^{L} & \stackrel{N\to\infty}{\sim} \prod_{n=1}^{N} \left( \frac{4n+1}{4n-1} \right) e^{-1/2n} \sim \frac{e^{-\gamma/2}}{\sqrt{N}} \prod_{n=1}^{N} \left( \frac{n+(1/4)}{n-(1/4)} \right) \\ & \sim \frac{e^{-\gamma/2}}{\sqrt{N}} \frac{\Gamma\left(\frac{3}{4}\right)}{\Gamma\left(\frac{5}{4}\right)} \frac{\Gamma\left(N+\frac{5}{4}\right)}{\Gamma\left(N+\frac{3}{4}\right)} \sim \frac{e^{-\gamma/2}}{\sqrt{N}} \frac{\Gamma\left(\frac{3}{4}\right)}{\Gamma\left(\frac{5}{4}\right)} \frac{\left( \frac{N + (5/4)}{e} \right)^{N+\frac{5}{4}}}{\left( \frac{N + (3/4)}{e} \right)^{N+\frac{3}{4}}} \\ & \sim e^{-\gamma/2} \frac{\Gamma\left(\frac{3}{4}\right)}{\Gamma\left(\frac{5}{4}\right)} = 4 e^{-\gamma/2} \frac{\pi \sqrt{2}}{\Gamma\left(\frac{1}{4}\right)^2}, \end{align*}$$
where we have used the Euler's reflection formula in the last line. Combining all the efforts, we obtain
$$-\beta'(0) = \log (2 \pi \sqrt{2}) - 2 \log \Gamma\left(\frac{1}{4}\right) .$$
Now taking logarithmic differntiation to the functional equation, we have
$$ \frac{\beta'(s)}{\beta(s)} = \log\left(\frac{\pi}{2}\right) - \psi_0 (1-s) - \frac{\pi}{2} \tan \left( \frac{\pi s}{2} \right) - \frac{\beta'(1-s)}{\beta(1-s)}. $$
Taking $s = 0$, we have
$$ \frac{\beta'(0)}{\beta(0)} = \log\left(\frac{\pi}{2}\right) + \gamma - \frac{\beta'(1)}{\beta(1)} \quad \Longrightarrow \quad \beta'(1) = \beta(1) \left[ \log\left(\frac{\pi}{2}\right) + \gamma - \frac{\beta'(0)}{\beta(0)} \right]. $$
But again by the functional equation, we have $\beta(0) = \frac{1}{2}$. Therefore
$$ \beta'(1) = \frac{\pi}{4} \left[ \gamma + 2 \log 2 + 3 \log \pi - 4 \log \Gamma\left(\frac{1}{4}\right) \right], $$
and hence
$$ \int_{0}^{1} \frac{\log \log (1/x)}{1+x^2} \; dx = \frac{\pi}{4} \left[ 2 \log 2 + 3 \log \pi - 4 \log \Gamma\left(\frac{1}{4}\right) \right], $$
which is identical to the proposed answer.
$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle #1 \right\rangle} \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace} \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left( #1 \right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $\ds{\int_{0}^{1}\ln\pars{\ln\pars{1 \over x}}\,{\dd x \over 1 + x^{2}} ={\pi \over 2}\,\ln\pars{\root{\vphantom{\large A}2\pi}\, {\Gamma\pars{3/4} \over \Gamma\pars{1/4}}}:\ {\Large ?}}$
With $\ds{x \to 1/x}$: $$ \color{#f00}{\int_{0}^{1}\ln\pars{\ln\pars{1 \over x}}\,{\dd x \over 1 + x^{2}}}= \int_{\infty}^{1}\ln\pars{\ln\pars{x}}\,\pars{-\,{\dd x/x^{2} \over 1 + 1/x^{2}}} =\int_{1}^{\infty}{\ln\pars{\ln\pars{x}} \over 1 + x^{2}}\,\dd x $$
With $x \equiv \expo{t}\quad\iff\quad t = \ln\pars{x}$: \begin{align} &\color{#f00}{\int_{0}^{1}\ln\pars{\ln\pars{1 \over x}}\,{\dd x \over 1 + x^{2}}}= \int_{0}^{\infty}{\ln\pars{t} \over 1 + \expo{2t}}\,\expo{t}\dd t =\int_{0}^{\infty}\ln\pars{t}\expo{-t}\,{1 \over 1 + \expo{-2t}}\,\dd t \\[3mm]&=\int_{0}^{\infty}\ln\pars{t}\expo{-t}\, \sum_{\ell = 0}^{\infty}\pars{-1}^{\ell}\expo{-2\ell t}\,\dd t =\sum_{\ell = 0}^{\infty}\pars{-1}^{\ell}\lim_{\mu \to 0}\partiald{}{\mu}\bracks{% \int_{0}^{\infty}t^{\mu}\expo{-\pars{2\ell + 1}t}\,\dd t} \\[3mm]&=\sum_{\ell = 0}^{\infty}\pars{-1}^{\ell}\lim_{\mu \to 0}\partiald{}{\mu}\bracks{% {1 \over \pars{2\ell + 1}^{\mu + 1}}\ \overbrace{\int_{0}^{\infty}t^{\mu}\expo{-t}\,\dd t}^{\ds{\Gamma\pars{\mu + 1}}}\ } \end{align} where $\Gamma\pars{z}$ is the Gamma Function.
\begin{align} &\color{#f00}{\int_{0}^{1}\ln\pars{\ln\pars{1 \over x}}\,{\dd x \over 1 + x^{2}}}= \sum_{\ell = 0}^{\infty}\pars{-1}^{\ell}\lim_{\mu \to 0}\partiald{}{\mu}\bracks{% {\Gamma\pars{\mu + 1} \over \pars{2\ell + 1}^{\mu + 1}}} \\[3mm]&= \lim_{\mu \to 0}\partiald{}{\mu}\bracks{% \Gamma\pars{\mu + 1}\sum_{\ell = 0}^{\infty} {\pars{-1}^{\ell}\over \pars{2\ell + 1}^{\mu + 1}}} \\[3mm]&= \lim_{\mu \to 0}\braces{% \Gamma'\pars{\mu + 1}\sum_{\ell = 0}^{\infty} {\pars{-1}^{\ell}\over \pars{2\ell + 1}^{\mu + 1}} +\Gamma\pars{\mu + 1}\partiald{}{\mu}\bracks{% \sum_{\ell = 0}^{\infty}{\pars{-1}^{\ell}\over \pars{2\ell + 1}^{\mu + 1}}}} \\[3mm]&=-\gamma\sum_{\ell = 0}^{\infty}{\pars{-1}^{\ell}\over 2\ell + 1} + \lim_{\mu \to 0}\partiald{}{\mu} \sum_{\ell = 0}^{\infty}{\pars{-1}^{\ell}\over \pars{2\ell + 1}^{\mu + 1}}\tag{1} \end{align} In this result, we used $\Psi\pars{1} = -\gamma$ and $\Gamma\pars{1} = 1$ where $\Psi\pars{z} \equiv \dd\ln\Gamma\pars{z}/\dd z$ is the Digamma Function and $\gamma$ is the Euler-Mascheroni constant.
The first $\ell$-sum in the right member of $\pars{1}$ is given by: \begin{align} &\sum_{\ell = 0}{\pars{-}^{\ell} \over 2\ell + 1}= \sum_{\ell = 0}\pars{{1 \over 4\ell + 1} - {1 \over 4\ell + 3}} ={1 \over 8}\sum_{\ell = 0}{1 \over \pars{\ell + 1/4}\pars{\ell + 3/4}} \\[3mm]&=-\,{1 \over 4}\bracks{\Psi\pars{1 \over 4} - \Psi\pars{3 \over 4}} = {\pi \over 4} \end{align} where we used the identities: \begin{align} \sum_{\ell = 0}^{\infty}{1 \over \pars{\ell + z_{0}}\pars{\ell + z_{1}}} &={\Psi\pars{z_{0}} - \Psi\pars{z_{1}} \over z_{0} - z_{1}}\tag{1.1} \\[3mm]\Psi\pars{z} - \Psi\pars{1 - z} &= -\pi\cot\pars{\pi z}\tag{1.2} \end{align} $$ \mbox{Then,}\quad \color{#f00}{\int_{0}^{1}\ln\pars{\ln\pars{1 \over x}}\,{\dd x \over 1 + x^{2}}}= -\,{1 \over 4}\,\gamma\pi + \lim_{\mu \to 0}\partiald{}{\mu} \sum_{\ell = 0}^{\infty}{\pars{-1}^{\ell}\over \pars{2\ell + 1}^{\mu + 1}}\tag{2} $$ Also, \begin{align} \sum_{\ell = 0}^{\infty}{\pars{-1}^{\ell}\over \pars{2\ell + 1}^{\mu + 1}} &=\sum_{\ell = 0}^{\infty}{1 \over \bracks{2\pars{2\ell} + 1}^{\mu + 1}} -\sum_{\ell = 0}^{\infty}{1 \over \bracks{2\pars{2\ell + 1} + 1}^{\mu + 1}} \\[3mm]&=2^{-2\mu - 2}\bracks{% \sum_{\ell = 0}^{\infty}{1 \over \pars{\ell + 1/4}^{\mu + 1}} -\sum_{\ell = 0}^{\infty}{1 \over \pars{\ell + 3/4}^{\mu + 1}}} \\[3mm]&=2^{-2\mu - 2}\bracks{% \zeta\pars{\mu + 1,{1 \over 4}} - \zeta\pars{\mu + 1,{3 \over 4}}} \end{align} where $\ds{\zeta\pars{s,q} \equiv \sum_{n = 0}^{\infty}{1 \over \pars{q + n}^{s}}}$ with $\Re\pars{s} > 1$ and $\Re\pars{q} > 0$ i s the Hurwitz Zeta Function or/and Generalizated Zeta Funcion .
So, \begin{align} &\lim_{\mu \to 0}\partiald{}{\mu}\sum_{\ell = 0}^{\infty}{\pars{-1}^{\ell}\over \pars{2\ell + 1}^{\mu + 1}} \\[3mm]&=-\,{1 \over 4}\,\ln\pars{2}\ \underbrace{\overbrace{\sum_{\ell = 0}^{\infty}{1 \over \pars{\ell + 3/4}\pars{\ell + 1/4}}} ^{\ds{2\bracks{\Psi\pars{3/4} - \Psi\pars{1/4}} = 2\pi}}} _{\ds{\mbox{See}\ \pars{1.1}\ \mbox{and}\ \pars{1.2}}} +\ {1 \over 4}\ \overbrace{\partiald{}{\mu}\bracks{% \zeta\pars{\mu,{1 \over 4}} - \zeta\pars{\mu,{3 \over 4}}}_{\mu\ =\ 1}} ^{\ds{-\gamma_{1}\pars{1/4} + \gamma_{1}\pars{3/4}}} \end{align} where $\gamma_{n}\pars{z}$ is a Generalizated Stieltjes Constant .
With this result, $\pars{2}$ is reduced to: \begin{align} \color{#f00}{\int_{0}^{1}\ln\pars{\ln\pars{1 \over x}}\,{\dd x \over 1 + x^{2}}} &=-\,{1 \over 4}\,\braces{% \pi\bracks{% \gamma + 2\ln\pars{2}} + \gamma_{1}\pars{1 \over 4} - \gamma_{1}\pars{3 \over 4}} \tag{3} \end{align} The difference $\gamma_{1}\pars{1/4} - \gamma_{1}\pars{3/4}$ is evaluated with the 1846 Carl Malmsten identity : $$ \gamma_{1}\pars{m \over n} - \gamma_{1}\pars{1 - {m \over n}} =-\pi\bracks{\gamma + \ln\pars{2\pi n}}\cot\pars{m\pi \over n} + 2\pi\sum_{\ell = 1}^{n - 1} \sin\pars{{2\pi m \over n}\,\ell}\ln\pars{\Gamma\pars{\ell \over n}} $$
With $m = 1$ and $n = 4$: \begin{align} &\gamma_{1}\pars{1 \over 4} - \gamma_{1}\pars{3 \over 4} \\[3mm]&=-\pi\bracks{\gamma + \ln\pars{8\pi}}\cot\pars{\pi \over 4} + 2\pi\sum_{\ell = 1}^{3}\sin\pars{\pi\,\ell \over 2} \ln\pars{\Gamma\pars{\ell \over 4}} \\[3mm]&=-\pi\bracks{\gamma + 2\ln\pars{2} + \ln\pars{2\pi}} \\[3mm]&\phantom{=} + 2\pi\bracks{\sin\pars{\pi \over 2}\ln\pars{\Gamma\pars{1 \over 4}} + \sin\pars{\pi}\ln\pars{\Gamma\pars{1 \over 2}} + \sin\pars{3\pi \over 2}\Gamma\pars{3 \over 4}} \\[3mm]&=-\pi\bracks{\gamma + 2\ln\pars{2} + \ln\pars{2\pi}} +2\pi\bracks{\ln\pars{\Gamma\pars{1 \over 4}} - \ln\pars{\Gamma\pars{3 \over 4}}} \\[3mm]&=-\pi\bracks{\gamma + 2\ln\pars{2}} - 2\pi\bracks{% \ln\pars{\root{2\pi}} +\ln\pars{\Gamma\pars{3 \over 4} \over \Gamma\pars{1 \over 4}}} \\[3mm]&=-\pi\bracks{\gamma + 2\ln\pars{2}} -2\pi\ln\pars{\root{2\pi}\,{\Gamma\pars{3 \over 4} \over \Gamma\pars{1 \over 4}}} \end{align}
By replacing this result in $\pars{3}$, we find: $$\color{#00f}{\large% \int_{0}^{1}\ln\pars{\ln\pars{1 \over x}}\,{\dd x \over 1 + x^{2}} ={\pi \over 2}\,\ln\pars{\root{\vphantom{\large A}2\pi}\, {\Gamma\pars{3/4} \over \Gamma\pars{1/4}}}} $$
As an 'extra-bonus' we can use the identity $\ds{\Gamma\pars{z} = {\pi \over \Gamma\pars{1 - z}\sin\pars{\pi z}}}$ to 'kill' one of the $\Gamma\,$'s functions: $\ds{\Gamma\pars{1 \over 4} = {\root{2}\pi \over \Gamma\pars{3/4}}}$ which yields: $$ \int_{0}^{1}\ln\pars{\ln\pars{1 \over x}}\,{\dd x \over 1 + x^{2}} ={\pi \over 2}\,\ln\pars{\Gamma^{\,2}\pars{3/4} \over \root{\pi}} $$
See also V. Adamchik's formula $$\int_0^1 \frac{x^{p-1}}{1+x^n}\log \log \frac{1}{x}dx = \frac{\gamma+\log(2n)}{2n}(\psi(\frac{p}{2n})-\psi(\frac{n+p}{2n}))+\frac{1}{2n}(\zeta'(1,\frac{p}{2n})-\zeta'(1,\frac{n+p}{2n}))$$ in http://dx.doi.org/10.1145/258726.258736 .
The above integral was long-time known as the Vardi's integral. Quite recently, an interesting research work devoted to this integral was published by Iaroslav Blagouchine. It appears that this integral was first evaluated by Carl Malmsten in 1842 (and not by Ilan Vardi in 1988). Blagouchine describes two different methods for its evaluation: the original Malmsten's method and the contour integration method. In the same paper, numerous integrals similar to the above one are also treated, and it is shown that many of them may be evaluated by the contour integration methods.
More recent paper by Blagouchine (mentioned above in the context of Malmsten's formula derived in 1846) is also integersting, but it is less directly related to the integral in question than the first one.
Let $y=\ln \frac{1}{x}$, then $x= e^{-y}$ and $dx=-e^{-y}dy$. $\displaystyle \begin{aligned}I & =\int_0^{\infty} \frac{\ln y}{1+e^{-2 y}} \cdot e^{-y} d y\\& =\sum_{n=0}^{\infty}(-1)^n \int_0^{\infty} e^{-(2 n+1) y} \ln y d y\\&=\left.\sum_{n=0}^{\infty}(-1)^n \frac{d}{d a}(I(a))\right|_{a=0}\end{aligned}\tag*{} $
where $\displaystyle \begin{aligned}I(a) & =\int_0^{\infty} y^a e^{-(2 n+1) y} d y \\& =\frac{\Gamma(a+1)}{(2 n+1)^{a+1}}\end{aligned}\tag*{} $ Differentiating $I(a)$ w.r.t. $a$ yields
$\displaystyle \begin{aligned}I^{\prime}(a) & =\frac{(2 n+1)^a \Gamma’(a+1)-\Gamma(a+1) \ln (2 n+1)(2 n+1)^a}{(2 n+1)(2 n+1)^{2 a}} \\& =\frac{\Gamma^{\prime}(a+1)-\Gamma(a+1) \ln (2 n+1)}{(2 n+1)^{a+1}}\end{aligned}\tag*{} $
Putting $a=0$ yields
$\displaystyle \begin{aligned}\int_0^{\infty} e^{-(2 n+1) y} \ln y d y &=I’(0) \\& =\frac{\Gamma^{\prime}(1)-\Gamma(1) \ln (2 n+1)}{2 n+1} \\& =-\frac{\gamma+\ln (2 n+1)}{2n+1}\end{aligned}\tag*{} $ $\displaystyle \begin{aligned}I & =\sum_{n=0}^{\infty}(-1)^{n+1} \frac{\gamma+\ln (2 n+1)}{2 n+1} \\& =\gamma \sum_{n=0}^{\infty} \frac{(-1)^{n+1}}{2 n+1}+\sum_{n=0}^{\infty} \frac{(-1)^{n+1} \ln (2 n+1)}{(2 n+1)^{a+1}}\end{aligned}\tag*{} $
Using the results $ \sum_{n=0}^{\infty} \frac{(-1)^{n+1}}{2 n+1}=-\frac{\pi}{4}$ and the result contributed by Carl Johan Malmsten - Wikipedia (https://en.wikipedia.org/wiki/Carl_Johan_Malmsten) in $1842$ that
$\displaystyle \sum_{n=0}^{\infty}(-1)^n \frac{\ln (2 n+1)}{2 n+1}=\frac{\pi}{4}(\ln \pi-\gamma)-\pi \ln \Gamma\left(\frac{3}{4}\right),\tag*{} $
we have $\displaystyle \begin{aligned}I & =\sum_{n=0}^{\infty}(-1)^{n+1} \frac{\gamma+\ln (2 n+1)}{2 n+1} \\& =\gamma \sum_{n=0}^{\infty} \frac{(-1)^{n+1}}{2 n+1}+\sum_{n=0}^{\infty} \frac{(-1)^{n+1} \ln (2 n+1)}{2 n+1} \\& =-\frac{\pi \gamma}{4}+\frac{\pi}{4}(\gamma-\ln \pi)+\pi \ln \left(\Gamma\left(\frac{3}{4}\right)\right) \\& =\frac{\pi}{2}\left[2 \ln \left(\Gamma\left(\frac{3}{4}\right)\right)-\ln \sqrt{\pi}\right] \\& =\frac{\pi}{2}\left[\ln \left(\frac{\Gamma^2\left(\frac{3}{4}\right)}{\sqrt \pi}\right)\right] \\& =\frac{\pi}{2}\left[\ln \left(\frac{\Gamma\left(\frac{3}{4}\right) \Gamma\left(\frac{1}{4}\right) \Gamma\left(\frac{3}{4}\right)}{\sqrt{\pi} \Gamma\left(\frac{1}{4}\right)}\right)\right]\cdots (*) \\& =\frac{\pi}{2} \ln \left(\frac{\sqrt{2 \pi} \Gamma\left(\frac{3}{4}\right)}{\Gamma\left(\frac{1}{4}\right)}\right)\end{aligned}\tag*{} $
where $(*)$ using the reflection property of Gamma function: $\Gamma(x) \Gamma(1-x)=\pi \csc (\pi x)$ for $x\not\in Z$.
Fourier Series of the Log-Gamma Function:
For $s \in(0,1)$, we have $$ \log \Gamma(s)=\left(\frac{1}{2}-s\right)(\gamma+\log 2)-\frac{1}{2} \log (\sin (\pi s))+(1-s) \log (\pi)+\frac{1}{\pi} \sum_{n=1}^{\infty} \frac{\sin (2 \pi n s) \log n}{n} $$
Proof: It suffices to do the following integrals: $$ \begin{aligned} \int_0^1 \log \Gamma(s) d s & =\frac{1}{2} \log (2 \pi) \\ \int_0^1 \log \Gamma(s) \cos (2 \pi n s) d s & =\frac{1}{4 n} \quad \forall n \in \mathbb{Z}^{+} \\ \int_0^1 \log \Gamma(s) \sin (2 \pi n s) d s & =\frac{\gamma+\log (2 n \pi)}{2 n \pi} \quad \forall n \in \mathbb{Z}^{+} \end{aligned} $$
The first integral can be evaluated with the help of Euler's reflection formula: $$ \begin{aligned} \int_0^1 \log \Gamma(s) d s & =\frac{1}{2} \int_0^1 \log \Gamma(s) d s+\frac{1}{2} \int_0^1 \log \Gamma(1-s) d s \\ & =\frac{1}{2} \int_0^1 \log \left(\frac{\pi}{\sin (\pi s)}\right) d s \\ & =\frac{\log (\pi)}{2}-\frac{1}{2} \int_0^1 \log (\sin (\pi s)) d s \\ & =\frac{\log (2 \pi)}{2} \end{aligned} $$
The second integral can also be dealt in a similar way. We have: $$ \begin{aligned} \int_0^1 \log \Gamma(s) \cos (2 \pi n s) d s & =\frac{1}{2} \int_0^1 \log \Gamma(s) \cos (2 \pi n s) d s+\frac{1}{2} \int_0^1 \log \Gamma(1-s) \cos (2 \pi n(1-s)) d s \\ & =\frac{1}{2} \int_0^1 \log \left(\frac{\pi}{\sin (\pi s)}\right) \cos (2 \pi n s) d s \\ & =-\frac{1}{2} \int_0^1 \log (2 \sin (\pi s)) \cos (2 \pi n s) d s \\ & =\frac{1}{2} \int_0^1\left(\sum_{k=1}^{\infty} \frac{\cos (2 \pi k s)}{k}\right) \cos (2 \pi n s) d s \\ & =\frac{1}{2} \sum_{k=1}^{\infty} \frac{1}{k} \int_0^1 \cos (2 \pi k s) \cos (2 \pi n s) d s \\ & =\frac{1}{4 n} \end{aligned} $$
In the above calculation, we used the well known Fourier series: $$ \log (2 \sin (\pi s))=-\sum_{k=1}^{\infty} \frac{\cos (2 \pi k s)}{k} \quad \forall s \in(0,1) $$
The third integral is the most troublesome of all since the trick involving Euler's reflection formula does not work. We will instead use the following integral representation of the LogGamma function: $$ \log \Gamma(s)=\int_0^{\infty}\left(\frac{s-1}{t e^t}-\frac{1-e^{t(1-s)}}{t\left(e^t-1\right)}\right) d t \quad \forall s>0 $$
One can easily verify the above equation via the differentiation under the integral technique. Therefore, upon changing the order of integration we get: $$ \begin{aligned} \int_0^1 \log \Gamma(s) \sin (2 \pi n s) d s & =\int_0^{\infty}\left(\frac{1}{t e^t} \int_0^1 s \sin (2 \pi n s) d s+\frac{1}{t\left(e^t-1\right)} \int_0^1 e^{t(1-s)} \sin (2 \pi n s) d s\right) d t \\ & =\int_0^{\infty}\left[\frac{1}{t e^t}\left(-\frac{1}{2 \pi n}\right)+\frac{1}{t\left(e^t-1\right)}\left(\frac{2 \pi n\left(e^t-1\right)}{t^2+(2 \pi n)^2}\right)\right] d t \\ & =\int_0^{\infty}\left(-\frac{1}{2 \pi n t e^t}+\frac{2 \pi n}{t\left(t^2+(2 \pi n)^2\right)}\right) d t \\ & =\frac{1}{2 \pi n} \int_0^{\infty}\left(-\frac{1}{t e^t}+\frac{1}{t}-\frac{t}{t^2+(2 \pi n)^2}\right) d t \\ & =\frac{1}{2 \pi n}\left(\gamma+\left[\log t-\log (t) e^{-t}-\frac{1}{2} \log \left(t^2+(2 \pi n)^2\right)\right]_{t=0}^{\infty}\right) \\ & =\frac{\gamma+\log (2 \pi n)}{2 \pi n} \end{aligned} $$
To get equation (1), one needs to piece together all these calculations. Equation (2) along with the result: $$ \pi\left(\frac{1}{2}-s\right)=\sum_{k=1}^{\infty} \frac{\sin (2 \pi k s)}{k} \quad \forall s \in(0,1) $$ are needed to perform simplifications.
Now, we turn our attention to the Vardi's integral: $$ \begin{aligned} I & =\int_0^1 \frac{\log \log \left(\frac{1}{x}\right)}{1+x^2} d x \\ & =\int_0^{\infty} \frac{\log (t) e^{-t}}{1+e^{-2 t}} d t \\ & =\int_0^{\infty} \log (t) \sum_{n=0}^{\infty}(-1)^n e^{-(2 n+1) t} d t \\ & =\sum_{n=0}^{\infty}(-1)^n \int_0^{\infty} \log (t) e^{-(2 n+1) t} d t \\ & =-\sum_{n=0}^{\infty}(-1)^n\left(\frac{\log (2 n+1)}{2 n+1}+\frac{\gamma}{2 n+1}\right) \\ & =-\frac{\gamma \pi}{4}-\sum_{n=1}^{\infty} \frac{\sin \left(\frac{\pi n}{2}\right) \log n}{n} \end{aligned} $$
To get the value of the sum, plug $s=\frac{1}{4}$ in equation (1). $$ \sum_{n=1}^{\infty} \frac{\sin \left(\frac{\pi n}{2}\right) \log n}{n}=\pi \log \Gamma\left(\frac{1}{4}\right)-\frac{\gamma \pi}{4}-\frac{\pi \log 2}{2}-\frac{3 \pi}{4} \log (\pi) $$
Substituting the above into equation (3) and performing some simplifications gives the desired result.
