A closed form for $\int_{0}^{\pi/2}\frac{\ln\cos x}{x}\mathrm{d}x$?

The following integrals are classic, initiated by L. Euler.

\begin{align} \displaystyle \int_{0}^{\pi/2} x^3 \ln\cos x\:\mathrm{d}x & = -\frac{\pi^4}{64} \ln 2-\frac{3\pi^2}{16} \zeta(3)+\frac{93}{128} \zeta(5), \\ \int_{0}^{\pi/2} x^2 \ln\cos x\:\mathrm{d}x & = -\frac{\pi^3}{24} \ln 2-\frac{\pi}{4} \zeta(3), \\ \int_{0}^{\pi/2} x^1 \ln\cos x\:\mathrm{d}x & = -\frac{\pi^2}{8} \ln 2-\frac{7}{16} \zeta(3), \\ \int_{0}^{\pi/2} x^0 \ln\cos x\:\mathrm{d}x & = -\frac{\pi}{2}\ln 2. \end{align}

We may logically consider the case when the first factor of the integrand is $\displaystyle x^{-1} = \frac 1x $ leading to the following non classic convergent integral.

$$ \int_{0}^{\pi/2} \frac{\ln\cos x}{x}\:\mathrm{d}x \qquad (*)$$

I do not have a closed form for this integral.

My question is does someone have some references/results about $(*)$?

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \begin{align}&\color{#66f}{\large% \int_{0}^{\pi/2}{\ln\pars{\cos\pars{x}} \over x}\,\dd x} =\int_{0}^{\pi/2}{\ln\pars{\sin\pars{x}} \over \pi/2 - x}\,\dd x \end{align}

We'll use the expansion

$$ \ln\pars{\sin\pars{x}} =-\ln\pars{2} - \sum_{k\ = 1}^{\infty}{\cos\pars{2kx} \over k} =\sum_{k\ = 1}^{\infty}{\pars{-1}^{k} - \cos\pars{2kx} \over k} $$

Then,

\begin{align}&\color{#66f}{\large% \int_{0}^{\pi/2}{\ln\pars{\cos\pars{x}} \over x}\,\dd x} =\sum_{k\ =\ 1}^{\infty}{1 \over k} \int_{0}^{\pi/2}{\pars{-1}^{k} - \cos\pars{2kx} \over \pi/2 - x}\,\dd x \\[5mm]&=\sum_{k\ =\ 1}^{\infty}{1 \over k} \int_{0}^{\pi/2}{\pars{-1}^{k} - \cos\pars{k\pi - 2kx} \over x}\,\dd x =\sum_{k\ =\ 1}^{\infty}{\pars{-1}^{k} \over k} \int_{0}^{\pi/2}{1 - \cos\pars{2kx} \over x}\,\dd x \end{align}

The integral in the RHS is given by

$$ \int_{0}^{\pi/2}{1 - \cos\pars{2kx} \over x}\,\dd x =\gamma + \ln\pars{\pi} + \ln\pars{k} - \,{\rm Ci}\pars{\pi k} $$

where $\ds{\gamma}$ is the Euler-Mascheroni Constant and $\ds{\,{\rm Ci}}$ is the Cosine Integral function. Moreover, $\ds{\sum_{k\ =\ 1}^{\infty}\pars{-1}^{k}\,{\ln\pars{k} \over k} =\gamma\ln\pars{2} - \half\,\ln^{2}\pars{2}}$ such that

\begin{align}&\color{#66f}{\large% \int_{0}^{\pi/2}{\ln\pars{\cos\pars{x}} \over x}\,\dd x} \\[5mm]&=-\bracks{\gamma + \ln\pars{\pi}}\ln\pars{2} +\bracks{\gamma\ln\pars{2} - \half\,\ln^{2}\pars{2}} -\sum_{k\ =\ 1}^{\infty}\pars{-1}^{k}\,{\,{\rm Ci}\pars{\pi k} \over k} \\[5mm]&=\color{#66f}{\large% -\ln\pars{\pi}\ln\pars{2} - \half\,\ln^{2}\pars{2} -\sum_{k\ =\ 1}^{\infty}\pars{-1}^{k}\,{\,{\rm Ci}\pars{\pi k} \over k}} \end{align}

We have $$ I = \int_{0}^{\pi/4}\frac{\log(1-2\sin^2\theta)}{\theta}\,d\theta = \int_{0}^{\pi/4}\frac{\log(\sin(2\theta))}{\pi/4-\theta}\,d\theta,\tag{1}$$ hence by considering the Taylor series of $\log(1-x)$ we end with a series of CosIntegral values, not so appealing. An interesting approach may be to represent both $$\frac{\log(1-2\sin^2\theta)}{\sin\theta}\qquad\text{and}\qquad\frac{\sin\theta}{\theta}$$ or $$\frac{\log(1-2\sin^2\theta)}{\sin2\theta}\qquad\text{and}\qquad\frac{\sin2\theta}{\theta}$$ as Fouries sine series (or the first function as a Fourier sine series and the second one as a Fourier transform), then integrate their product through the orthogonality relations.

This may lead to a re-writing of $I$ as a well-known series.

Also notice that $(1)$ gives: $$ I = \sum_{j=0}^{+\infty}\left(\frac{2}{\pi}\right)^{j+1}\int_{0}^{\pi/2}x^j\log(\sin x)\,dx = \sum_{j=0}^{+\infty}\left(\frac{2}{\pi}\right)^{j+1}\int_{0}^{\pi/2}(\pi/2-x)^j\log(\cos x)\,dx,$$ hence $I$ can be written as a series of powers of $\frac{2}{\pi}$ times binomial coefficients times the values of the Euler integrals.

[Continues]