TITLE: A closed form for $\int_{0}^{\pi/2}\frac{\ln\cos x}{x}\mathrm{d}x$?
QUESTION [17 upvotes]: 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 $(*)$?

REPLY [4 votes]: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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\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}