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 $(*)$?

I've established some related explicit formulae.

Theorem 1. 
  Let $n$ be any positive integer. 
Set
  $$  I_{2n}:=\int_{0}^{\pi/2}\! \! x^{2n} \ln \cos x \: \mathrm{d}x $$
  Then
  $$
I_{2n}  = - \frac{\pi^{2n+1}\ln 2}{2^{2n+1}(2n+1)} - (-1)^{n}\frac{(2n)!}{2^{2n+1}}\sum_{p=1}^{n} \frac{(-1)^p}{(2p-1)!}\pi^{2p-1}\zeta(2n-2p+3) \tag1
$$ 
  Set
  $$  I_{2n+1}:=\int_{0}^{\pi/2}\! \! x^{2n+1} \ln \cos x \:\mathrm{d}x $$
  Then
  $$
\begin{align}
 I_{2n+1}=-  \frac{\pi^{2n+2}\ln 2}{2^{2n+2}(2n+2)} - (-1)^n\left(1-\frac{1}{2^{2n+2}}\right)\frac{(2n+1)!}{2^{2n+2}}\zeta(2n+3) \\\\ - (-1)^n\frac{(2n+1)!}{2^{2n+2}} \sum_{p=0}^{n} \frac{(-1)^p}{(2p)!}\pi^{2p}\zeta(2n-2p+3) \tag2
\end{align}
$$ 

[To be continued]
