Ramanujan log-trigonometric integrals I discovered the following conjectured identity numerically while studying a family of related integrals.
Let's set
$$
R^{+}:=  \frac{2}{\pi}\int_{0}^{\pi/2}\sqrt[\normalsize{8}]{x^2 + \ln^2\!\cos x}
\sqrt{ \frac{1}{2}+\frac{1}{2}\sqrt{
\frac{1}{2}+ \frac{1}{2}  \sqrt{ \frac{\ln^{2}\!\cos x}{
x^2 + \ln^2\! \cos x}}}}\,\mathrm{d}x, \tag1
$$
$$
R^{-}:=  \frac{2}{\pi}\int_{0}^{\pi/2}\frac{1}{\sqrt[\normalsize{8}]{x^2 + \ln^2\!\cos x}}
\sqrt{ \frac{1}{2}+\frac{1}{2}\sqrt{
\frac{1}{2}+ \frac{1}{2} \sqrt{ \frac{\ln^{2}\!\cos x}{
x^2 + \ln^2\! \cos x}}}}\,\mathrm{d}x. \tag2
$$
We may numerically observe with at least 500 digits of precision that
$$ 
\begin{align}
& R^{+}R^{-} \stackrel{?}{=}1 \tag 3 \\\\
& R^{+} \stackrel{?}{=} \sqrt[\normalsize{4}]{\ln 2} \tag4 \\\\
& R^{-} \stackrel{?}{=}\frac{1}{\sqrt[\normalsize{4}]{\ln 2}}. \tag5
\end{align}
$$
How can we prove it?
A version of this has been sent to Eric Weisstein, these integrals are on Mathworld as Ramanujan log-trigonometric integrals.
 A: Oloa was right, I was on the right track last night (it makes me wonder if he has already solved it and just didn't tell us?). The final trick is to realize $(ix-\ln(\cos x))^{1/4}=\ln(1+i\tan(x))^{1/4}$ and then make the substitution $u=1+i\tan x$. This gives $R^+=\frac{2}{\pi}\text{Re}\lbrace\frac{i}{2}\Gamma(5/4) \text{PolyLog}(5/4, 2)\rbrace=(\ln 2)^{1/4}$. Details to follow in an edit in the next few minutes.
Edit: details follow
So, this may not be the best way but at least it's a way:
To get to the compact form I posted in the above comment, let $\theta(x)$ be such that $$\cos(\theta(x))=\frac{-\ln(\cos x)}{\sqrt{x^2+\ln^2(\cos x)}}.$$ Considering the right triangle with legs $x$ and $-\ln(\cos x)$ we notice that the hypotenuse is given by $h(x)=\sqrt{x^2+\ln^2(\cos x)}$ and further that $\sin(\theta(x))=\frac{x}{h(x)}$. This gives, using the correct identity $\sqrt{1/2+1/2\cos x}=\cos(x/2)$:
$$R^+:=\frac{2}{\pi}\int_0^{\pi/2}\left(\frac{x}{\sin(\theta[x])}\right )^{1/4}\cos\left (\frac{\theta[x]}{4}\right )dx\\R^-:=\frac{2}{\pi}\int_0^{\pi/2}\left(\frac{x}{\sin(\theta[x])}\right )^{-1/4}\cos\left (\frac{\theta[x]}{4}\right )dx. $$
The trick is now to turn that $\theta/4$ into some kind of 4th root. We use the exponential definition of cosine:
$\cos(\theta/4)=\frac{1}{2}\left ( e^{i\theta/4}+e^{-i\theta/4} \right )=\frac{1}{2}\left ( \left [\cos\theta+i\sin\theta \right ]^{1/4}+\left [\cos\theta-i\sin\theta \right ]^{1/4} \right )$. Then we have $$\left(\frac{x}{\sin(\theta[x])}\right )^{1/4}\cos\left (\frac{\theta[x]}{4}\right )=\frac{1}{2}\left ( [x\cot\theta+ix]^{1/4}+[x\cot\theta-ix]^{1/4} \right ).$$ Considering the triangle again this gives $$\left(\frac{x}{\sin(\theta[x])}\right )^{1/4}\cos\left (\frac{\theta[x]}{4}\right )=\frac{1}{2}\left ( [-\ln(\cos x)+ix]^{1/4}+[-\ln(\cos x)-ix]^{1/4} \right )\\=\text{Re}\left \lbrace [-\ln(\cos x)+ix]^{1/4} \right \rbrace. $$ Similar methods show $$\left(\frac{x}{\sin(\theta[x])}\right )^{-1/4}\cos\left (\frac{\theta[x]}{4}\right )=\text{Re}\left \lbrace [-\ln(\cos x)-ix]^{-1/4} \right \rbrace.$$ Now we use the addition property of logarithms and the exponental form of cosine again:$$-\ln(\cos x)+ix=-\ln(e^{ix}/2+e^{-ix}/2)+\ln(e^{ix})=\ln\left ( \frac{2e^{ix}}{e^{ix}+e^{-ix}} \right ) \\=\ln(1+i\tan x).$$ Similarly, $-\ln(\cos x)-ix=\ln(1-i\tan x).$ Okay, putting this together gives, using the fact that the integral over a real part is the real part of an integral: $$R^+=\frac{2}{\pi}\text{Re}\left \lbrace \int_0^{\pi/2}\ln(1+i\tan x)^{1/4}dx \right  \rbrace \\ R^-=\frac{2}{\pi}\text{Re}\left \lbrace \int_0^{\pi/2}\ln(1-i\tan x)^{-1/4}dx \right  \rbrace.$$ It seems that Mathematica still couldn't solve this, but with a simple substitution it works. For $R^+$ let $u=1+i\tan x$. Then $$ R^+=\frac{2}{\pi}\text{Re}\left \lbrace i\int_1^{1+i\infty}\frac{\ln(u)^{1/4}}{u^2-2u}du \right  \rbrace\\=\frac{2}{\pi}\text{Re}\lbrace\frac{i}{2}\Gamma(5/4) \text{PolyLog}(5/4, 2)\rbrace\\=(\ln 2)^{1/4}.$$ I assume someone skilled in contour integration can do whatever Mathematica did to get this answer. For $R^-$ let $u=1-i\tan x$. Then $$ R^-=\frac{2}{\pi}\text{Re}\left \lbrace i\int_1^{1-i\infty}\frac{\ln(u)^{-1/4}}{u^2-2u}du \right  \rbrace\\=\frac{2}{\pi}\text{Re}\lbrace\frac{i}{2}\Gamma(3/4) \text{PolyLog}(3/4, 2)\rbrace\\=(\ln 2)^{-1/4}.$$ I had to fudge a minus sign on this one, probably something with direction of contour integration or something. Anyway, conjecture confirmed. I assume similar methods can be used to confirm the nth order Ramanujan Log-Trigonometric Integral as mentioned by Oloa. Specifically, it seems we have $$R_n^+=\frac{2}{\pi}\text{Re}\left \lbrace \int_0^{\pi/2}\ln(1+i\tan x)^{1/(2^n)}dx \right  \rbrace \\ R_n^-=\frac{2}{\pi}\text{Re}\left \lbrace \int_0^{\pi/2}\ln(1-i\tan x)^{-1/(2^n)}dx \right  \rbrace.$$
Edit:
It was silly for me to choose different representations for the real part. Better to chose the same one in which case we have:
$$R_n^\pm=\frac{2}{\pi}\text{Re}\left \lbrace \int_0^{\pi/2}\ln(1+i\tan x)^{\pm 1/(2^n)}dx \right  \rbrace.$$
