Conjecture $\int_0^1\frac{\ln\left(\ln^2x+\arccos^2x\right)}{\sqrt{1-x^2}}dx\stackrel?=\pi\,\ln\ln2$
$$\int_0^1\frac{\ln\left(\ln^2x+\arccos^2x\right)}{\sqrt{1-x^2}}dx\stackrel?=\pi\,\ln\ln2$$ Is it possible to prove this?
Solution 1:
Substitute $x \mapsto \cos x$ to obtain an equivalent formulation:
$$ \int_{0}^{\frac{\pi}{2}} \log(x^{2} + \log^{2}\cos x) \, dx = \pi \log \log 2. $$
You can find my solution here.
Solution 2:
Hint: Note that $$\ln^2\cos t+t^2=\ln^2\cos t-\ln^2 e^{it}=(\ln e^{it}\cos t)(\ln e^{-it}\cos t)$$ Then let $ e^{2it}=s$ to obtain a contour integral surrounding the simple pole at $s=0$