# Conjecture : $\int_{0}^{\pi\over 2}\mathrm dt{\ln[2\cos(t)]\over t^2+\ln^2[2\cos(t)]}={\pi\over 4}?$

$$\int_{0}^{\pi}\mathrm dt{t^2\over t^2+4\ln^2[2\cos({t\over 2})]}={\pi\over 4}[1-\gamma+\ln(2\pi)]\tag1$$

Surprisingly the second integral has a very simple closed form (not proven yet only an assumption)

$$\int_{0}^{\pi\over 2}\mathrm dt{\ln[2\cos(t)]\over t^2+\ln^2[2\cos(t)]}={\pi\over 4}\tag2$$

Integral $(1)$ is definitely correct, it is proven from the link. But integral $(2)$ is hasn't been proven.

I would like to know what method are used to prove $(2)$ or disprove it is ${\pi\over 4}$.

• is [a] the floor part of a? – Zacky Apr 12 '18 at 20:27
• @Zacky No, they are just brackets – John Doe Apr 12 '18 at 20:28

$$I=\text{Re}\int_0^{\pi/2}\frac{dx}{\log(1+e^{2 i x})}=\frac{1}{4}\text{Re}\int_0^{2 \pi}\frac{dx}{\log(1+e^{2 i x})}\underbrace{=}_{\color{blue}{ e^{ix}\rightarrow z}}\frac{1}{4}\text{Re}\oint_C\frac{dz}{z \log(1+z^2)}=\frac{\pi}{4}$$
where $C$ denotes the unit circle and the last equality follows by the residue theorem (note that $\frac1{z\log(1+z^2)}=\frac1{z^3}+\frac1{2z}+\mathcal{O}(z)$ as $z\rightarrow 0$)
the algebra for the initial transformation goes as follows ($x\in(0,\pi/2))$
$$\frac{1}{\log(1+e^{2 i x})}=\frac{1}{i x+\log(e^{ i x}+e^{-i x})}=\frac{-i x+\log(2 \cos(x))}{ x^2+\log^2(2 \cos(x))}$$