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Here is the link concerning the integral to be ask about,

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

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 ${\pi\over 4}$.

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  • $\begingroup$ is [a] the floor part of a? $\endgroup$ – Zacky Apr 12 '18 at 20:27
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    $\begingroup$ @Zacky No, they are just brackets $\endgroup$ – John Doe Apr 12 '18 at 20:28
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The integral in question might be rewitten (see below) as

$$ I=\text{Re}\int_0^{\pi/2}\frac{dx}{\log(1+e^{2 i x})}=\frac{1}{4}\text{Re}\int_0^{2 \pi}\frac{dx}{\log(1+e^{2 i x})}\underbrace{=}_{\color{blue}{ e^{ix}\rightarrow z}}\frac{1}{4}\text{Re}\oint_C\frac{dz}{z \log(1+z^2)}=\frac{\pi}{4} $$

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 ($x\in(0,\pi/2))$

$$ \frac{1}{\log(1+e^{2 i x})}=\frac{1}{i x+\log(e^{ i x}+e^{-i x})}=\frac{-i x+\log(2 \cos(x))}{ x^2+\log^2(2 \cos(x))} $$

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    $\begingroup$ [+1] Splendid ! $\endgroup$ – Jean Marie Apr 12 '18 at 22:27

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