irreducible polynomial

$\text{Problem 12184-05}$

American Mathematical Monthly, May 2020

$\text{Proposed by P. Perfetti, Italy}$

Prove $\int_1^{\infty}\frac{\ln(x^4-2x^2+2)}{x\sqrt{x^2-1}}dx=\pi \ln(2+\sqrt 2)$

$\text{Proposed solution by:}$ Narendra Bhandari, National Academy of Science and technology, Pokhara University,Bajura,Nepal

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We make substitution of $x^2-1=y\Rightarrow x=\sqrt{y+1}$ giving us $\int_0^{\infty}\frac{\ln((\sqrt{y+1})^4-2y)}{ \sqrt{y+1}\sqrt{y}}\frac{dy}{2 \sqrt{y+1}}=\frac{1}{2}\int_0^{\infty}\frac{\ln(y^2+1)}{(y+1)\sqrt{y}}dy$ further we substitute $y=\tan^2 \theta\Rightarrow dy=2\tan \theta \sec^2\theta d\theta$ and the latter integral becomes $\int_0^{\frac{\pi}{2}}\ln(1+\tan^4\theta) d\theta =\underbrace{\int_0^{\frac{\pi}{2}}\ln(\sin^4\theta + \cos^4\theta ) d\theta }_{I_1}-\underbrace {\int_0^{\frac{\pi}{2}}\ln(\cos^4\theta)d\theta}_{I_2}$ Note that for all $a,b>0$ we have the Lemma and which I proved here is MSE $\int_0^{\frac{\pi}{2}} \ln(b^2\sin^2\theta+a^2\cos^2\theta) d\theta=\pi\ln\frac{a+b}{2}$ and $\sin^4\theta +\cos^4\theta =1-2\sin^2\theta \cos^2\theta=1-\frac{\sin^22\theta}{2} =\frac{1}{2}(2\cos^2 2\theta + \sin^2 2\theta)$ thus integral $I_1=$ $\int_0^{\frac{\pi}{2}}\ln\left(\cos^22\theta+\frac{1}{2}\sin^22\theta\right)d\theta=\frac{1}{2}\int_0^{\pi}\ln\left(\cos^2\phi + \frac{1}{2}\sin^2\phi\right)d\phi=\pi\ln\frac{1+\sqrt{2}}{2 \sqrt{2}}$ $\cdots (1)$ the latter result is obtained using the Lemma above

And for $I_2$ is easy to compute the integral using Fourier series of $\ln(\cos x)$ for $0 \leq \theta < \frac{\pi}{2}$ . That is $\int_0^{\frac{\pi}{2}}\ln(\cos \theta) d\theta=\int_0^{\frac{\pi}{2}}\left(-\ln 2 -\sum_{k=1}^{\infty}\frac{(-1)^k}{k}\cos (2k\theta) \right)d\theta=-\frac{\pi}{2}\ln 2$ $-\sum_{k=1}^{\infty}\frac{(-1)^k}{k}\int_0^{\frac{\pi}{2}}\cos(2k\theta)d\theta =-\left[\sum_{k=1}^{\infty}\frac{(-1)^{k}}{k^2}\sin(2k\theta)\right]_0^{\frac{\pi}{2}}=-\frac{1}{2}\sum_{k=1}^{\infty}\frac{(-1)^k\sin(2\pi k)}{k^2}$ the obtained series is $0$ as $\sin(2\pi k) = 0$ for positive integers $k$ and hence $\displaystyle I_2=4\int_0^{\frac{\pi}{2}}\ln(\cos\theta)d\theta =-\frac{4\pi}{2}\ln2 =-\pi\ln(4)\cdots (2)$ from $(1)$ and $(2)$ we have $I_1-I_2=\pi\ln\left(\frac{1+\sqrt{2}}{2\sqrt 2}\right)+\pi\ln(4)=\pi\ln\left(2+\sqrt{2}\right)$ which completes the proof.

Without fourier cosine series . Also it is easy to see that $\int_0^{\frac{\pi}{2}}\ln(\cos x) dx= \frac{1}{2}\operatorname{Cl}_2\left(\pi -\pi\right)-\frac{\pi}{2}\ln 2=-\frac{\pi}{2}\ln 2$ where $\operatorname{Cl}_2(x)$ is Clausen Function of order 2 and $\operatorname{Cl}_2(0)=0$ .

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