Simple closed form given by this complicated integral
$$\int_{0}^{\infty}{{\cos\left({2x\over \pi}\right)-\cos^2\left({2x\over \pi}\right)}\over x^2}\cdot\ln(x)\,\mathrm dx=-\ln(2)\tag1$$
Making an attempt:
Splitting $(1)$ results in the integral to be diverges.
$$I(a)=\int_{0}^{\infty}{{\cos\left({2x\over \pi}\right)-\cos^2\left({2x\over \pi}\right)}\over x^a}\cdot\ln(x)\,\mathrm dx\tag2$$
$$I'(a)=\int_{0}^{\infty}{{\cos\left({2x\over \pi}\right)-\cos^2\left({2x\over \pi}\right)}\over x^a}\,\mathrm dx\tag3$$
$$I'(2)=\int_{0}^{\infty}{{\cos\left({2x\over \pi}\right)-\cos^2\left({2x\over \pi}\right)}\over x^2}\,\mathrm dx\tag4$$
$u={2x\over \pi}$
$$I'(2)={2\over \pi}\int_{0}^{\infty}{{\cos\left({u}\right)-\cos^2\left({u}\right)}\over u^2}\,\mathrm du=I_1-I_2\tag5$$ Recall from table of integral, I was thinking of using $(6)$ for $I_1$ but it is only valid for $0\le p\le 1$ $$\int_{0}^{\infty}{\cos x\over x^p}\,\mathrm dx={\pi\over 2\Gamma(p)\cos(p\pi/2)}\tag6$$
How may we prove (1)?