TITLE: How may we show that $\int_{0}^{\infty}{{\cos\left({2x\over \pi}\right)-\cos^2\left({2x\over \pi}\right)}\over x^2}\cdot\ln(x)\,\mathrm dx=-\ln(2)?$
QUESTION [9 upvotes]: 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)?

REPLY [8 votes]: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\int_{0}^{\infty}
{\cos\pars{2x/\pi} - \cos^{2}\pars{2x/\pi} \over x^{2}}\,\ln\pars{x}\,\dd x
\\[5mm] = &\
\int_{0}^{\infty}
{1 - \cos^{2}\pars{2x/\pi} \over x^{2}}\,\ln\pars{x}\,\dd x -
\int_{0}^{\infty}
{1 - \cos\pars{2x/\pi} \over x^{2}}\,\ln\pars{x}\,\dd x
\\[5mm] = &\
\int_{0}^{\infty}
{\sin^{2}\pars{2x/\pi} \over x^{2}}\,\ln\pars{x}\,\dd x -
\int_{0}^{\infty}
{2\sin^{2}\pars{x/\pi} \over x^{2}}\,\ln\pars{x}\,\dd x
\\[5mm] = &\
{2 \over \pi}\int_{0}^{\infty}
{\sin^{2}\pars{x} \over x^{2}}\,\ln\pars{\pi x \over 2}\,\dd x -
{2 \over \pi}\int_{0}^{\infty}
{\sin^{2}\pars{x} \over x^{2}}\,\ln\pars{\pi x}\,\dd x
\\[5mm] = &\
{2 \over \pi}\int_{0}^{\infty}
{\sin^{2}\pars{x} \over x^{2}}\,\bracks{\ln\pars{1 \over 2}}\,\dd x =
\bbx{\ds{-\ln\pars{2}}}
\end{align}