$\int_\limits{0}^\infty \frac{(1-\cos(x))\cos(x)}{x^2}dx=0$ I know that  $\int_\limits{0}^\infty \frac{\sin(x)}{x}dx=\int_\limits{0}^\infty \frac{\sin^2(x)}{x^2}dx=\int_\limits{0}^\infty \frac{1-\cos(x)}{x^2}dx=\frac{\pi}{2}$ and I can show the first equality using integration by parts. I can show that they each separately equal $\frac{\pi}{2}$, but I want to find an elementary way to show the second equality. This is equivalent to proving that $\int_\limits{0}^\infty \frac{(1-\cos(x))\cos(x)}{x^2}dx=0$. I feel there must be some simple trick I can use to show that this, but I cant seem to find it. Any help would be appreciated. Thanks! :) 
 A: I am not sure that this is the answer you expect.
Consider $$I=\int \frac{(1-\cos(x))\cos(x)}{x^2}dx$$ and integrate by parts $$u=(1-\cos(x))\cos(x)\implies u'=\sin (x) \cos (x)-\sin (x) (1-\cos (x))=-\sin(x)+\sin(2x)$$ $$v'=\frac 1{x^2}\implies v=-\frac1 x$$ $$I=-\frac{(1-\cos(x))\cos(x)}x-\int\frac{\sin(x)-\sin(2x)} {x }\,dx$$ Now $$\int\frac{\sin(x)-\sin(2x)} {x }\,dx=\int\frac{\sin(x)} {x }\,dx-\int\frac{\sin(2x)} {x }\,dx$$ For the last integral, change variable $x=\frac y2$ which gives $$\int\frac{\sin(2x)} {x }\,dx=\int\frac{\sin(y)} {y }\,dy$$ which makes $$\int_0^\infty\frac{\sin(x)-\sin(2x)} {x }\,dx=0$$ Now, use the bounds for the only left term $$-\frac{(1-\cos(x))\cos(x)}x$$ to get the result.
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,}
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\begin{align}
&\int_{0}^{\infty}{\bracks{1 - \cos\pars{x}}\cos\pars{x} \over x^{2}}\,\dd x =
\int_{0}^{\infty}{\cos\pars{x} - \cos^{2}\pars{x} \over x^{2}}\,\dd x
\\[5mm] = &\
\int_{0}^{\infty}{\cos\pars{x} - \bracks{1 + \cos\pars{2x}}/2 \over x^{2}}
\,\,\dd x
\\[5mm] = &
\underbrace{\int_{0}^{\infty}{\cos\pars{x} - 1 \over x^{2}}
\,\dd x}_{\ds{\mc{J}}}\,\,\, -\,\,\, \underbrace{%
{1 \over 2}\int_{0}^{\infty}{\cos\pars{2x} - 1 \over x^{2}}\,\dd x}
_{\ds{\mbox{With}\ \pars{~2x\ \mapsto\ x~},\ \mbox{it's}\ =\ \mbox{to}\ \,\mc{J}}}
\,\,\, =\,\,\, \bbx{\ds{0}}
\end{align}
A: Rewrite the integral as
$$\int_{-\infty}^{\infty} dx \frac{\sin^2{(x/2)}}{x^2} e^{i x} = \frac12 \int_{-\infty}^{\infty} dx \frac{\sin^2{x}}{x^2} e^{i 2 x} $$
Note the FT relation
$$\int_{-\infty}^{\infty} dx \frac{\sin^2{x}}{x^2} e^{i k x} = \begin{cases} \pi \left (1-\frac{|k|}{2} \right ) & |k| \le 2 \\ 0 & |k| \gt 2 \end{cases}$$
Thus when $k=2$, the integral is zero.
