Prove $\frac{2}{\pi}\int_0^{\infty}\frac{\sin^2u}{u^2}\cos(2ux)\,du=\begin{cases} 1-x,&\mbox{$x\in[0,\,1]$}\\ 0,&\mbox{$x>1$}. \end{cases}$ Prove 
$$
\frac{2}{\pi}\int_0^{\infty}\frac{\sin^2u}{u^2}\cos(2ux)\,du=\begin{cases}
1-x,&\mbox{$x\in[0,\,1]$}\\
0,&\mbox{$x>1$}.
\end{cases}
$$
Let 
$$
I(x)=\frac{2}{\pi}\int_0^{\infty}\frac{\sin^2u}{u^2}\cos(2ux)\,du.
$$
It's easy to check that $I(x)$ is uniformly convergent on $[0,\,\infty)$.
I want to evaluate $I(x)$ through $I'(x)$, but $I'(x)$ is still has a complicated form. So how to do this?
 A: Define
$$ I(a)=\frac{2}{\pi}\int_0^{\infty}\frac{\sin^2(au)}{u^2}\cos(2ux)\,du, a\in[0,1] $$
and then
$$ I'(a)=\frac{2}{\pi}\int_0^{\infty}\frac{\sin(2au)}{u}\cos(2ux)\,du=\frac{1}{\pi}\int_0^{\infty}\frac{\sin(2(a+x)u)}{u}\,du+\frac{1}{\pi}\int_0^{\infty}\frac{\sin(2(a-x)u)}{u}\,du. $$
Using
$$ \int_0^{\infty}\frac{\sin(\alpha u)}{u}\,du=\text{sgn}(\alpha)\frac\pi2,$$
one has
$$ I'(a)=\frac12[\text{sgn}(a+x)+\text{sgn}(a-x)]=\frac12[1+\text{sgn}(a-x)]. $$
If $x>1$, then $a<x$ and hence $\text{sgn}(a-x)=-1$. So
$$ I'(a)=0 $$
which implies $I(a)=C$. So $I(1)=I(0)=0$.
If $x\in[0,1]$, then 
$$ I(1)=\frac12\bigg(1+\int_0^1\text{sgn}(a-x)da\bigg)=\frac12\bigg(1+\int_0^x(-1)da+\int_x^1da\bigg)=1-x.$$
A: $$2\sin u\cos(2ux)=\sin(1+2x)u+\sin(1-2x)u$$
$$4\sin^2u\cos(2ux)=2\cos(2xu)-\cos(2(1+x)u)-{\cos(2(1-x)u)}$$
So the proposed integral is equal to
$$\eqalign{I(x)&=\frac{2}{\pi}\int_{0}^\infty\frac{\sin^2u}{u^2}\cos(2ux)du\cr
&=F(1+x)+F(1-x)-2F(x)}$$
Where
$$\eqalign{F(x)&=\frac{1}{2\pi}\int_{0}^\infty\frac{1-\cos(2ux)}{u^2}du\cr
&=|x|\frac{1}{2\pi}\int_{0}^\infty\frac{1-\cos(2t)}{t^2}dt=c|x|
}$$
with $c=\frac{1}{2\pi}\int_{0}^\infty\frac{1-\cos(2t)}{t^2}dt$.
It follows that
$$I(x)=c(|x+1|+|x-1|-2|x|)$$ but it is well-known that $I(0)=1$ so $c=1/2$ that is
$$I(x)=\frac{|1+x|+|1-x|}{2}-|x|$$ which agrees with the proposed answer for positive $x$.
