How to prove $\int_0^\infty e^{-x}\frac{\sin^2 x}{x}\text{ dx}=\frac{\text{log }5}{4}$ I'm asked to prove
      $$\displaystyle\int_0^\infty e^{-x}\frac{\sin^2 x}{x}\text{ dx}=\frac{\text{log }5}{4}\tag{$\ast$}$$
by integration of $e^{-x}\text{sin}(2xy)$ over an suitable measurable subset of $\mathbb{R}^2$. I've no idea how to do this. Can anyone tell me how one would choose such a subset and why integration of $e^{-x}\text{sin}(2xy)$ proofs or helps to proof $(\ast)$?
 A: $\newcommand{\+}{^{\dagger}}%
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$\ds{{\cal I}\pars{\mu}\equiv
\int_{0}^{\infty}\expo{-x}\,{\sin^2\pars{\mu x} \over x}\,\dd x\,,
\qquad{\cal I}\pars{0} = 0\,,\qquad{\cal I}\pars{1}:\ {\large ?}}$

\begin{align}
{\cal I}'\pars{\mu}&=\int_{0}^{\infty}\expo{-x}\,\sin\pars{2\mu x}\,\dd x
=\Im\int_{0}^{\infty}\expo{-x}\,\expo{2\ic\mu x}\,\dd x
=\Im\pars{-1 \over -1 + 2\ic\mu} = {2\mu \over 4\mu^{2} + 1}
\end{align}

$$\color{#00f}{\large%
\int_{0}^{\infty}\expo{-x}\,{\sin^2\pars{x} \over x}\,\dd x}
=
{\cal I}\pars{1}
=\int_{0}^{1}{2\mu \over 4\mu^{2} + 1}\,\dd\mu
=\left.{1 \over 4}\,\ln\pars{4\mu^{2} + 1}\right\vert_{0}^{1}
=\color{#00f}{\large{1 \over 4}\,\ln\pars{5}}
$$
A: The idea is to obtain $\dfrac{\sin^2 x}{x}$ as an integral
$$\int_u^v \sin (2xy)\,dy.$$
Now, it's not hard to guess the upper bound $1$, and if we check, we see
$$\frac{d}{dy} \frac{\sin^2 (xy)}{x} = \frac{2\sin(xy)\cos (xy)\cdot x}{x} = 2\sin(xy)\cos(xy) = \sin(2xy).$$
So write
$$\frac{\sin^2 x}{x} = \int_0^1 \sin(2xy)\,dy$$
for $x > 0$.
Then change the order of integration (justify why you can do that),
$$\int_0^\infty e^{-x}\int_0^1 \sin (2xy)\,dy \,dx = \int_0^1 \int_0^\infty e^{-x}\sin (2xy)\,dx\,dy.$$
The inner integral here has a known closed form, and that can be explicitly integrated to obtain the result.
