How to prove $\int_0^\infty e^{-x}\frac{\sin^2 x}{x}\text{ dx}=\frac{\text{log }5}{4}$

Question

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 $(\ast)$?

Answer 1

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.

Answer 2

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\down}{\downarrow}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $\ds{{\cal I}\pars{\mu}\equiv \int_{0}^{\infty}\expo{-x}\,{\sin^2\pars{\mu x} \over x}\,\dd x\,, \qquad{\cal I}\pars{0} = 0\,,\qquad{\cal I}\pars{1}:\ {\large ?}}$

\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}

$$\color{#00f}{\large% \int_{0}^{\infty}\expo{-x}\,{\sin^2\pars{x} \over x}\,\dd x} = {\cal I}\pars{1} =\int_{0}^{1}{2\mu \over 4\mu^{2} + 1}\,\dd\mu =\left.{1 \over 4}\,\ln\pars{4\mu^{2} + 1}\right\vert_{0}^{1} =\color{#00f}{\large{1 \over 4}\,\ln\pars{5}} $$