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 $(\ast)$?

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    $\begingroup$ Introduce a parameter $\lambda$ in the exponential and differentiate w.r.t $\lambda$. $\endgroup$ – Julien Godawatta Jan 13 '14 at 14:25
  • $\begingroup$ @JulienGodawatta - What's the idea behind this? Do you want me to differentiate $\frac{d}{d\lambda}\text{exp}(-\lambda x)$? This leads to $-xe^{-\lambda x}$. How does this help? $\endgroup$ – 0xbadf00d Jan 13 '14 at 14:59
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    $\begingroup$ @0xbadf00d $$\frac{d}{d\lambda}\left(e^{-\lambda x}\frac{\sin^2x}{x}\right) = -e^{-\lambda x}\sin^2 x.$$ $\endgroup$ – Daniel Fischer Jan 13 '14 at 15:05

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.

  • $\begingroup$ For istance, $$\int_{0}^{+\infty}e^{-x}\sin(2xy)dx=\frac{2y}{1+4y^2},$$ from which the conclusion is straightforward. $\endgroup$ – Jack D'Aurizio Jan 13 '14 at 15:11
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    $\begingroup$ I think we can change the order of integration because of the Fubini–Tonelli theorem. $\endgroup$ – 0xbadf00d Jan 13 '14 at 15:52
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    $\begingroup$ Right, @0xbadf00d, that justifies the change. (Since everything is smooth here, one can also justify the change more elementarily, but if one has Fubini and Tonelli available, that's of course the simplest way.) $\endgroup$ – Daniel Fischer Jan 13 '14 at 15:56

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

\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}

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


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