Integration using Theory of Fourier Transform

Question

I encounter the following integral which I have attempted but couldn't get anything. $$\int_0^\infty \frac{\sin^2x}{x^2(1+x^2)}dx$$ I attempted to use the Plancherel formula to the convolution of the window function ($\chi_{[-1,1]}$) and $e^{-|x|}$. Unfortunately, this gives me $(1+x^2)^2$ in the denominator. I have no idea how to do this integral. Do you guys have any suggestions? Thank you.

Thanks for your amazing hint, Alex! The first integral can be done by Plancherel formula on the window function. The second integral can be done by writing $$\sin^2 x=\frac{1-\cos 2x}{2}$$ and then use Fourier Inversion Formula to $e^{-|x|}$! Correct me if there is any error.

Answer 1

I do not really get where you are stuck at: if you know that the Fourier transform of $\left(\frac{\sin x}{x}\right)^2$ is the tent-function (convolution of two window functions) and the Fourier transform of $\frac{1}{1+x^2}$ is the Laplace distribution, the computation of your integral immediately boils down to the computation of $\int_{I} p(x) e^{-x}\,dx $ where $p(x)$ is a linear polynomial and $I$ is a finite interval.

In explicit terms: $$ \int_{0}^{+\infty}\left(\frac{\sin x}{x}\right)^2\frac{dx}{1+x^2}\stackrel{\text{parity}}{=}\frac{1}{2}\int_{\mathbb{R}}\left(\frac{\sin x}{x}\right)^2\frac{dx}{1+x^2}=\frac{\pi}{8}\int_{-2}^{2}(2-|s|)e^{-|s|}\,ds$$ and since $\int_{0}^{2}(2-s)e^{-s}\,ds = 1+\frac{1}{e^2}$ we have $$ \int_{0}^{+\infty}\left(\frac{\sin x}{x}\right)^2\frac{dx}{1+x^2} = \color{blue}{\frac{\pi}{4}\left(1+\frac{1}{e^2}\right)}.$$

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \int_{0}^{\infty}{\sin^{2}\pars{x} \over x^{2}\pars{1 + x^{2}}}\,\dd x & = {1 \over 2}\int_{-\infty}^{\infty}{\sin^{2}\pars{x} \over x^{2}\pars{1 + x^{2}}} \,\dd x \\[5mm] & = {1 \over 2}\int_{-\infty}^{\infty}{1 \over 1 + x^{2}} \pars{{1 \over 2}\int_{-1}^{1}\expo{\ic yx}\,\dd y} \pars{{1 \over 2}\int_{-1}^{1}\expo{-\ic zx}\,\dd z}\,\dd x \\[5mm] & = {1 \over 8}\int_{-1}^{1}\int_{-1}^{1}\ \overbrace{\int_{-\infty}^{\infty}{\expo{\ic\,\verts{y - z}x}\over 1 + x^{2}}\,\dd x}^{\ds{\pi\expo{-\verts{y - z}}}}\,\ \dd y\,\dd z \\[5mm] & = {\pi \over 8}\int_{-1}^{1}\int_{-1}^{1}\expo{-\verts{y - z}}\dd y\,\dd z = {\pi \over 8}\int_{-1}^{1}\!\pars{\expo{-z}\!\!\int_{-1}^{z}\!\!\expo{y}\dd y + \expo{z}\!\!\int_{z}^{1}\!\!\expo{-y}\dd y}\dd z \\[5mm] & = {\pi \over 8}\int_{-1}^{1}\!\! \pars{1 - \expo{-z}\expo{-1} - \expo{z}\expo{-1} + 1}\dd z = \bbx{\!\!\pars{1 + {1 \over \expo{2}}}{\pi \over 4}\!\!} \approx 0.8917 \end{align}