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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.

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    $\begingroup$ There is a nice video on youtube about this integral! youtube.com/watch?v=WP8JMVGuOD8&t=191s $\endgroup$ – Tim Dikland Apr 25 '18 at 15:28
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    $\begingroup$ Hint: you can use partial fraction decomposition to turn this into $\int\limits_0^{\infty}\frac{\sin^2x}{x^2}-\frac{\sin^2x}{1+x^2}dx$ $\endgroup$ – Alex Pavellas Apr 25 '18 at 15:30
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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 $$ \int_{0}^{+\infty}\left(\frac{\sin x}{x}\right)^2\frac{dx}{1+x^2} = \color{blue}{\frac{\pi}{4}\left(1+\frac{1}{e^2}\right)}.$$

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  • $\begingroup$ I can't see why your last line holds. The Fourier Inversion Formula doesn't seem to lead me to your result mentioned in the last line. Is there is a more general version for the Plancherel identity? (What I have learnt about Plancherel Identity is the one which applies to a single function only.) $\endgroup$ – Jerry Apr 25 '18 at 16:03
  • $\begingroup$ @Jerry: you just have to exploit $$ \int_{\mathbb{R}}f(x)g(x)\,dx = C\int_{\mathbb{R}} (\mathscr{F}f)(s)\cdot(\mathscr{F}^{-1}g)(s)\,ds $$ where the constant $C$ depends on the way $\mathscr{F}$ has been defined. Answer expanded. $\endgroup$ – Jack D'Aurizio Apr 25 '18 at 16:21
  • $\begingroup$ Is there any name for this formula? $\endgroup$ – Jerry Apr 25 '18 at 16:27
  • $\begingroup$ I think I know how to prove this formula now. $\endgroup$ – Jerry Apr 25 '18 at 16:38
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    $\begingroup$ @mathreadler: I used \mathscr{F}. $\endgroup$ – Jack D'Aurizio Apr 25 '18 at 17:21
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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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}

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