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I have come across another improper integral I wish to evaluate via residues.

The integral is:

$$\int_{-\infty}^\infty{\frac{\sin(x)^2}{x^2}}dx$$

$\sin(z)$ behaves in an uneasy way so I tried using the function $\frac{{e^{iz}}^2}{z^2}$ with a half circle on the upper complex plane with radius R and a half-circle of radius 1/R which arcs below $0$.

The problem is the small semi-circles integral does not go to $0$ and in fact doesn't exist.

What other types of contours or function substitutions should be used here?

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Is it $\,\sin^2x\,$ (headline) or $\,\sin x\,$ (message)? –  DonAntonio Oct 17 '12 at 5:00
    
$sin(x)^2$, I wrote the function down incorrectly but then edited it. –  Mike Oct 17 '12 at 5:02
    
Is that $\exp(iz^2)$ or $\exp(iz)^2=\exp(2iz)$? I don't see how they're linked with $\sin(x)^2$ anyway. –  Philippe Malot Oct 26 '13 at 18:28
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2 Answers

up vote 2 down vote accepted

Note that $ \cos(2x)=1-2\sin(x)^2 $, this suggest to consider the integral

$$ \int_{C} \frac{ {\rm e}^{2 i z} - 1 }{ z^2} dz \,.$$

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Yes but how does one get rid of the $\frac{1}{z^2}$ term? –  Mike Oct 17 '12 at 16:10
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$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\dd}{{\rm d}}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\ic}{{\rm i}}% \newcommand{\imp}{\Longrightarrow}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert #1 \right\vert}% \newcommand{\yy}{\Longleftrightarrow}$ Since $\ds{% \sin\pars{x} \over x} = {1 \over 2}\int_{-1^{-}}^{1^{+}}\expo{\ic kx}\,\dd k$, we have: \begin{align} {\large\int_{-\infty}^{\infty}{\sin^{2}\pars{x} \over x^{2}}\,\dd x} &= \int_{-\infty}^{\infty}\dd x\,{1 \over 2}\int_{-1^{-}}^{1^{+}}\expo{\ic kx}\,\dd k\, {1 \over 2}\int_{-1^{-}}^{1^{+}}\expo{\ic qx}\,\dd q \\[3mm]&= {\pi \over 2}\int_{-1_{-}}^{1^{+}}\dd k\int_{-1_{-}}^{1^{+}}\dd q \int_{-\infty}^{\infty}\expo{\ic\pars{k + q}x}\,{\dd x \over 2\pi} = {\pi \over 2}\int_{-1^{-}}^{1^{+}}\dd k \int_{-1_{-}}^{1^{+}}\dd q\,\delta\pars{k + q} \\[3mm]&= {\pi \over 2}\int_{-1^{-}}^{1^{+}}\Theta\pars{1 - \verts{k}}\,\dd k = {\pi \over 2}\int_{-1^{-}}^{1^{+}}\dd k = {\large \pi} \end{align}

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