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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?

MJD
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Mike
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4 Answers4

2

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

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

Mhenni Benghorbal
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Since there are no singularities of $\frac{\sin^2(z)}{z^2}$ in the rectangle $[-R,R]\times[0,-i]$ and the function vanishes along the ends near $-R$ and $R$, $$ \begin{align} \int_{-\infty}^\infty\frac{\sin^2(x)}{x^2}\,\mathrm{d}x &=\int_{-\infty}^\infty\frac{2-e^{-2iz}-e^{2iz}}{4z^2}\,\mathrm{d}z\\ &=\int_{-\infty-i}^{\infty-i}\frac{2-e^{-2iz}-e^{2iz}}{4z^2}\,\mathrm{d}z\\ \end{align} $$ Next we can use two $D$-shaped contours $$ \gamma_+=[-R-i,R-i]\cup Re^{[0,\pi]i}-i $$ and $$ \gamma_-=[-R-i,R-i]\cup Re^{-[0,\pi]i}-i $$ Since the integral along the large semicircles vanishes and the singularity is only circled once counterclockwise by $\gamma_+$ and not by $\gamma_-$, we get $$ \begin{align} \int_{-\infty-i}^{\infty-i}\frac{2-e^{-2iz}-e^{2iz}}{4z^2}\,\mathrm{d}z &=\int_{\gamma_-}\frac{2-e^{-2iz}}{4z^2}\,\mathrm{d}z -\int_{\gamma_+}\frac{e^{2iz}}{4z^2}\,\mathrm{d}z\\ &=0-2\pi i\cdot\frac{i}{2}\\[9pt] &=\pi \end{align} $$ since the residue of $$ \frac{e^{2iz}}{4z^2}=\frac1{4z^2}+\color{#C00000}{\frac{2i\color{#000000}{z}}{4\color{#000000}{z^2}}}+\frac{-2z^2}{4z^2}+\dots $$ at $z=0$ is $\frac i2$.

robjohn
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The function $\mathrm{sinc}: x\mapsto \frac{\sin x}x$ is the Fourier transform of the gate function $\Pi:x\mapsto \mathbf{1}_{[-1,1]}(x)$ $$\mathcal{F}(\Pi)=2\mathrm{sinc}.$$ Parseval's identity for this Fourier pair yields $$\int\Pi^2=\frac{1}{2\pi}\int (2\mathrm{sinc})^2.$$ Since $\int\Pi^2=2$, we get the result $$ \int_{-\infty}^\infty \frac{\sin^2x}{x^2}\mathrm dx=\pi.$$

Tom-Tom
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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}

Felix Marin
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