Integrating $ \int \limits_{-\infty}^{\infty} \frac{\sin^2(x)}{x^2} \operatorname d\!x $

Question

I'm trying to evaluate $\displaystyle \int \limits_{-\infty}^{\infty} \dfrac{\sin^2(x)}{x^2} \operatorname d\!x $.

My first though was to use residue calculus, since we've got the pole of order 2 there at the origin. Given that the pole actually lies on the real axis though, I'd guess I'd have to use some sort of keyhole contour to include the pole and then limit. Maybe it's just one of those days, but I'm not seeing an obvious contour to use here. Any help would be appreciated.

Answer 1

Suppose that $\int_0^\infty \frac{\sin t}{t} dt = \frac{\pi}{2}$. By an integration by parts, $$\int_0^\infty \frac{\sin^2 x}{x^2} dx = -\frac{\sin^2 x}{x}|_0^\infty + \int_0^\infty \frac{\sin (2x)}{x} dx .$$

Check that $\lim_{x\to \infty}\frac{\sin^2 x}{x} = \lim_{x\to 0} \frac{\sin^2 x}{x}=0$. By a change of variable $t=2x$, $\int_0^\infty \frac{\sin (2x)}{x} dx = \int_0^\infty \frac{\sin t}{t} dt = \pi/2$ using our supposition. Hence $\int_0^\infty \frac{\sin^2 x}{x^2} dx = \frac{\pi}{2}$.

Therefore $\int_{-\infty}^\infty \frac{\sin^2 x}{x^2} dx = \pi$.

Answer 2

This may be a starting point, it is not a complete answer. Notice that $\frac{\sin(x)}{x} =: \text{sinc}(x)$ is an even function. So $\int_{-\infty}^{\infty} \text{sinc}(x)dx = 2\int_0^{\infty}\text{sinc}(x)dx$. Now we can make use of Laplace transforms and the fact that

$$\int_0^t f(\tau)d\tau = \frac{F(s)}{s}$$

where $F(s)$ is the Laplace transform of $f(x)$.

$$F(s) = \int_0^\infty \frac{\sin x}{x}e^{-sx}dx$$

Now, $$F'(s) = -\int_0^{\infty}\sin(x)e^{-sx}dx = -\frac{d}{ds}\tan^{-1}s$$

So, $F(s) = -\tan^{-1}s + const.$. We know that $F(\infty) = 0$ so it should be clear that $F(s) = \frac{\pi}{2}-\tan^{-1}s = \cot^{-1}s$.

If we can take the inverse laplace transform of $F(s)/s$ it may be of use.

Answer 3

I reckon there are many ways to go about it, but if you want to use contours you could perhaps consider

$\displaystyle f(z)=\frac{e^{iaz}-e^{ibz}}{z^{2}}$ around a standard semi circle in the UHP indented at the origin.

Calculate the Laurent series at 0 and we have:

$\displaystyle f(z)=\frac{1}{z^{2}}\left[(1+iaz-a^{2}z^{2}+O(z^{3}))-(1+ibz-b^{2}z^{2}+O(z^{3}))\right]$

$\displaystyle=\frac{i(a-b)}{z}-(a^{2}-b^{2})+O(z)$

The pole at the origin is simple and has residue $i(a-b)$.

Thus, $\displaystyle\int_{-\infty}^{\infty}\frac{e^{iax}-e^{ibx}}{x^{2}}dx=\pi (b-a)$.

But, $\displaystyle\sin^{2}(x)=1/2(1-\cos(2x))=1/2(\cos(0x)-\cos(2x))$.

So, from the above we have $\displaystyle\int_{-\infty}^{\infty}\frac{\sin^{2}(x)}{x^{2}}dx=\frac{\pi}{2}(2-0)=\pi$

Answer 4

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle #1 \right\rangle} \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace} \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left( #1 \right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ \begin{align} \color{#00f}{\large\int_{-\infty}^{\infty}{\sin^{2}\pars{x} \over x^{2}}\,\dd x}&= \int_{-\infty}^{\infty}\half\int_{-1}^{1}\expo{\ic kx}\,\dd k \half\int_{-1}^{1}\expo{-\ic qx}\,\dd q \\[3mm]&=\half\,\pi\int_{-1}^{1}\dd k\int_{-1}^{1}\dd q \int_{-\infty}^{\infty}\expo{\ic\pars{k - q}x}\,{\dd x \over 2\pi} =\half\,\pi\int_{-1}^{1}\dd k\int_{-1}^{1}\dd q\,\delta\pars{k - q} \\[3mm]&=\half\,\pi\int_{-1}^{1}\dd k = \half\,\pi\times 2 = \color{#00f}{\Large\pi} \end{align}