## Monday Math 30

We can find the definite integral $\int_0^{\infty}\frac{\sin^2x}{x^2}\,dx$ in three different ways:

1. Long: Differentiation under the integral sign.
In my previous post on this topic, I showed that $\int_{0}^{\infty}\frac{\sin{x}}{x}\,dx=\frac{\pi}{2}$. We can use a similar method on $\int_{0}^{\infty}\frac{\sin{kx}}{x}\,dx$, namely, defining $f(a)=\int_{0}^{\infty}e^{-ax}\frac{\sin{kx}}{x}\,dx$, and then finding
$\begin{array}{rcl}f'(a)&=&\int_{0}^{\infty}\frac{\partial}{\partial{a}}\left(e^{-ax}\frac{\sin{kx}}{x}\right)\,dx\\&=&-\int_{0}^{\infty}e^{-ax}\sin{kx}\,dx\\&=&\left[\frac{e^{-ax}(k\cos{kx}+a\sin{kx})}{a^2+k^2}\right]_{x=0}^{\infty}\\&=&-\frac{k}{a^2+k^2}\end{array}$.
For k=0, this gives $f(a)=0$, as expected.
For non-zero k, we integrate with respect to a, and find
$f(a)=C-\arctan\left(\frac{a}{k}\right)$.
Again, we note that as $a\to\infty$, we see that for all x>0, $e^{-ax}\frac{\sin{kx}}{x}\to0$, and thus $\lim_{a\to\infty}f(a)=0$.
For k>0, we have $\lim_{a\to\infty}\arctan\left(\frac{a}{k}\right)=\frac{\pi}{2}$, giving $C=\frac{\pi}{2}$, and so $f(a)=\frac{\pi}{2}-\arctan\left(\frac{a}{k}\right)$.
For k<0, we instead have $\lim_{a\to\infty}\arctan\left(\frac{a}{k}\right)=-\frac{\pi}{2}$, giving $C=-\frac{\pi}{2}$, and thus $f(a)=-\frac{\pi}{2}-\arctan\left(\frac{a}{k}\right)$.
Thus $\int_{0}^{\infty}\frac{\sin{kx}}{x}\,dx=f(0)= \begin{cases}\frac{\pi}{2},&\;k>{0}\\{0},&\;k={0}\\-\frac{\pi}{2},&\;k<{0}\end{cases}$.
Now, to our original integral $\int_0^{\infty}\frac{\sin^2x}{x^2}\,dx$.
Define $I(p)=\int_0^{\infty}\frac{\sin^2(px)}{x^2}\,dx$, for p≥0
Then
$\begin{array}{rcl}I'(p)&=&\int_0^{\infty}\frac{\partial}{\partial{p}}\left(\frac{\sin^2(px)}{x^2}\right)\,dx\\&=&\int_0^{\infty}\frac{2\sin(px)\cos(px)}{x}\,dx\\&=&\int_0^{\infty}\frac{\sin(2px)}{x}\,dx\end{array}$,
and now we know, from our previous work, that for p>0, this is $\frac{\pi}{2}$. Integrating, $I(p)=\frac{\pi}{2}p+C$, and as $I(0)=0$, we see $I(p)=\frac{\pi}{2}p$, and so our integral is
$\int_0^{\infty}\frac{\sin^2x}{x^2}\,dx=I(1)=\frac{\pi}{2}$.

2. Difficult: Trig and Double Integral.
We can use the trigonometric identity $\sin^2x=\frac{1-\cos(2x)}{2}$ to rewrite our integral:
$\int_0^{\infty}\frac{\sin^2x}{x^2}\,dx=\int_0^{\infty}\frac{1-\cos(2x)}{2x^2}\,dx$.
Substituting $u=2x$, we have
$\int_0^{\infty}\frac{1-\cos(2x)}{2x^2}\,dx=\int_0^{\infty}\frac{1-\cos{u}}{\frac{u^2}{2}}\,\frac{du}{2}=\int_0^{\infty}\frac{1-\cos{u}}{u^2}\,du$
Now, here is the hard part: we can recognize that $\int_0^{\infty}xe^{-kx}\,dx=\frac{1}{k^2}$. Thus, $\frac{1}{u^2}=\int_0^{\infty}ve^{-uv}\,dv$, and our integral can be rewritten as the double integral
$\int_0^{\infty}\frac{1-\cos{u}}{u^2}\,du=\int_0^{\infty}\int_0^{\infty}(1-\cos{u})ve^{-uv}\,dv\,du$.
We can reverse order of integration to solve this:
$\begin{array}{rcl}\int_0^{\infty}\frac{1-\cos{u}}{u^2}\,du&=&\int_0^{\infty}\int_0^{\infty}(1-\cos{u})ve^{-uv}\,du\,dv\\&=&\int_0^{\infty}\left[\left(\frac{v^2\cos{u}}{v^2+1}-\frac{v\sin{u}}{v^2+1}-1\right)e^{-uv}\right]_{u=0}^{\infty}\,dv\\&=&\int_0^{\infty}\frac{1}{1+v^2}\,dv\\&=&\left[\arctan{v}\right]_{v=0}^{\infty}\\&=&\frac{\pi}{2}\end{array}$

3. Simple (but needing some background): Fourier Transform.
Defining the sine cardinal as $\mathrm{sinc}(x)=\frac{\sin(\pi{x})}{\pi{x}}$. It is known that the Fourier transform transforms between the sinc function and the rectangle function: namely, $\mathcal{F}(\mathrm{sinc}(t))(f)={\Pi}(f)$, where the function Π(x) is the rectangle function ${\Pi}(x)=\begin{cases}1,&|x|\le{\normalsize\frac{1}{2}}\\{0},&|x|>{\normalsize\frac{1}{2}}\end{cases}$
The convolution theorem tells us that:
$\begin{array}{rcl}\mathcal{F}(\mathrm{sinc}^2(t))(f)&=&\mathcal{F}(\mathrm{sinc}(t)\cdot\mathrm{sinc}(t))(f)\\&=&\mathcal{F}(\mathrm{sinc}(t))(f)*\mathcal{F}(\mathrm{sinc}(t))(f)\\&=&{\Pi}(f)*{\Pi}(f)\end{array}$,
where the * represents convolution. The convolution of the rectangle function with itself is the triangle function:
${\Pi}(x)*{\Pi}(x)=\Lambda(x)\equiv\begin{cases}{0},&|x|\ge{1}\\1-|x|,&|x|<1\end{cases}$.
And so $\mathcal{F}(\mathrm{sinc}^2(t))(f)={\Lambda}(f)$,
Writing this out with the definition of the Fourier transform, we have:
$\int_{-\infty}^{\infty}\mathrm{sinc}^2(t)e^{-2\pi{i}ft}\,dt={\Lambda}(f)$.
Plugging in f=0, and using Λ(0)=1,
$\int_{-\infty}^{\infty}\mathrm{sinc}^2(t)\,dt=1$.
The sinc function is even, so we have
$\int_{0}^{\infty}\mathrm{sinc}^2(t)\,dt=\frac{1}{2}$
Using the subsistution xt, we have
$\begin{array}{rcl}\int_{0}^{\infty}\frac{\sin^2(\pi{t})}{(\pi{t})^2}\,dt&=&\frac{1}{2}\\\int_{0}^{\infty}\frac{\sin^2{x}}{x^2}\,\frac{dx}{\pi}&=&\frac{1}{2}\\\int_{0}^{\infty}\frac{\sin^2{x}}{x^2}\,dx&=&\frac{\pi}{2}\end{array}$
QED.