Solving integral of $\frac{\sin^2x}{x^2}$ with distributions

Question

I tried to evaluate $\int_{\mathbb{R}}\frac{\sin^2x}{x^2}dx$, using the theory of distribuions. I guess I should evaluate it by thinking of a distribution generated by $T=\frac{\sin^2x}{x^2}$ and testing it on a $\phi \in \mathcal{D}(\mathbb{R})$: \begin{equation} (T, \phi)= \int_{\mathbb{R}}dx\frac{\sin^2x}{x^2} \phi(x)= \int_{supp(\phi)} dx\frac{\sin^2x}{x^2} [\phi(x)-\phi(0)] + \int_{supp(\phi)}\frac{\sin^2x}{x^2} \phi(0) \end{equation}

Where the first integral on the right side of the equation goes to zero for Riemann-Lebesgue (at least, I suppose). I tried to rewrite the last integral as follows, supposing that, being $\phi \in \mathcal{D}(\mathbb{R})$, exists $L>0$ such that $supp(\phi) \subseteq [-L,L]$: \begin{equation} \phi(0) \int_{-L}^{L}dx\frac{\sin^2x}{x^2} \end{equation} I know that, as comes with complex anlysis, the result should be $\pi$ (and thus I should obtain $\pi \phi(0)$), but from now on I cannot find a way to get through it.. any help/suggestion/correction is very welcome!

Answer 1

We have this property of the Fourier transform:

$$g(t)\times h(t) \xrightarrow{\mathscr{F}} G(f) \ast H(f) \qquad, \qquad g(t)\ast h(t) \xrightarrow{\mathscr{F}} G(f) \times H(f)$$

where $\ast$ is the convolution operator and $\mathscr{F}$ means taking Fourier transform and thus $\mathscr{F}^{-1}$ is taking inverse Fourier transform and we have $h(t) \xrightarrow{\mathscr{F}} H(f)$ and $g(t) \xrightarrow{\mathscr{F}} G(f)$ are Fourier pairs and $t$ and $f$ are each domain variables (if it help think of $t$ as time and $f$ as frequency). It means multiplication in each domain is equivalent to convolution in the other domain. We are gonna use this property soon to compute our integral $\int_{\mathbb{R}}\frac{\sin^2x}{x^2}$ but we need some preliminaries. Now we define the following function (rectangular pulse) ($j=\sqrt{-1}$ is the imaginary unit) $$\Pi(t) = \begin{cases} 0, & \text{if} \qquad |t|>\frac{1}{2} \\[2ex] \frac{1}{2}, & \text{if} \qquad t= \pm \frac{1}{2} \\[2ex] 1, & \text{if} \qquad |t|<\frac{1}{2} \end{cases}$$

because Fourier transform of a non continuous function at the point of non continuity converges to the mean of both sides. The Fourier transform of $\Pi(t)$ is

$$\int_{-\infty}^{\infty} \Pi(t) e^{-2\pi j ft}dt = \frac{e^{-2\pi j ft}}{-2\pi j f} \Biggl|_{t=-\frac{1}{2}}^{t=+\frac{1}{2}} = \frac{1}{\pi f} \left( \frac{e^{\pi j f}-e^{-\pi j f}}{2j} \right)= \frac{\sin(\pi f)}{\pi f} = sinc(f)$$

Also there is another property of the Fourier transform that states

$$h(t) \xrightarrow{\mathscr{F}} H(f) \Rightarrow h(at) \xrightarrow{\mathscr{F}} \frac{1}{|a|}H(\frac{f}{a})$$

and since Fourier transform is a linear operator we conclude that

$$\Pi(t) \rightarrow \frac{\sin(\pi f)}{\pi f} \Rightarrow \pi \times \Pi(\pi t) \rightarrow \frac{\sin( f)}{ f}$$

Therefore by using the first property of Fourier transform and setting $g(t)=h(t) = \pi \times \Pi(\pi t)$ we get

$$\left( \pi \times \Pi(\pi t)\right) \ast \left( \pi \times \Pi(\pi t)\right) \xrightarrow{\mathscr{F}} \left(\frac{\sin( f)}{ f}\right) \times \left(\frac{\sin( f)}{ f}\right) = \frac{\sin^2f}{f^2}$$ and since these two are Fourier pair, by taking inverse Fourier transform we have

$$\int_{-\infty}^{\infty} \frac{\sin^2f}{f^2} e^{2\pi j f t} df = \pi^2 \left( \Pi(\pi t) \ast \Pi(\pi t)\right) = \pi^2 \Lambda(\pi t)$$

where $\Lambda(t)$ is the triangular pulse with following definition (it can be obtained by calculating convolution integral, it's not hard actually but a little ambiguous if you encounter the first time)

$$\Lambda(t) = \begin{cases} t+1, & \text{if} \qquad -1<t<0 \\[2ex] 1-t, & \text{if} \qquad 0<t<1 \\[2ex] 0, & \text{if} \qquad |t|>1 \end{cases}$$

So here you have it:

$$\int_{-\infty}^{\infty} \frac{\sin^2f}{f^2} df = \pi^2 \Lambda(\pi t) \Biggl|_{t=0} = \pi^2 \times \frac{1}{\pi} = \pi$$

You can use this link for further explanation on the rect and tri pulses http://s-mat-pcs.oulu.fi/~ssa/ESignals/em2_22-1.htm

Also there are other ways of evaluating this integral. To make it a little more interesting try to introduce a new parameter like $a$ and define $\int_{-\infty}^{\infty} \frac{\sin^2(ax)}{x^2} df = k(a)$. Differentiate one time with respect to $a$ and turn the integral to a differential equation. There you get a non-homogeneous differential equation which can be solved and $k(1)$ is the answer of our integral.

Answer 2

Here let $$ f(t) = \int_{-\infty}^{\infty} \frac{\sin^2(xt)}{x^2}\:dx $$ Then if your integral is $I$ that $I = f(1)$. The approach I will take is to use Fubini's Theorem with Laplace Transforms. In order for $f(t)$ to first be used here, we must appeal to the Dominated Convergence Theorem which is thankfully satisfied.

To proceed we take the Laplace Transform of $f(t)$ (this uses Fubini's Theorem)

Thus,

\begin{align} \mathscr{L}\left[f(t)\right] &= \mathscr{L}\left[\int_{-\infty}^{\infty} \frac{\sin^2(xt)}{x^2}\:dx \right] = \int_{-\infty}^{\infty} \frac{\mathscr{L}\left[\sin^2(xt)\right]}{x^2}\:dx = \int_{-\infty}^{\infty} \frac{2x^2}{s\left(s^2 + 4x^2\right)} \cdot \frac{1}{x^2}\:dx \\ &= \frac{2}{s}\int_{-\infty}^{\infty} \frac{1}{s^2 + 4x^2}\:dx = \frac{2}{s} \left[\frac{1}{2s}\tan^{-1}\left(\frac{2x}{s} \right) \right]_{-\infty}^{\infty} \end{align} As $s > 0$, we have $$ \mathscr{L}\left[f(t)\right] = \frac{2}{s}\left[\frac{1}{2s}\cdot \frac{\pi}{2} - \frac{1}{2s}\cdot -\frac{\pi}{2} \right] = \frac{1}{s^2}\pi $$ Now take the inverse Laplace Transform:

$$ f(t) = \mathscr{L}^{-1}\left[\frac{1}{s^2}\pi \right] = \pi t $$ And so $$ I = f(1) = \pi \longrightarrow \int_{-\infty}^{\infty} \frac{\sin^2(x)}{x^2}\:dx = \pi $$

Answer 3

Take the function $$ f(x) = \left\{\begin{array}{ll} 1 & \text{for } |x|<1\\ \frac12 & \text{for } |x|=1 \\ 0 & \text{for } |x|>1 \end{array} \right.$$

The Fourier transform of this function is $$ \tilde f(k) = \int_{-\infty}^\infty f(x) e^{-ikx} dx = \int_{-1}^1 e^{-ikx} dx = \frac{2\sin k}{k} $$

The Parseval's theorem for Fourier transform states that $$ \int_{-\infty}^\infty |\tilde f(k)|^2 \frac{dk}{2\pi} = \int_{-\infty}^\infty |f(x)|^2 dx $$

That means that $$ \int_{-\infty}^\infty \frac{4 \sin^2k}{k^2} \frac{dk}{2\pi} = \int_{-1}^1 dx = 2$$ $$ \int_{-\infty}^\infty \frac{\sin^2k}{k^2} dk = \pi$$