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

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.
