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!

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