Contour Integration: non-convergent integral The question is 
$$I=\int_{-\infty}^{\infty} \frac{\sin^2{x}}{x^2} dx$$
My attempt:
$$I=-\frac{1}{4}\int_{-\infty}^{\infty} \frac{e^{2ix}-2+e^{-2ix}}{x^2} dx$$
$$I=-\frac{1}{4} \Big[ \int_{-\infty}^{\infty} \frac{e^{2ix}}{x^2} dx - \int_{-\infty}^{\infty} \frac{2}{x^2} dx +\int_{-\infty}^{\infty} \frac{e^{-2ix}}{x^2} dx \Big] $$
Now I use contours as mentioned in this question.
If we write 
$$I=-\frac{1}{4}\Big[A+B+C\Big]$$ where $A,B,C$ are the corresponding integrals.
For, $A$, and $C$ they can be worked out using Jordan's lemma and the residue theorem.
I run into a problem with $B$, as it does not converge.
Any ideas?

We can use Parseval's theorem to solve the integral. 
The rectangle function is, 
$$\Pi(x) = \begin{cases} 1 \quad |x|<1/2 \\ 0 \quad \text{otherwise} \end{cases}$$
The Fourier transform of the rectangle function is the $\mathrm{sinc}$ function. 
$$ \hat{\Pi}(k) = \mathrm{sinc}(k) = \frac{1}{\sqrt{2\pi}}\frac{\sin(k)}{k}.$$
Parseval's theorem tells us that, 
$$\frac{1}{2\pi}\int_{-\infty}^\infty \mathrm{sinc}^2(k) = \int_{-\infty}^\infty \Pi^2(x) dx = \frac12,$$
which gives us,
$$ \boxed{\int_{-\infty}^\infty \mathrm{sinc}^2(k) = \pi}.$$
