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?

Write
$$ \sin^2{x} = \frac{1}{2}(1-\cos{2x}), $$
then
$$ I = \int_{-\infty}^{\infty} \frac{1-\cos{2x}}{2x^2} \, dx. $$
Now, you can either consider this as the real part of $(1-e^{2ix})/x^2$, and then do the contour integral as usual with $(1-e^{2iz})/z^2$ around a circle in the upper half-plane, taking care with the simple pole at $z=0$, or you can integrate by parts:
$$ \int_{-\infty}^{\infty} \frac{1-\cos{2x}}{2x^2} \, dx = \left[ \frac{1-\cos{2x}}{-2x} \right]_{-\infty}^{\infty}-\int_{-\infty}^{\infty} \frac{\sin{2x}}{-x} \, dx, $$
which you've no doubt seen how to do before.
(It's surprising how often people forget you can do that when they're taught to do real integrals using residue calculus! I had a group of students the other day with much the same problem.)
