Hi

I need helping showing that: $\displaystyle \int_{-\infty}^{\infty} \frac{\sin^2(ax)}{x^2} dx = 2\pi$ using contour integration. So far what I have is:

this has a removable singularity at 0, and has pole of order 2 and $\displaystyle x=0$,then

$\displaystyle \frac{\sin^2(ax)}{x^2} = \lim_{R\rightarrow \infty} \lim_{\varepsilon \rightarrow \0}Re\bigg{(}\int_{-R}^{-\varepsilon} \frac{1-cos(2ax)}{x^2} dx +\int_{\varepsilon}^{R} \frac{1-cos(2ax)}{x^2} dx\bigg{)}$

= $\displaystyle -\int_{C_\varepsilon} \frac{1-e^{2iaz}}{z^2} dz -\int_{C_R} \frac{1-e^{2iaz}}{z^2} dz $

I have shown that the integral $\displaystyle \int_{C_R} \frac{1-e^{2iaz}}{z^2} dz = 0$, but I'm having trouble with $\displaystyle \int_{C_\varepsilon} \frac{1-e^{2iaz}}{z^2} dz $, I would like to convert it to polar and solve but the $\displaystyle e^{2iaz}$ in the numerator is giving me trouble