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