Complex contour integral

needmathhelp2012

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

Prove It

MHF Helper
Which contour(s) are you integrating over?

needmathhelp2012

the contour around the origin, since that is where the discontinuity lies, so $$\displaystyle C_\varepsilon$$