# Complex contour integral

• Oct 26th 2012, 11:30 AM
needmathhelp2012
Complex contour integral
Hi
I need helping showing that: $\int_{-\infty}^{\infty} \frac{\sin^2(ax)}{x^2} dx = 2\pi$ using contour integration. So far what I have is:

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

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

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

I have shown that the integral $\int_{C_R} \frac{1-e^{2iaz}}{z^2} dz = 0$, but I'm having trouble with $\int_{C_\varepsilon} \frac{1-e^{2iaz}}{z^2} dz$, I would like to convert it to polar and solve but the $e^{2iaz}$ in the numerator is giving me trouble
• Oct 26th 2012, 05:03 PM
Prove It
Re: Complex contour integral
Which contour(s) are you integrating over?
• Oct 26th 2012, 09:40 PM
needmathhelp2012
Re: Complex contour integral
the contour around the origin, since that is where the discontinuity lies, so $C_\varepsilon$