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Math Help - Complex contour integral

  1. #1
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    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
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  2. #2
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    Re: Complex contour integral

    Which contour(s) are you integrating over?
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  3. #3
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    Re: Complex contour integral

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