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Math Help - Evaluating a real integral

  1. #1
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    Evaluating a real integral

    The aim is to evaluate \int \frac {sin^{2}x}{x^2}dx. The strategy is to do a contour integral of the function f(z)=\frac{e^{2iz}-1}{z^{2}+a^{2}} for real a>0 around the contour \gamma consisting of two parts: \gamma_{1} the semi-circle { z \in \mathbb{C} | |z|=R, Im(z)>0}, and \gamma_{2} the straight line segment from -R to R.

    (i) Explain why \int_{\gamma_{1}}f(z)dz \to 0 as R \to \infty.

    \int_{\gamma_{1}}f(z)dz=\int_{0}^{\pi}f(\gamma_{1}  (t))\gamma_{1} '(t)dt =\int_{0}^{\pi} \frac{e^{2iRe^{it}}-1}{R^{2}e^{2it}+a^{2}} iRe^{it} dt

    Now this is where I get stuck.

    (ii) Use Cauchy's integral formula to show that \int_{\gamma}f(z)dz=\frac{\pi}{a}(e^{-2a}-1).

    Now I've tried to use partial fractions so that \frac{e^{2iz}-1}{z^{2}+a^{2}}=\frac{e^{2iz}-1}{(z+ai)(z-ai)}=\frac{A}{(z+ai)}+\frac{B}{(z-ai)}=\frac{A(z-ai)+B(z+ai)}{(z+ai)(z-ai)}

    Equating the numerators gives e^{2iz}-1=A(z-ai)+B(z+ai) but I'm not sure how to solve this equation for A and B.

    (iii) Show that e^{2ix}-1=-2sin^{2}x+isin2x and hence that \int_{\gamma_{2}}f(z)dz is real.
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  2. #2
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    For (i):

    \left|\int_{0}^{\pi} \frac{e^{2iRe^{it}}-1}{R^{2}e^{2it}+a^{2}} iRe^{it} dt\right|\leq \int_0^{\pi} \left|\frac{R(e^{2iRe^{it}}-1)}{R^2e^{2it}-1}\right|dt

    \leq \int_0^{\pi}\left| \frac{e^{2iR(\cos(t)+i\sin(t)}-1}{R(e^{it}-\frac{1}{R^2})}\right|dt

    and as R\to\infty, the integral tends to the form of:

    \frac{1}{R}\int_0^{\pi}e^{-2R\sin(t)}dt

    which tends to zero for unbounded R.

    That however may need a little cleaning up but it's a good start I think.

    For (ii), I don't see why you doing all that. Just use Cauchy's Integral Formula directly:

    \oint \frac{e^{2iz}-1}{z+ia}\frac{1}{z-ia}dz=2\pi i\left( \frac{e^{2iz}-1}{z+ia}\right)_{z=ia}=\frac{\pi}{a}(e^{-2a}-1)

    For (iii), \sin(x) is an odd function so the imaginary component from (-k,k) is zero right?
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