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Now, I want to integrate $\int_{0}^{\infty} \dfrac{\cos (2x) -1}{x^2} \mathrm{d}x$, now I attempted to set $f(z)=\dfrac{e^{i2z}-1}{z^2}$ and then integrate around a similar contour to the classical one used for $\dfrac{\sin x}{x}$ i.e. a semicircle with a smaller semicircle cut out around 0. The difficulty I'm having is the -1 term, I can't get the contour integrals into the right form to use to evaluate the real integral. Anyone have any advice on how to proceed?

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  • $\begingroup$ Dear David, one question what do you then set the complex function as? $e^{iz^2}$? $\endgroup$ – user2958701 Jun 15 '14 at 9:35
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Integrating $f(z)=(e^{2iz}-1)/z^2$ along the real axis from $\rho$ to $R$ gives $$J_1=\int_\rho^R\frac{e^{2ix}-1}{x^2}\,dx\ .$$ Integrating from $-R$ to $-\rho$ gives $$J_3=-\int_R^\rho \frac{e^{-2ix}-1}{x^2}\,dx =\int_\rho^R\frac{e^{-2ix}-1}{x^2}\,dx\ .$$ Along the upper semicircle from $R$ to $-R$ we have $$|e^{2iz}|=|e^{2i(x+iy)}|=e^{-2y}\le1$$ and so $$|J_2|\le\pi R \frac{1+1}{R^2}$$ which tends to $0$ as $R\to\infty$. Since $f$ has a simple pole at the origin, we have for the integral along the small semicircle $$\lim_{\rho\to0}J_4=-\pi i{\rm Res}\bigl(f(z),z=0\bigr)=2\pi\ .$$ Since $f$ has no singularities inside the contour we have $$J_1+J_2+J_3+J_4=0\ ,$$ and taking $R\to\infty$ and $\rho\to0$ gives $$\int_0^\infty\frac{e^{2ix}-1}{x^2}+\frac{e^{-2ix}-1}{x^2}\,dx+2\pi=0\ .$$ Therefore $$\int_0^\infty\frac{\cos 2x-1}{x^2}\,dx=-\pi\ .$$

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