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Just as the title say, consider the integral: $$I=\int_0^{+\infty} \frac{\sin^2 x}{x^2} \,\mathrm{d} x=\frac{1}{2}\int_{-\infty}^{+\infty} \frac{\sin^2 x}{x^2} \,\mathrm{d} x,$$ how to apply the residue theorem to get the answer?

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    Why do you have to apply the residue theorem? Because $\dfrac{\sin^2 z}{z^2}$ is entire, it may suffice to use Cauchy's theorem with some contour. Here's a related question, but the answers there probably don't use methods you're currently looking for: http://math.stackexchange.com/questions/13344/proof-for-an-integral-involving-sinc-function2012-12-17
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    From this https://math.stackexchange.com/questions/5248/evaluating-the-integral-int-0-infty-frac-sin-x-x-dx-frac-pi-2 We know that , $$\frac{\pi}{2} =\int_0^\infty\frac{\sin x}{x} dx = \int_0^\infty\frac{\sin 2u}{2u} d(2u) =\int_0^\infty\frac{\sin 2u}{u} du\\ = \underbrace{\left[\frac{\sin^2 u}{u}\right]_0^\infty}_{=0} +\int_0^\infty\frac{\sin^2u}{u^2} du =\color{blue}{\int_0^\infty\frac{\sin^2u}{u^2} du} $$ Given that, $\sin2x = 2\sin x\cos x=(\sin^2x)'$ and $\lim_{x\to 0}\frac{sin^2 x}{x^2} = 1$2017-11-27

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