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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:… – Jonas Meyer Dec 17 '12 at 8:18


$$ \begin{align} \int_0^\infty\frac{\sin^2(x)}{x^2}\,\mathrm{d}x &=\frac12\int_{-\infty}^\infty\frac{\sin^2(x)}{x^2}\,\mathrm{d}x\\ &=-\frac18\int_{-i-\infty}^{-i+\infty}\frac{e^{2ix}+e^{-2ix}-2}{x^2}\,\mathrm{d}x\\ &=-\frac18\left(\int_{\gamma_+}\frac{e^{2ix}-1}{x^2}\,\mathrm{d}x +\int_{\gamma_-}\frac{e^{-2ix}-1}{x^2}\,\mathrm{d}x\right)\tag{1} \end{align} $$ where $\gamma_+$ goes from $-i-R$ to $-i+R$ then circles back counterclockwise on $-i+Re^{i\theta}$ for $\theta\in[0,\pi]$ as $R\to\infty$, and where $\gamma_-$ goes from $-i-R$ to $-i+R$ then circles back clockwise on $-i+Re^{-i\theta}$ for $\theta\in[0,\pi]$ as $R\to\infty$.

Each of the integrals in $(1)$ can easily be handled with the residue theorem.

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$$\int_{0}^{\infty}\frac{(\sin x)^2}{x^2}dx=-\int_{0}^{\infty}(\sin x)^2{d\frac{1}{x}}=-\frac{(\sin x)^2}{x}\Bigg|^{\infty}_{0}+\int_{0}^{\infty}\frac{(2\sin x\cos x)}{x}dx=\int_{0}^{\infty}\frac{(\sin2x)}{2x}d2x=\ldots$$

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Really? How does the first equality follow? In any event, you should give at least some verbal explanation... – tomasz Dec 17 '12 at 9:14
@tomasz Integration by parts. – Ragib Zaman Dec 17 '12 at 9:29
How to calculate $\int_0^\infty \sin x/x d x$ then? – van abel Dec 17 '12 at 12:30
OK, I saw it in the older… – van abel Dec 17 '12 at 12:32

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