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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?

van abel
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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-function – Jonas Meyer Dec 17 '12 at 08:18
  • 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 $\lim_{x\to 0}\frac{sin^2 x}{x^2} = 1$

     – Guy Fsone Nov 27 '17 at 15:46

2 Answers2

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Hint:

$$ \begin{align} \int_0^\infty\frac{\sin^2(x)}{x^2}\,\mathrm{d}x &=\frac12\int_{-\infty}^\infty\frac{\sin^2(x)}{x^2}\,\mathrm{d}x\\ &=-\frac18\int_{-i-\infty}^{-i+\infty}\frac{e^{2ix}+e^{-2ix}-2}{x^2}\,\mathrm{d}x\\ &=-\frac18\left(\int_{\gamma_+}\frac{e^{2ix}-1}{x^2}\,\mathrm{d}x +\int_{\gamma_-}\frac{e^{-2ix}-1}{x^2}\,\mathrm{d}x\right)\tag{1} \end{align} $$ 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.

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

Stefan Hansen
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Tao
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