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This question already has an answer here:

I am trying to compute the following integral using the Residue Theorem but am quite stuck: $$\int_0^\infty \frac{\sin^2x}{x^2}dx$$ I have tried applying Jordan's lemma, having written $\sin(x)$ as $\dfrac{\mathrm{e}^{ix}-\mathrm{e}^{-ix}}{2i}$, to not much avail. I have also tried using a rectangular integration path but it didn't get me far. I'd be grateful for an insightful advice.

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marked as duplicate by Guy Fsone, José Carlos Santos, Arnaud D., Dando18, Nosrati Nov 7 '17 at 16:39

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Hint. First note that $$ \sin^2x=\frac{1}{2}\mathrm{Re}\, \big(1-\mathrm{e}^{2xi}\big), $$ and hence your integral equals $$ \frac{1}{4}\int_{-\infty}^{\infty}\frac{1-\mathrm{e}^{2xi}}{x^2}dx $$ Then define the curve $\gamma_{\varepsilon,R}$ to be the union of:

  1. $\gamma_R(t)=R\mathrm{e}^{it}, \,\,t\in[0,\pi]$.

  2. $\gamma_-(t)=t, \,\,t\in[-R,\varepsilon],$

  3. $\gamma_\varepsilon(\pi-t),\,\,t\in[0,\pi],$

  4. $\gamma_+(t)=t,\,\,t\in[\varepsilon,R]$.

Then use Residue Theorem (here the function DOES NOT have poles in the interior of $\gamma_{\varepsilon,R}$), and let $\varepsilon\to0$, $R\to\infty$.

ANSWER. $\dfrac{\pi}{2}$.

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Hint: Write $\sin^2(x) = \frac{1}{2}Re(1 - e^{2ix})$.

Take the integration as $$\int \frac{1 - e^{2iz}}{z^2} dz$$

Take the contour s.t. $0 + 0i$ is out of your closed region.

Find residue and solve.

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In this answer, it is shown using contour integration and the binomial theorem that $$ \int_0^\infty\left(\frac{\sin(z)}{z}\right)^n\mathrm{d}z =\frac{\pi}{2^n(n-1)!}\sum_{k=0}^{\lfloor n/2\rfloor}(-1)^k\binom{n}{k}(n-2k)^{n-1} $$ Plugging in $n=2$ gives $\dfrac\pi2$.

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