# Calculating $\int_0^\infty \frac {\sin^2x}{x^2}dx$ using the Residue Theorem. [duplicate]

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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.

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

• Which contour should you be taking? – Lost1 Jan 21 '14 at 11:45
• What do you mean by "should be taking"? Presumably either a rectangular or a circular path. – Grtv Jan 21 '14 at 11:49
• – Martín-Blas Pérez Pinilla Jan 21 '14 at 11:51
• From this math.stackexchange.com/questions/5248/… 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 15:54

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

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.

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 $\dfrac\pi2$.