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This question already has an answer here:

I am trying to integrate $\displaystyle \int_0^\infty \frac{\sin^2(x)}{x^2} dx$ by method of contour. I am considering the following contour but I am not being able to. Also I am not sure if it's right approach. contour

$$\int_\Gamma f(z) dz + 2 \int_0^\infty f(z) dz + \int_\gamma f(z)dz = 0$$ The first integral tends to zero as $R \to \infty $, but letting $\epsilon \to 0$, for the last integral I am getting. $$\int_{\pi }^0 \frac{\sin^2({\epsilon e^{i\theta}})}{\epsilon^2 e^{i2\theta }} i \epsilon e^{i\theta}d\theta = 0$$

ADDED:: Taking the above contour, we do not have any pole inside the contour. $$\int_{-\infty}^\infty \frac{1 - e^{i2z}}{2z^2}dz + \int_\Gamma \frac{1 - e^{i2z}}{2z^2} dz + \int_\gamma \frac{1 - e^{i2z}}{2z^2} dz = 0 \hspace{1 cm }(1)$$

$\displaystyle \int_\Gamma \frac{1 - e^{i2z}}{2z^2} dz \to 0$ due to Jordan Lemma. To evaluate $\displaystyle \int_\gamma \frac{1 - e^{i2z}}{2z^2} dz $ let the radius of the small semi circle be $\epsilon \to 0 $.

$$\lim_{\epsilon \to 0}\int_\pi^0 \frac{1 - e^{i2\epsilon e^{i\theta}}}{2\epsilon ^2 e^{i2\theta}} i \epsilon e^{i\theta}d\theta =\lim_{\epsilon \to 0} \int_\pi^0 \frac{1 - 1 - 2i\epsilon e^{i\theta} + O(\epsilon^2)}{2\epsilon e^{i\theta}} i = \lim_{\epsilon \to 0} \int_\pi^0 1+ O(\epsilon) d\theta = -\pi \hspace{1 cm }(2)$$

From $(2)$, $(1)$ reduces to $$\int_{-\infty}^\infty \frac{1 - e^{i2z}}{2z^2}dz - \pi = 0 \implies \Re \int_{-\infty}^\infty \frac{1 - e^{i2z}}{2z^2}dz = \int_{-\infty}^{\infty}\frac{\sin(x)^2}{x^2}dx = \pi$$

Including singularity at $z = 0$, we will have that small inner circle on lower plane. $$\int_{-\infty}^\infty \frac{1 - e^{i2z}}{2z^2}dz + \int_\Gamma \frac{1 - e^{i2z}}{2z^2} dz + \int_\gamma \frac{1 - e^{i2z}}{2z^2} dz = 2 \pi i \text{Residue}[f(z), z = 0] = 2\pi \hspace{1 cm }(3)$$

As for $\displaystyle \int_\gamma \frac{1 - e^{i2z}}{2z^2} dz = \lim_{\epsilon \to 0}\int_{-\pi}^0 \frac{1 - e^{i2\epsilon e^{i\theta}}}{2\epsilon ^2 e^{i2\theta}} i \epsilon e^{i\theta}d\theta = \pi \hspace{1 cm }(4)$

From $(3)$ and $(4)$ we get the same result.

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marked as duplicate by Guy Fsone, Arnaud D., Nosrati, Aqua, user223391 Nov 9 '17 at 16:54

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  • $\begingroup$ No need (informally) to take that detour around zero, since $$\lim_{z\to 0}\frac{\sin^2z}{z^2}=1$$so we're talking of a removable singularity... $\endgroup$ – DonAntonio May 13 '13 at 16:27
  • $\begingroup$ @DonAntonio I had encountered this type of contour in $\int_0^{\infty}\frac{\sin x}{x} dx = \frac{\pi}{2}$, so I considered it might work := $\endgroup$ – Mula Ko Saag May 13 '13 at 16:29
  • $\begingroup$ Of course it works but not with your function...:) Read my answer. $\endgroup$ – DonAntonio May 13 '13 at 16:36
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    $\begingroup$ @user17762: I disagree. I am sure if you scour the M.SE posts you will find something similar, but that isn't it. Besides, where in that post is this particular question being asked? And, even so, it is unfair to the OP, who may not be as sophisticated as you, to call this a duplicate just because the problem is attacked more generally. $\endgroup$ – Ron Gordon May 13 '13 at 16:48
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    $\begingroup$ This question shouldn't be closed since the other option is wa too advanced and general for a beginner. I'm voting to reopen. $\endgroup$ – DonAntonio May 13 '13 at 17:36
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What you try to do won't work since your function is (almost) analytic insde the path you take and thus won't help you to evaluate the real integral.

Let us try the following:

$$\cos 2x=1-2\sin^2x\implies \sin^2x=\frac{1-\cos2x}{2} \;\text{define}\;\;f(z):=\frac{1-e^{2iz}}{2z^2}:$$

$$\text{Res}_{z=0}(f)=\lim_{z\to 0}\,zf(z)=\lim_{z\to 0}\frac{1-e^{2iz}}{2z}\stackrel{\text{l'Hospital}}=-i$$

Question: The above implies $\,z=0\,$ is a simple pole...why is this so and not a double one?

Taking your contour, taking the limits and etc. and using the lemma and, specially, its corollary in the 2nd. answer here , we get after comparing real and imaginary parts

$$\int\limits_{-\infty}^\infty\frac{\sin^2x}{x^2}dx=\pi\;\;\ldots\ldots$$

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  • $\begingroup$ Actually, it won't work because $\sin^2{z}$ does not converge on the outer contour. $\endgroup$ – Ron Gordon May 13 '13 at 16:38
  • $\begingroup$ @RonGordon, are you talking of my answer or of the intent the OP made with the function $\,\frac{\sin^2z}{z^2}\,$ ? $\endgroup$ – DonAntonio May 13 '13 at 16:40
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    $\begingroup$ I am speaking of the first line in your answer: "What you try to do won't work since your function is (almost) analytic insde the path you take and thus won't help you to evaluate the real integral." $\sin^2{z}$ has a component of $e^{-i 2 z}$ which does not behave nicely on the outer contour of radius $r$ as $r \to \infty$. That is why using the function $\sin^2{z}/z^2$ won't work. $\endgroup$ – Ron Gordon May 13 '13 at 16:42
  • $\begingroup$ Well, perhaps...I'm not that sure, yet the complex integral of this function on that contour is zero and that also doesn't usually help to solve real integrals by means of complex integration... $\endgroup$ – DonAntonio May 13 '13 at 16:44
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    $\begingroup$ If by "this function" you mean $\sin^2{z}/z^2$, you are wrong: the integral of this function over the above contour blows up as $r \to \infty$. You need to take the piece you specify in the upper half plane, and another piece $(1-e^{-i 2 z})/z^2$ in the lower half-plane. $\endgroup$ – Ron Gordon May 13 '13 at 16:51
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There is a more easier way From this 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 $\lim\limits_{x\to 0}\frac{\sin^2 x}{x^2} = 1$

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