Integrate: $\int_0^\infty \frac{\sin^2(x)}{x^2}dx$ 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.

$$\int_\Gamma f(z) dz + 2 \int_0^\infty f(z) dz + \int_\gamma f(z)dz = 0$$
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.
 A: 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
$$\int\limits_{-\infty}^\infty\frac{\sin^2x}{x^2}dx=\pi\;\;\ldots\ldots$$
A: 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 $\lim\limits_{x\to 0}\frac{\sin^2 x}{x^2} = 1$

