integration to infinite

The integral can be solved using differentiation under the integral sign .

$\begin{aligned} I(a) & = \int_{-\infty}^\infty \frac {\sin^2 (ax)}{x^2} dx, & \small \blue{\text{Since the integrand is even,}} \\ & = 2 \int_0^\infty \frac {\sin^2 (ax)}{x^2} dx \\ \frac {\partial I(ax)}{\partial a} & = 2 \int_0^\infty \frac {2\cancel x \sin (ax) \cos (ax)}{x^{\cancel 2}} dx \\ & = 2 \int_0^\infty \frac {\sin (2ax)}x dx & \small \blue{\text{Let }u = 2ax \implies du = 2a \ dx} \\ & = 2 \int_0^\infty \frac {\sin u}u du \\ & = 2 \text{ Si }(\infty) & \small \blue{\text{where Si }(\cdot) \text{ denotes the sine integral.}} \\ & = 2 \cdot \frac \pi 2 = \pi & \small \blue{\text{and Si }(\infty) = \frac \pi 2} \\ \implies I(a) & = \pi a + C & \small \blue{\text{where }C \text{ denotes the constant of integration.}} \\ I(0) & = C = 0 \\ \implies I(a) & = \pi a \end{aligned}$

Therefore, $\displaystyle I(1) = \int_{-\infty}^\infty \frac {\sin^2 x}{x^2} dx = \pi \approx \boxed{3.14}$